1. Engineering Mechanics Full Material

Introduction – Forces, Equilibrium Introduction In Physics, the branch which deals with the study of the state of rest or motion caused by the action of forces on the bodies is called as mechanics. Engineering mechanics applies the principles and laws of mechanics to solve problems of common engineering elements. Newtonian mechanics: Newtonian mechanics or classical mechanics deals with the study of the motion of macroscopic objects under the action of a force or a system of forces. Branches of Newtonian Mechanics 1. Statics: It is the study of forces and conditions of equilibrium of material bodies, at rest, subjected to the action of forces. 2. Dynamics: It is the branch of mechanics which deals with the study of motion of rigid bodies and the co-relation with the forces causing and affecting their motion. Dynamics is divided into Kinematics and Kinetics. 3. Kinematics: Kinematics deals with space time relationship of a given motion of body and not at all with the forces that cause the motion. 4. Kinetics: The study of the laws of motion of material bodies under the action of forces or kinetics is the study of the relationship between the forces and the resulting motion. Some of the definitions of the idealizations used in engineering mechanics are as follows: 1. Continuum: It is defined as continuous non special whole which has no empty spaces and no part is distinct from the adjacent parts. Considering objects in this way ignores that the matter present in the object is made up of atoms and molecules. 2. Particle: A particle is a body which has finite mass but the dimensions can be neglected. 3. System of particles: When a group of particles which are inter-related are dealt together for studying the behavior, it is called as a system of particles. 4. Rigid body: A solid body which does not undergo any deformations under the application of forces is called a rigid body. In reality, solid bodies are not rigid but are assumed as rigid bodies. 5. Matter: It is anything which occupies space. 6. Mass: It is a measure of inertia. The mass of body is the quantity of matter contained in it and is the sum of the masses of its constituent’s mass points. Deformation of Body A body which changes its shape or size under action of external forces is called deformable body. Action and Reaction Action and reaction occurs when one body exerts a force on another body, the later also exerts a force on the former. These forces are equal in magnitude and opposite in direction. ☞ Introduction ☞ Newtonian Mechanics ☞ Branches of Newtonian Mechanics ☞ Deformation of Body ☞ Action and Reaction ☞ Resolution of a Force into a Force and a Couple ☞ Resultant of a System of Coplanar Forces ☞ Resultant of Multiple Forces Acting at a Point ☞ Collinear Forces ☞ Coplanar Concurrent Force System ☞ Coplanar Parallel Force System ☞ Coplanar Non-concurrent, Non-parallel Force System ☞ Distributed Force System CHAPTER HIGHLIGHTS Chapter 1

3.6 |  Part III  •  Unit 1  •  Engineering Mechanics Tension It is the pulling force which is acting through a string when it is tight. It acts in the outward direction. Tension Thrust It is acting in the inward direction and it is the pushing force transferred through a light rod. Thrust Force Force may be defined as any action that tends to change the state of rest or uniform motion on of a body on which it is applied. The specifications or characteristics of a force are: 1. Magnitude 2. Point of application 3. Direction, (force is a vector quantity). 4. Line of action Force is a vector quantity since it has amagnitude and a direction (scalar quantities have only magnitudes and no directions). The direction of a force is the direction, along a straight line passing through its point of application, in which the force tends to move the body on which it is applied. The straight line is called the line of action of the force. For the force of gravity, the direction of the force is vertically downward. 1. System of forces: A system of forces or a force system is the set of forces acting on the body or a group of bodes of interest. Force system can be classified, according to the orientation of the lines of action of the constituting forces, as follows: Coplanar Collinear Concurrent Concurrenet Non-concurrenet Non-concurrenet Non-coplanar (Space forces) Parallel Like parallel Unlike parallel Non-parallel general Like parallel Unlike parallel Parallel General System of forces 2. Coplanar and non-coplanar (spatial) force Systems: In a coplanar force system (Figure (a)), the constituting forces have their lines of action lying in the same plane. If all the lines of action do not lie in the same plane, then the corresponding forces constitute a non-coplanar force system (Figure (b)). y x z F1 F2 F3 F4 y x z F1 F2 F3 F4 (a) (b) 3. Collinear force system: In a collinear force system (figure), the lines of action, of the entire constituting forces, will be along the same line. y x z F 4. Concurrent and non-concurrent force systems: If the lines of action of all the forces in a force system pass through a single point, then the force system is called as a concurrent force system (Figure (a)) else it is called as a non-concurrent force system (Figure (b)). y x z F1 F F2 3 F4 y x z F1 F2 F3 F4 (a) (b) 5. Parallel and non-parallel (general) force systems: In a parallel force system (Figure (a)), the lines of action, of the entire constituting forces, are parallel to each other. If the line of action of at least one constituting force is not parallel to the line of action of another constituting force in a force system, then the force system is called ‘non-parallel force system’ (Figure (b)). y x z F1 F2 F3 F4 y x z F1 F2 F3 F4 (a) (b)

Chapter 1  •  Introduction – Forces, Equilibrium  | 3.7 6. Like parallel and unlike parallel force systems: In a like parallel force system (Figure (a)), the lines of action, of the entire constituting forces, are parallel to each other and act in the same direction. In an unlike parallel force system (figure), the lines of action, of the entire constituting forces, are parallel to each other where some of them act in different directions. y x z F1 F2 F3 F4 System of Forces Force System Examples 1. Collinear Forces on a rope in a tug of war 2. Coplanar parallel System of forces acting on a beam subjected to vertical loads including reactions. 3. Coplanar like parallel Weight of a stationary train on the rail when track straight. 4. Coplanar concurrent Forces of a rod resting against a wall. 5. Coplanar, non–concurrent forces (ii) (b) Forces on a ladder resting against a wall when a person stands on a rung which is not at its centre of gravity. 6. Non–coplanar parallel The weight of the benches in a classroom 7. Non–coplanar concurrent forces (II) (b) Forces on a tripod carrying 8. Non–coplanar non–concurrent forces (iv) Forces acting on a moving bus. Force systems can also be classified, according to the magnitude of the constituting forces as: (a) System of equal forces—all the constituting forces has the same magnitude; (b) System of unequal forces—all the constituting forces do not have the same magnitude. NOTE 7. Representation of a force: Graphically, a force may be represented by the segment of a straight line with arrow head at one end of the line segment. The straight line represents the line of action of the force 1 kN and its length represents its magnitude. The direction of force is indicated by placing an arrow head on this straight line. The arrow head at one end of the straight line segment indicate the direction of the force along the line segment. Either the head or tail may be used to indicate the point of application of the force. Note that all the forces involved must be represented consistently as shown in figures below. 0 0 30° 30° (a) (b) Resolution of a Force into a Force and a Couple A given force ‘P’ applied to a body at any point ‘A’ can always be replaced by an equal force applied at another point ‘B’ together with a couple which will be statically equivalent to the original force. To prove this let the given force ‘P’ act at ‘A’ as shown below. Then at B, we introduce two oppositely directed collinear forces each of magnitude ‘P’ and parallel to the line of action of the given force P at A. P P P a A P A B P (a) (b) (c) It follows from the law of superposition that the system in figure (b) is statically equivalent to that in figure (a). However, we may now regard the original force P at ‘A’ and the oppositely directed force ‘P’ at B as a couple of moment M = Pa. Since this couple may now be transformed in any manner in its plane of action so long as its moment remains unchanged, we may finally represent the system as shown in the figure (c), where the couple is simply indicated by a curved arrow and the magnitude of its moment. It will be noted that the moment of the couple introduced in the above manner will always be equal to the product of the original force ‘P’ and the arbitrary distance ‘a’ that we decide to move its line of action. This resolution of a force into a force and couple is very useful in many problems of statics. Resultant of a System of Coplanar Forces Parallelogram Law of Forces When two concurrent forces P and Q are represented in magnitude and direction by the two adjacent sides of a parallelogram as shown in fig., the diagonal of the parallelogram concurrent with the two forces, P and Q represents the resultant R of the forces in magnitude and direction.

3.8 |  Part III  •  Unit 1  •  Engineering Mechanics If P and Q are two forces making in angle θ with each other, then R P = + Q Q + P 2 2 2 cos θ Q a q q Q R q − a (180 − q) P tan sin cos α θ θ = + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ Q P Q α θ θ = + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ - tan sin cos 1 Q P Q Q R P sin s α θ in sin(θ α) = = - Resultant of Multiple Forces Acting at a Point Let ΣH = algebraic sum of resolved part of the forces along the x-axis. ΣV = algebraic sum of resolved part of the forces along the y-axis. R H = ∑( ) + ∑( ) V 2 2 x F2 F1 0 y 0 y R x ΣH ΣV tan θ = ∑ ∑ V H where θ is the angle which the resultant vector R makes with the x-axis. Triangle Law The resultant of two forces can be obtained by the triangle law of forces. The law states that if two forces acting at a point are represented by the two sides of a triangle, taken in order, the remaining side taken in an opposite order will give the resultant. 180 − a 180 − b 180 − g P b a Q R g b P g a Q R Q P R sin s α γ in sin β = = Coplanar Force System It can be classified into collinear, concurrent, parallel, nonconcurrent, and non-parallel type of force system. The resultant of a general coplanar system of forces may be (a) single force, (b) a couple in the system’s plane or in a parallel plane or (c) zero. Collinear Forces The resultant of a collinear force system (R) can be determined by algebraically adding the forces. R = ΣF = F1 + F2 + F3 A F1 B F2 C F3 D Coplanar Concurrent Force System y P Q x q3 q1 q2 S The analytical method procedure consists of resolving the forces into components that coincide with the two arbitrarily chosen axes. ∑ = F P x cos c θ θ 1 2 + + Q S os cos θ3 ∑ = F P y sin s Q Q 1 2 + - in θ θ S sin 3 And the resultant R � = + ( ) F � x y ( ) F 2 2 . Its angle with respect to the x-axis is given by α = - ∑ ∑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ tan 1 Fy Fx . Coplanar Parallel Force System A O B C O xs xQ xp x R P Q S Resultant of the parallel forces P, Q and S are R = ΣF = P + Q + S Now, ΣM0 = Rx

Chapter 1  •  Introduction – Forces, Equilibrium  | 3.9 So, ΣM0 → sum of the moments of the forces P, Q and S, at point 0. Rx = Px p + QxQ + Sxs Coplanar non-concurrent, non-parallel force system: As in the case of an unlike parallel force system, the resultant may be a single force, a couple in the plane of the system or zero. The resultant is given by R fx f = ∑( ) + ∑( ) y 2 2 and its angle α with the x-axis is given by tan α = ∑ ∑ P F y x o a b c d x P Q S y Distributed force system: Distributed forces (or loads) are forces which act over a length, area or volume of a body. On the other hand, a concentrated force (point load) is a force which acts of a point. Solved Examples Example 1: Three forces P, 2P and 3P are exerted along the direction of the three sides of an equilateral triangle as shown in the following figure. Determine the resultant force. Solution: ΣFx = P - 2P cos 60° - 3P cos 60° = -1 5 = - 3 2 . P P ΣF y = 2P sin 60° − 3P sin 60° = -0 866 - 3 2 . P P R F = ∑ x y + ∑ F 2 2 R P = + P 9 4 3 4 2 2 12 4 3 2 P P = A B C 3P 2P P Example 2: The resultant of two concurrent forces 3P and 2P is R. If the first force is doubled, the resultant is also doubled. Determine the angle between the forces. Solution: R P = + [(3 2 ) ( P P ) c + ×2 3 × × 2P os ] 2 2 1 α 2 = × P [ ] 13+12 1 cos α 2 (1) Being the angle between the forces. 2 6 2 2 6 2 2 2 1 R P = + [( ) ( P P ) c + × × × P os α]2 = + [ c 40 24 os ] 1 α 2 (2) From equations (1) and (2) we have 2 13 12 40 24 1 2 1 P P [ c + = os α α ] [ + cos ]2 or 2 1[ c 3 1 + = 2 4 os α α ] [ 0 2 + 4cos ] cos , α α = - = ° 1 2 120 Example 3: A weight ‘W’ is supported by two cables. At what value of ‘θ’ the tension is cable making be minimum? q 60° W Solution: q 60° W T1 T2 ΣFj = 0 T1 sin θ + T2 sin 60° = W T1 cos θ = T2 cos 60° T T 2 T 1 1 60 = 2 ° = cos cos cos θ θ = T1 sin θ + 2T1 cos θ⋅sin 60° = W = T1 sin θ + 2T1 cos θ⋅sin 60° = W T T 1 1 sin c θ θ + = 3 os W dT d T T 1 0 3 1 1 0 θ = = cos ( θ θ + -sin ) = T T 1 1 cos s θ θ = 3 in tan θ = 1 3 θ = 30°. Example 4: An electric fixture weighing 18 N hangs from a point C by two strings AC and BC as shown in the following figure. The string AC is inclined to the vertical wall at 40° and BC is inclined to the horizontal ceiling at 50°. Determine the forces in the strings.

3.10 |  Part III  •  Unit 1  •  Engineering Mechanics 40° 50° 40 A E B D 18 N C T1 T1 T2 Solution: It can be deducted that ∠DCA = 40 and ∠BCD = 40° so that ∠ACB = 80°. Now, ∠ACE = 180° - 40 = 140°; ∠BCE = 180° - 40 = 140° Using sine rule; T T 1 2 40 40 18 sin sin sin80 = = T1 18 40 80 = 11 75 × = sin sin . N T2 18 40 80 = 11 75 × = sin sin . N Example 5: Determine the resultant of the coplanar concurrent force system shown in the following figure. 70° 45° 30° 20° C X 200 N 100 N O B 250 N 150 N Solution: ΣFx = 100 cos 20 + 250 cos 45 - 200 cos 70 - 150 cos 30 = 72.44 N ΣF y = 100 sin 20 - 250 sin 45 - 200 sin 70 + 150 sin 30 = -255 N Now, Resultant R F = ∑( ) x y + ∑( ) F 2 2 = + 72 44 255 5 = 265 2 2 . . N Its inclination α = ∑ ∑ - tan 1 F F y x = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ° - tan . . 1 255 5 72 44 74 Since Σy is negative, the angle falls in the fourth quadrant. ∴ Angle made with x-axis is 360 - 74 = 286° (Counter clock wise). Direction for solved examples 6 and 7: R P Q 600 N 60° 30° 30° 60° TPR TQR 600 N 60° 30° Example 6: The tension in the wire QR will be (A) 519.6 N (B) 625 N (C) 630 N (D) 735 N Solution: TQR TPR sin(180 60) sin( ) 180 30 sin 600 - 90 = - = TQR TPR sin s 60 in 30 sin 600 90 = = The tension in the wire QR, TQR = = 300 3 519. N6 Example 7: The tension in the wire ‘PR’ will be (A) 575 N (B) 300 N (C) 275 N (D) 400 N Solution: The tension in the wire PR, TPR = 600 sin 30 = 300 N Example 8: A point is located at (−6, 2, 16) with respect to the origin (0, 0, 0). Specify its position. (i) In terms of the orthogonal components. (ii) In terms of the direction cosines. (iii) In terms of its unit vector. Solution: A(−6, 2, 16) O(0, 0, 0) The components of the vector OA are (−6 − 0) = −6 along the x-axis (2 − 0) = 2 along the y-axis (16 − 0) = 16 along the z-axis x y O z −6 16 2 i j k A

Chapter 1  •  Introduction – Forces, Equilibrium  | 3.11 Position vector, r = −6i + 2j + 16k (in terms of the orthogonal components.) Magnitude r = -( ) 6 2 + +16 = 17.2 2 2 2 Direction cosines are  = = cos - = - . θ . x 6 17 2 0 3488 m = = cos y = . θ . 2 17 2 0 1163 n = = cos z = . θ . 16 17 2 0 9302 r = (17.2 ) i + (17.2 m)j + (17.2 n) (in terms of the direction cosines) Example 9: Consider a truss ABC with a force P at A as shown in the following figure, B 45° 30° C A P The tension in member CB is (A) 0.5P (B) 0.63P (C) 0.073P (D) 0.87P Solution: Consider point A. For equilibrium, resolving the forces, 4 6 45° 3 TBC TAB P B C A P TAB cos 45° + TAC cos 60 - P = 0 TAB sin 45 = TAC sin 60 Solving, T P AB = + 2 3 ( ) 6 2 Balancing of forces at point B give TAB cos 45 = TBC T P BC = P + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 6 6 2 0. . 633 Example 10: A man sitting on a wheel chair tries to roll up a step of 8 cm. The diameter of the wheel is 50 cm. The wheel chair together with the man weighs 1500 N. What force he will have to apply on the periphery of the wheel? F P C r T W b h 2r − h r − h h = 8 cm Solution: The free body diagram given above shows the horizontal force applied by the man, the weight W acting at the centre of the wheel and the reaction R at the point P. (The reaction at O will be zero at the instant the wheel being lifted up). From the geometry b2 = r2 - (r - h)2 = 2rh - h2 = 2 × 0.25 × 0.08 - (0.08)2 ∴ b = 0.1833 m Taking moments about the point P, -F (2r - h) + wb = 0 Or, F Wb r h = 2 - Where W = load on one wheel = 1500 2 F = Force applied on one wheel. ∴ = × × - F = = 1500 2 0 1833 2 0 25 0 08 137 475 0 42 327 32 . . . . . . . N Moment of a Force The product of a force and the perpendicular distance of the line of action of the force from a point or axis is defined as the moment of the force about that point or axis. P r O In the figure, the moment of force P about the point O or about the y axis is p × r. Moment may be either clockwise or anti clockwise. In the Figure 1.15 the, moment tends to rotate the body in anti clockwise direction. The right hand rule is a convenient tool to identify the direction of a moment. The moment M about an axis may be represented as a vector pointing towards the direction of the thumb of the right hand, while the other fingers show the direction of turn, the force offers about the axis (clockwise or anti-clockwise)

3.12 |  Part III  •  Unit 1  •  Engineering Mechanics M y d2 d1 F2 F1 F x o y d Varignon’s Theorem of Moments B C O P Q A R X It can be proved that ΔOXA + ΔOXB = ΔOXC. This illustrates Varignon’s theorem of moments. Moment of the force Q about X = Twice the area of ΔOXA. Moment of force P about X = Twice the area of ΔOXB. R is the resultant of P and Q. Moment of the resultant about X is = Twice the area of the triangle OXC. The theorem may be stated as follows—the moment of a force about any point is equal to the sum of the moments of components about the same point. Example: As shown in figure resultant force F and its components F1 and F2; d1 and d2 are the respective normal distances between F, F1 and F2 from any point O. Varignon’s theorem states that Fd = F1d1 + F2d2 . Moment of a Couple Two parallel forces having the same magnitude and acting in the opposite directions form a couple. Moment of the couple is the algebraic sum of moment of the forces involved in it about a point. d o A B d1 d2 F2 F1 Moment of the force F1 1 about O OA F    = × Moment of the force F2 2 about O OB F OB F2         = × ( ) - = - × Algebraic sum of the moments M OA F OB F      = × 1 2 - × (But F1 = F2 for a couple F1 = F2 = F) = - ( ) OA OB × F     (1) But OA AB BO        + + = 0 OA OB AB       - = ∴ (1) Becomes M AB F     = × . B O F A F The resultant force is zero, but the displacement ‘d’ of the force couple creates a couple moment. Moment about some arbitrary point is 0. M = F1d1 + F2d2 = F1d1 − F1d2 = F1 (d1 − d2). If point O is placed in the line of action of force Fz (or F1), then M = Fid (or F2d). Orthogonal components (scalar components) of force F along the rectangular axises, x, y and z axis’s, are Fx , F y and Fz respectively. Fx = |F| cos θx , F y = |F| cos θ y , Fz = |F| cos θ2, where cos θx (zl ), cos θ y (zm) and cos θ2(zn) are the direction cosines of the force F and |F| is the magnitude of the force F. Now, F F = + x y F F + z ( ) ( ) ( ) 2 2 2 We have, F F = + x y i F j F + = z n k F| | (cos θ θ i j + + cos c y z os θ k) = |F| (l I + mj + nk). where i, j and k are vectors of unit length along the positive x, y and z directions. Unit vector corresponding to the force vector F F F F , | | = ⋅If nˆ is a unit vector in the direction of the force F, then F = |F| nˆ. Equilibrium of Force Systems A body is said to be acted upon by a system of forces in equilibrium if the force system cannot change the body’s stationary or constant velocity state. When the resultant is neither a force nor a couple, i.e., ΣF = 0 (1) ΣM = 0 (2) Then, ΣF → vector sum of all forces of the system. ΣM → vector sum of the moments (relative to any point) of all the forces of the system.

Chapter 1  •  Introduction – Forces, Equilibrium  | 3.13 Scalar equation equivalent to vector equation (1), in a rectangular coordinate system, are ΣFx = 0 ΣF y = 0 ΣFz = 0 Scalar equation equivalent to the vector equation (2) as ΣMx = 0 ΣM y = 0 ΣMz = 0 Now, ΣFx , ΣF y and ΣFz → algebraic sum of forces in the x, y and z directions respectively. ΣMx , ΣM y and ΣMz → algebraic sum of moments in the x, y and z directions respectively. Equilibrium Equations for Different Coplanar Force Systems 1. Concurrent coplanar force system ΣFx = 0, ΣF y = 0 2. Concurrent non co-planar force system ΣFx = 0, ΣF y =0, ΣFz = 0 3. Non-concurrent coplanar force system ΣFx = 0, ΣF y = 0 and ΣM = 0 at any suitable point. 4. Non-concurrent non-coplanar force system ΣFx = 0, ΣF y = 0; ΣFz = 0 and ΣMx = 0, ΣM y = 0, ΣMz = 0 Analysis of a System of Forces in Space A spatial force system may consist of a set of concurrent forces, parallel forces or non-concurrent non parallel forces. The resultant of a spatial force system is a force R and a couple C, where R = Σ (forces) and C = Σ (moments) Concurrent Spatial Force System Resultant R is given by R F = ∑ x y + ∑ F F + ∑ z ( ) ( ) ( ) 2 2 2 With the direction cosines given by cos θx x i F R = ∑ cos θy y i F R = ∑ cos θz z i F R = ∑ ΣFx , ΣF y and ΣFz are algebraic sums of the components of all the forces in the x, y, and z directions and θx , θ y , θz are the angles which the resultant vector R makes with the x, y, and z axes respectively. Parallel Spatial Force System The resultant R F R M R M x x z z = ∑ = ∑ = ∑ 1 where x and z are the perpendicular distances of the resultant vector from the xy and yz plane respected and ΣMz , ΣMx are algebraic sums of the moments of forces of the force system about the x and z axes respectively. If ΣF = 0, the resultant couple can be evaluated As C M = ∑( ) x z 2 2 + ∑( ) M So, tan ϕ = ∑ ∑ M M z x Where φ is the angle made by the couple. Non-concurrent Non Parallel Force System The resultant, R F = ∑( x y ) + ∑( ) F F + ∑( z ) 2 2 2 and corresponding direction cosines are cos , θ θ cos x x y F y R F R = ∑ = ∑ and, cos θz Fz R = ∑ Now, C M = ∑ x y + ∑ M M + ∑ z ( ) ( ) ( ) 2 2 2 and the corresponding direction cosines are cos , θ θ cos x x c y M C y C M C = = ∑ 1 cos θz Mz C = ∑ 1 Where θx , θ y , and θz are the angles which the vector representing the couple C makes with the x, y and z axes respectively.

3.14 |  Part III  •  Unit 1  •  Engineering Mechanics Exercises Practice Problems 1 Direction for questions 1 to 10: Select the correct alternative from the given choices. 1. Concurrent forces in a plane will be in equilibrium if (A) Sum of the forces is zero (B) Algebraic sum is zero (C) Sum of resolved parts zero (D) Sum of the resolved parts in any two perpendicular direction are zero. 2. A free body diagram is a representation of (A) The forces on the body (B) The reactions on the body (C) Both the active and reactive forces (D) Neither the active nor the reactive forces 3. When a driver operates the gear shift lever it is (A) Coplanar force (B) Non-coplanar force (C) Moment (D) Couple 4. Three coplanar forces acting at a point can be in equilibrium only if (A) All the three forces are parallel (B) All the three forces are concurrent (C) All the three forces are parallel or concurrent (D) All the three forces are spatial 5. Lami’s theorem gives the following when three concurrent forces acting on a body kept in equilibrium (A) Force divided by tan of angle is zero (B) Force is proportional to tan θ (C) Force/cos θ is constant (D) Each force is proportional to the sine of angle between the other two 6. For a particle to be in equilibrium under the action of two forces, the forces must be (A) Concurrent and parallel (B) Unequal non concurrent (C) Equal parallel non collinear (D) Equal, opposite and collinear 7. A number of co-planar forces will be in equilibrium when (A) ΣFx = 0 and ΣF y = 0 (B) ΣFM0 = 0 (C) ΣFx = 0 (D) ΣFx = 0, ΣF y = 0, ΣM0 = 0 8. Two cylinders having diameter 50 mm and 100 mm respectively are held in a container, bigger one at bottom and the smaller on top. Container has a width of 110 mm. What will be the reaction at the point of contact of bigger cylinder to the container wall (neglect friction). Weight of small cylinder is W and weight of big cylinder is 2 W. (A) 0.472 W (B) 0.528 W (C) 0.163 W (D) 0.725 W 9. The resultant of coplanar non-concurrent forces constitutes (A) Force (B) Couple (C) Moment or couple (D) Force or couple 10. Turning the cap of a pen is example of (A) Moment (B) Force (C) Couple (D) Impulse Practice Problems 2 Direction for questions 1 to 10: Select the correct alternative from the given choices. 1. Two forces are acting at a point O as shown in figure O O a 15° 30° 200 N 100 N P Determine the resultant in magnitude and direction. (A) 84.644 kN, 35° (B) 90.0 kN, 33° (C) 100 kN, 28° (D) 120 kN, 25° 2. Two equal forces are acting at a point with an angle of 60° between them. If the resultant force, is equal to 40 × ⋅ 3N Find the magnitude of each force. (A) 20 N (B) 40 N (C) 50 N (D) 30 N 3. A weight of 1900 N is supported by two chains of lengths of 4m and 3m as shown in figure. Determine the tension in each chain 4 m 5 m T1 T2 a b q1 q2 A B 3 m Chain No 1 Chain No 2 1900 N (A) 1200 N, 1300 N (B) 1100 N, 100 N (C) 1100 N, 1200 N (D) 1520, 1140 N Direction for questions 4 and 5: Three forces in magnitude 80 kN, 30 kN, 40 kN are acting at a point O as shown. The angles made by 80 kN, 30 kN and 40 kN forces with x axis are 60°, 120° and 240° respectively.

Chapter 1  •  Introduction – Forces, Equilibrium  | 3.15 4. Determine the magnitude of the resultant force. 60° 120° 240° 30 kN 80 kN O X Y 40 kN (A) 60.82 kN (B) 50.8 kN (C) 30.5 kN (D) 62.8 kN 5. Find the direction of the resultant force. (A) 85.28° (B) 60° (C) 84.38° (D) 15.5° Direction for questions 6 and 7: Four forces of magnitude 20 kN, 30 kN, 40 kN and 80 kN are acting at a point O as shown in figure. The angles made by 20 kN, 30 kN, 40 kN and 80 kN with x-axis are 30°, 60°, 90° and 120° respectively. 6. Find the magnitude of the resultant force. (A) +5 kN (B) −6 kN (C) −7.679 kN (D) +8 kN 7. Find the direction of the resultant force. (A) −86.97° (B) 98° (C) 97.5° (D) −85° 8. A beam simply supported at A and B of span 10 m is carrying a point load of 10 kN at a distance of 4 m from A. Determine the reactions at the supports (A) 7 kN, 8 kN (B) 4 kN, 6 kN (C) 5 kN, 4 kN (D) 10 kN, 8 kN 9. Four forces of magnitudes 20 N, 40 N, 60 N and 80 N are acting respectively along the four sides of a square ABCD as shown in figure. Determine magnitude of resultant. D A C B 40 N 80 N 20 N 60 N (A) 40 2 N (B) 50 2 N (C) 45 2 N (D) 60 2 N 10. Find the direction of the resultant referred to the direction of 20 N forces (A) 50° (B) 45° (C) 60° (D) 80° Previous Years’ Questions 1. If point A is in equilibrium under the action of the applied forces, the value of tensions TAB and TAC are respectively. [2006] 600 N TAB TAC 60° A 30° (A) 520 N and 300 N (B) 300 N and 520 N (C) 450 N and 150 N (D) 150 N and 450 N 2. A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in Newton) required to maintain equilibrium of the ladder is ____ [2014] 2.5 m 2.5 m 4 m 3 m W P B A 3. A rigid ball of weight 100 N is suspended with the help of a string. The ball is pulled by a horizontal force F such that the string makes an angle of 30° with the vertical. The magnitude of force F (in N) is _______. [2016] 30º F 100 N

3.16 |  Part III  •  Unit 1  •  Engineering Mechanics Answer Keys Exercises Practice Problems 1 1. D 2. C 3. D 4. C 5. D 6. D 7. D 8. B 9. C 10. C Practice Problems 2 1. A 2. B 3. D 4. A 5. A 6. C 7. A 8. B 9. A 10. B Previous Years’ Questions 1. A 2. 399 to 401 3. 57.5 – 58.0

Free Body Diagram Free body diagram (FBD) is a sketch of the isolated body, which shows the external forces on the body and the reactions exerted on it by the removed elements. A general procedure for constructing a free body diagram is as follows: 1. A sketch of the body is drawn, by removing the supporting surfaces. 2. Indicate on the sketch all the applied or active forces, which tend to set the body in motion, such as those caused by weight of the body, etc. 3. Also indicate on this sketch all the reactive forces, such as those caused by the constraints or supports that tend to prevent motion. 4. All relevant dimensions and angles; reference axes are shown on the sketch. A smooth surface is one whose friction can be neglected. Smooth surface prevents the displacement of a body normal to both the contacting surfaces at their point of contact. The reaction of a smooth surface or support is directed normal to both contacting surfaces at their point of contact and is applied at that point. Some of the examples are shown in the following figures. W W A RA W W A W S RA We isolate the body from its supports and show all forces acting on it by vectors, both active (gravity force) and reactive (support reactions) forces. We then consider the conditions this system of forces must satisfy in order to be in equilibrium, i.e., in order that they will have no resultant. W W A B H H RA RB Free Body Diagrams – Trusses ☞ Free Body Diagram ☞ Composition and Resolution of Forces ☞ Resolution of a Force ☞ Equilibrium Law ☞ Internal and External Forces ☞ Superposition and Transmissibility ☞ Equilibrium of Concurrent Forces in a Plane ☞ Analysis of Roof Trusses ☞ Plane Truss ☞ Nodes ☞ Perfect Frame ☞ Supports ☞ Assumptions: Analysis of Trusses ☞ Method of Members: Analysis of Plane Frames CHAPTER HIGHLIGHTS Chapter 2

3.18 |  Part III  •  Unit 1  •  Engineering Mechanics A C D E B A C D E B Figure 1 Beam with roller support at one end A C E D B A M Figure 2 Beam with hinged end and fixed end Composition and Resolution of Forces The reduction of a given system of forces to the simplest system that will be its equivalent is called the problem of composition of forces. If several forces F1, F2, F3, applied to a body at one point, all act in the same plane, then they represent a system of forces that can be reduced to a single resultant force. It then becomes possible to find this resultant by successive application of the parallelogram law. Let us consider, for example, four forces F1, F2, F3, and F4 acting on a body at point A, as shown in the following figure. To find their resultant, we begin by obtaining the resultant AC of the two forces F1 and F2. Combining this resultant with force F3 we obtain the resultant AD which must be equivalent to F1, F2, and F3. Finally, combining the forces AD and F4, we obtain the resultant ‘R’ of the given system of forces F1, F2, F3, and F4, This procedure may be carried on for any number of given forces acting at a single point in a plane. F2 F3 F4 F1 A B C D E R F2 F3 F4 F1 A B C D E R Resolution of a Force The replacement of a single force by several components, which will be equivalent in action to the given force, is called the problem of resolution of a force. In the general case of resolution of a force into any number of coplanar components intersecting at one point on the line of action, the problem will be indeterminate unless all but two of the components are completely specified in both their magnitudes and directions. Equilibrium Law Two forces acting at a point can be in equilibrium only if they are equal in magnitude, opposite in direction and collinear in action. Let us consider the equilibrium of a body in the form of a prismatic bar on the ends of which two forces are acting as shown in the figure below: A B S m n S Neglecting their own weights, it follows from the principle just stated that the bar can be in equilibrium only when the forces are equal in magnitude, opposite in direction and collinear in action which means that they must act along the line joining the points of application. Considering the equilibrium of a portion of the bar ‘AB’ to the left of a section ‘mn’, we conclude that to balance, the external force S at A the portion to the right must exert on the portion to the left an equal, opposite and collinear force S as shown in the above figure. The magnitude of this internal axial force which one part of a bar in tension exerts on another part called the tensile force in the bar or simply the force in the bar, since in general it may be either a tensile force or a compressive force. Such an internal force is actually distributed over the cross sectional area of the bar and its intensity, i.e., the force per unit of cross section area is called the stress in the bar. Internal and External Forces Internal forces are the forces which hold together the particles of a body. For example, if we try to pull a body by applying two equal, opposite and collinear forces an internal force comes into play to hold the body together. Internal forces always occur in pairs and equal in magnitude, opposite in direction and collinear. Therefore, the resultant of all of these internal forces is zero and does not affect the

Chapter 2  •  Free Body Diagrams – Trusses  | 3.19 external motion of the body or its state of equilibrium. External forces or applied forces are the forces that act on the body due to contact with other bodies or attraction forces from other separated bodies. These forces may be surface forces (contact forces) or body forces (gravity forces). Let us consider the equilibrium of a prismatic bar on each end of which two forces are acting as shown below. A A B RB RB RB = −RA RA F1 F2 F4 F3 Other examples of two force members held in equilibrium are shown below. A B RA A B RB = RA A force which is equal, opposite, and collinear to the resultant of the two given forces is known as equilibrant of the given two forces. Superposition and Transmissibility When two forces are in equilibrium (equal, opposite and collinear) and their resultant is zero and their combined action on a rigid body is equivalent to that—there is no force at all. A generalization of this observation gives us the third principle of statics. Sometimes,it is called the law of superposition. Law of Superposition The action of a given system of forces on a rigid body will in no way be changed if we add to or subtract from them another system of forces in equilibrium Let us consider now a rigid body ‘AB’ under the action of a force ‘P’ applied at ‘A’ and acting along BA as shown in the figure (a). From the principles of superposition we conclude that the application at point ‘B’ of two oppositely directed forces, each equal to and collinear with P will in no way alter the action of the given force P. That is, the action on the body by the three forces shown in figure (b) is same as the action on the body by the single force P shown in figure (a). P B A P P1 P11 B A (a) (b) This proves that the point of application of a force may be transmitted along the line of action without changing the effect of the force on any rigid body to which it may be applied. This statement is called the theorem of transmissibility of a force. Equilibrium of Concurrent Forces in a Plane If a body known to be in equilibrium is acted upon by several concurrent coplanar forces then these forces or rather their free vectors, when geometrically added must form a closed polygon. This statement represents the condition of equilibrium for any system of concurrent forces in a plane. In the figure (a), we consider a ball supported in a vertical plane by a string ‘BC’ and a smooth wall ‘AB’. The free body diagram in which the ball has been isolated from its supports and in which all forces acting upon it both active and reactive, are indicated by vectors as shown in figure (b). A W B C a W S O a RA (a) (b) RA W S O a RA W S O a (c) (d) The three concurrent forces W, S and RA are a system of forces in equilibrium and hence their free vectors must build a closed polygon, in this case, a triangle as shown in figure (c). If numerical data are not given, we can still sketch the closed triangle of forces and then express: RA = Wtana and S = Wseca Lami’s Theorem If three concurrent forces are acting on a body, kept in equilibrium, then each force is proportional to the sine of the angle between the other two forces and the constant of proportionality is the same. Consider forces P, Q and R acting

3.20 |  Part III  •  Unit 1  •  Engineering Mechanics at a point ‘O’ as shown in Figure (a). Mathematically Lami’s theorem is given by the following equation. P Q R k sin s a b in sinγ = = = Since the forces are in equilibrium, the triangle of forces should close. Draw the triangle of forces DABC, as shown in Figure (b), corresponding to the forces P, Q, and R acting at a point ‘O’. From the sine rule of the triangle, we get P Q R sin(p a- ) sin( ) p b sin(p γ) = - = - o R g g b b a a Q P P Q A R B C sin(p - a) = sina sin(p - b) = sinb sin(p - γ) = sinγ When feasible, the trigonometric solution, or Lami’s theorem is preferable to the graphical solution since it is free from the unavoidable small errors associated with the graphical constructions and scaling. Analysis of Roof Trusses Definitions Truss A ‘truss’ or ‘frame’ or ‘braced structure’ is the one consisting of a number of straight bars joined together at the extremities. These bars are members of the truss. Plane truss If the centre line of the members of a truss lies in a plane, the truss is called a plane truss or frame. If the centre line is are not lying in the same plane, as in the case of a shear leg, the frame is called a space frame. Figure 3 Plane trusses Strut and tie A member under compression is called a strut and a member under tension is called a tie. Loads A load is generally defined as a weight or a mass supported. Trusses are designed for permanent, intermittent or varying loads. Nodes The joints of a frame are called as nodes. A frame is designed to carry loads at the nodes. Perfect frame A pin joined frame which has got just the sufficient number of members to resist the loads without undergoing appreciable deformation in shape is called a perfect frame. Supports A truss or a framed structure is held on supports which exert reaction on the truss or framed structure that they carry. Reactions are to be considered for finding the stresses in the various members of the structures. The types of supports commonly used are 1. Simple supports 2. Pin joint and roller supports 3. Smooth surfaces 4. Fixed on encaster and fixtures The reactions of the supports are analytically or graphically evaluated. 1. In a simply supported truss, the reactions are always vertical at the supports. 2. At a pin joint support, the reaction passes through the joint. 3. At a roller surface, the support reaction is vertically upwards at the surface. 4. The reaction at a support which has a smooth surface is always normal to the surface. Assumptions: Analysis of Trusses Each truss is assumed to be composed of rigid members to be all lying in one plane. This means that co-planar force systems are involved. Forces are transmitted from one member to another through smooth pins fitting perfectly in the members. These are called two force members. Weights of the members are neglected because they are negligible in comparison to the loads. A C D Pin joint B W1 W2 Roller support Figure 4 Pin joint and roller support A Pin joint B Smooth surface W1 W2 Figure 5 Pin joint and smooth surface W1 W2 W3 Figure 6 Fixed support

Chapter 2  •  Free Body Diagrams – Trusses  | 3.21 Free-body Diagram of a Truss and the Joints C D F E B A L Figure 7 Truss C D F E B A L R1 R2 Figure 8 Free-body diagram of truss as a whole Tension compression R1 Figure 9 Free-body diagram of point A AF AF AB R1 AB R1 Joint A BF BF BE BE BC BC AB L L AB Joint B CE = 0 BC CD Joint C DE CD R2 DE R2 CD Joint D BE CE = 0 EF DE EF BE DE Joint E BF AF EF AF BF EF Joint F F E B C A D R1 R2 L Figure 10 Free-body diagram of joints Solution by method of joints To use this technique, draw a free-body diagram of any pin in the truss provided no more than two unknown forces act on that pin. This limitation is imposed because the system of forces is a concurrent one for which of course, only two equations are available for a solution. From one pin to another, until the unknown is found out, the procedure can be followed. Working Rule 1. Depending on the nature of support, provide the reaction components. (a) For hinged support, provide horizontal and vertical reaction components. (b) For roller support, provide vertical reaction components only. 2. Considering the external loads affecting the truss only, apply the laws of statics at equilibrium to evaluate the support reactions. 3. Give the values of the support reactions at appropriate joints. 4. Take the joint which contains the minimum number of members (minimum number of unknowns) and apply the conditions of equilibrium to evaluate the forces in the members. For example: 3m A C 2 m 2 m B AV BV AH Q D P 3m

3.22 |  Part III  •  Unit 1  •  Engineering Mechanics The reaction components at A are AH and AV (because A is a hinged joint). The reaction component at B is BV only (no horizontal reaction since it is a roller). Now evaluate the reaction components considering the external loads only i.e., AV + BV = P + Q AH = 0 Taking algebraic sum of the moments of all forces about ‘A’ and equating to zero, we get two more equations. These three equations are sufficient to evaluate the support reactions. Once the support reactions are evaluated, joints can be considered one by one, to evaluate the force in the members. Method of Members: Analysis of Plane Frames Frames differ from trusses principally in one aspect, i.e., the action of forces are not limited to their ends only and so the members are subjected to bending also with tension or compression. In this method the members are isolated as a free body and analyzed with the forces acting on them by vectors. Consider the folding stool with the dimensions as shown in the figure resting on a horizontal floor and a force F is acting at a distance of xℓ form the end A. A E B RD RC /2 x /2 The floor is considered as smooth floor therefore the reactions at C and D are vertical. ∴ By taking moments at C and D. RD = (1 – x) F and Rc = x F. Now we separate the members AC and BD and analyze the forces acting on each member. YB YA XB XA XE XE YE YE E E D B A C RD = (1 − x) F RC = x F For the free body diagram BD, taking moments about B Y X E E x F    2 2 + - ( ) 1 0 - = For the free body diagram AC, taking the moments about A Y X E E X F    2 2 - + = 0. ∴ XE = F and YE = (1 –2x) F. The resultant is R X E E = +Y F E = + - x 2 2 2 1 1( ) 2 ∴ By knowing the numerical data for ℓ, x and the load F the unknowns RD, RC and RE can be found. To find the reactions at A and B i.e., RA and RB take the moments about the point E for both the free body diagrams and solve for RA and RB. Solved Examples Example 1: The magnitude and nature of stresses in the member ED of the truss, loaded as shown below, is (A) 30.4 kN (T) (B) 20.2 kN (T) (C) 18.69 kN (C) (D) 15.7 kN (T) 60° 60° 60° 60° 20 kN 40 kN 3 m B C A D E 3 m 1.5 m 1.5 m Solution: 60° 60° 60° 60° 20 kN 40 kN 20.2 kN R2 = 35 kN 40.41 kN 3 m B C A D 1.5 m E 1.5 m Free-body diagram Taking moments about (A) for equilibrium, ΣmA = 0 -20 × 1.5 - 40 × 4.5 + R2 × 6 = 0 6R2 = 30 + 180 6R2 = 210 R2 = 35 kN But R1 + R2 = 60 ∴ R1 = 25 kN

Chapter 2  •  Free Body Diagrams – Trusses  | 3.23 Take the joint D, force on the member CD, FCD = 40.41 kN because FCD sin60° = 35 kN ∴ Force on ED = FCD cos60° = 20.2 kN(T) Example 2: All the members of the truss shown below are of equal length and the joints are pinned smooth. It carries a load F at S whose line of action passes through V. The reaction at V is P R T V S Q U F q q (A) Zero (B) Vertically upwards and equal to F/4 (C) Vertically upwardsand equal to F/2 (D) Vertically upwards and equal to F Solution: P a R W S T V RV Q U F F sinq F cosq 60° q q Let a = length of one member From the above figure, sin 60° = SW SR . SW = = a SR a 3 2 ( ) ∵ Also cos60 , 2 ° = ⇒ = RW SR RW a TW RT RW a a a = - ∴ - = 2 2 VW VT TW a a a = + = + = 2 3 2 tanθ = = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = SW VW a a 3 2 3 2 1 3 θ = 30° Taking moments about P for equilibrium, ΣMP = 0 - × F - × + × = a F a RV sin c θ θ os 3 2 3 2 30 0 F a × × + × F a × = R a v × 1 2 3 2 3 2 3 2 3 Rv = F/2 Example 3: The force in the member RQ of the truss, as given in the figure below, is (A) 27 kN (Tensile) (B) 15 kN (Compressive) (C) 20 kN (Compressive) (D) 7 kN (Tensile) P W 15 kN 12 m V Q S R U 20 kN T 10 kN 7 kN 5 kN (4 × 3) Solution: 60° 60° 60° Consider the junction R. It must be in equilibrium. The force 7 kN can be balanced only by the member QR. ∴ The force in the member QR FQR = 7 kN (Tensile). Example 4: The figure is a pin jointed plane truss loaded at the point C by hanging a weight of 1200 kN. The member DB of the truss is subjected to a load of m A B C D E 1200 kN (A) Zero (B) 500 kN in compression (C) 1200 kN in compression (D) 1200 kN in tension Solution: Member DB is perpendicular to AC. Resolving the vertical component of the forces at B, we observe that no force can be present in member DB.

3.24 |  Part III  •  Unit 1  •  Engineering Mechanics Example 5: Find the force in the member EC of the truss shown in the figure B A E 4 kN 10 m 10 m 2 kN 2 kN C D (A) 8 kN (B) 4 kN (C) 3.5 kN (D) 2 kN Solution: B A E 4 kN 4 kN 30° VD HC 11.547 6.928 10 m 10 m 2 kN 2 kN C D 2√3 2√3 Equating vertical forces VD = 8 Taking moment about D, -2 × 20 - 4 × 10 + HC × 11.54 = 0 80 = 11.54 HC ∴ HC = 6.928 kN Consider the equilibrium of joint A FAE sin30° = 2 FAE = 4 kN (T) FAE cos30° = FBA 4 3 2 = 2 3FBA FBE = 0, FCB = 2 3 Consider point C Net horizontal force = - 6.928 2 3 = 3.463 kN It is to be balanced by the force on EC FEC cos30 = 3.463 ∴ FEC = 4 kN. Example 6: The type of truss, shown in the figure below, is D C A E F B (A) Perfect (B) Deficient (C) Redundant (D) None of above Solution: The number of joints, J = 6 The number of member, n = 10 Then, 2j – 3 = 2 × 6 – 3 = 9 Since n > (2j – 3), it is a redundant truss. Example 7: A weight 200 kN is supported by two cables as shown in the figure. A 200 kN T1 T2 B C q 60° The tension in the cable AB will be minimum when the angle θ is: (A) 0° (B) 30° (C) 90° (D) 120° Solution: T T 1 2 150 90 200 sin sin( ) sin ( 180 60) = ° + = θ θ ° - + kN T1 200 30 120 = ° ° - sin sin( θ) = ° ° - ° = ° 200 30 120 30 150 sin sin[ ] ( sin sin ) θ ∵ T1 is minimum when 1 sin(120° - θ) is minimum, i.e., sin(120° - θ) is maximum. sin(120° - θ) = 1 120° - θ = 90° θ = 30°. Example 8: A 200 kN weight is hung on a string as shown in the figure below. The tension T is: (A) 200 kN (B) 300 kN (C) 160 kN (D) 207.1 kN T 200 15° O Q Solution: The three forces T, RB and 200 kN are in equilibrium at point O. T W 15° Q O RQ

Chapter 2  •  Free Body Diagrams – Trusses  | 3.25 T W sin s 90 in( ) 90 15 = ° + ° T W= = × = sin sin . . 90 105 200 1 0 965 207 1 kN Example 9: A uniform beam PQ pinned at P and held by a cable at Q. If the tension in the cable is 5 kN then weight of the beam and reaction at P on the beam are respectively (A) 10 kN and 8.86 kN (B) 5 kN and 13 kN (C) 10 kN and 10 kN (D) 15 kN and 13 kN P Q 60° 60° 30° R RP T W Solution: Let w be the weight of the beam and RP be the reaction in the beam at P. Since it is the case of 3 forces forming a system in equilibrium all the three forces must be either concurrent or parallel; in this case, they are to be concurrent. W T RP sin s 90 in150 sin120 = = W T RP 1 0 5 0 886 = = . . W = × = 5 1 0 5 10 . kN RP = 8.86 kN. Example 10: RH 150 mm 300 mm q P Q R 3002 + 1502 √ = 335 mm 40 N T Ry A mass of 40 N is suspended from a weight less bar PQ which is supported by a cable QR and a pin at P. At P on the bar, the horizontal one vertical component of the reaction, respectively, are (A) 80 N and 0 N (B) 75 N and 0 N (C) 60 N and 80 N (D) 55 N and 80 N Solution: θ = = ° - cos . 1 150 335 63 4 Resolving the vertical components of the forces at Q, 40 = T cos 63.4° T T = = 40 63 4 89 336 cos . . N Resolving the vertical components of the forces at P, R y = 0 Resolving the horizontal component of the forces at RQP, RH = Tsin 63.4 = 80 N Example 11: A truck of weight M g is shown in figure below. A force F (pull) is applied as shown. The reaction at the front wheels at location P is F b P a a Q Mg (A) M a g Fb 2 2 + (B) M a g Fb + 2 (C) M Fb a g 2 2 + (D) M a g F 2 2 + Solution: Taking moment about Q, ΣMQ = 0 = RP × 2a - M g × a - F × b R M a F b a M Fb a P g g = × + × = + 2 2 2 . Direction for questions 12 to 13: All the forces acting on a particle situated at the point of origin of a two dimensional reference frame. One force has magnitude of 20 N acting in the positive x direction. Where as the other has a magnitude of 10 N at an angle of 120° with force directed away from the origin with respect to the positive direction to the direction of 20 N: 60° S Q 10 N 120° a O 20 N P R Example 12: The value of the resultant force, R, will be (A) 18 N (B) 20 N (C) 15 N (D) 21 N Solution: R = + 20 10 + ×2 20 5 × × 120° 2 2 cos = - 500 100 = 20 N Example 13: The value of a made by the resultant force with the horizontal force will be: (A) 25.65 (B) 13 (C) 14.5 (D) 15 Solution: From triangle OQP 10 60 20 0 866 23 09 sin sin . . a = = = R sin . a = = . 10 23 09 0 433 a = 25.65.

3.26 |  Part III  •  Unit 1  •  Engineering Mechanics Exercises Practice Problems 1 Direction for questions 1 to 10: Select the correct alternative from the given choices. 1. If point P is in equilibrium under the action of the applied forces, then the values of the tensions TPQ and TPR are respectively TPQ TPR P 500 N 60° 30° (A) 250 N and 250 3 N (B) 250 3 N and 250 N (C) 300 3 N and 300 N (D) 280 N and 280 3 N 2. For the loading of a truss shown in the figure below (length of member PU = length of member UT = length of member TS), the reaction at S (RS) is Q R U 2 kN 2 kN T S P (A) 2 kN (B) 3 kN (C) 2.57 kN (D) 4 kN 3. A truss consists of horizontal members (PU, UT, TS, QR) and vertical members (UQ, TR) all having a length B each. Q R U T B B B B B B P S The members PQ, TQ and SR are inclined at 45° to the horizontal. If an uniformly distributed load ‘F’ per unit length is present on the member QR of the truss shown in the figure above, then the force in the member UT is (A) FB 2 (B) FB (C) 0 (D) 2 3 FB 4. Consider the truss ABC loaded at A with a force F as shown in the figure below. F A 45° 30° B C The tension in the member BC is (A) 0.5 F (B) 0.63 F (C) 0.73 F (D) 0.87 F Direction for questions 5 and 6: Two steel members PQ and QR each having cross sectional area of 200 mm2 are subjected to a horizontal force F as shown in figure. All the joints are hinged. Q P F R 45° 60° 5. If F= 1 kN the magnitude of the vertical reaction force developed at the point R in kN is (A) 0.543 kN (B) 2 kN (C) 0.634 kN (D) 1 kN 6. The maximum value of the force F in kN that can be applied at P such that the axial stress in any of the truss members does not exceed 100 Pa is (A) 22.3 kN (B) 54 kN (C) 43.6 kN (D) 43.28 kN 7. Bars PQ and QR, each of negligible mass support a load F as shown in the figure below. In this arrangement, it can be deciphered that R P Q F (A) Bar PQ is subjected to bending but bar QR is not subjected to bending. (B) Bars PQ and QR are subjected to bending (C) Neither bar PQ nor bar QR is subjected to bending. (D) Bar QR is subjected to bending but bar PQ is not subjected to bending.

Chapter 2  •  Free Body Diagrams – Trusses  | 3.27 8. P pin P X X Y Y 200 mm 100 mm 100 N The figure shows a pair of pin jointed gripper tongs holding an object weighing 1000 N. The co-efficient of friction (m) at the gripping surface is 0.1. xx is the line of action of the input force P and yy is the line of application of gripping force. If the pin joint is assumed to be frictionless, then the magnitude of the force P required to hold the weight is: (A) 500 N (B) 1000 N (C) 2000 N (D) 2500 N 9. A cantilever truss of 4 m span is loaded as shown in the figure. 60° 60° B D A 4 m 20 kN C Determine the farce in member BC (A) 32.1 kN (B) 11.55 kN (C) 46.2 kN (D) 23.1 kN 10. In the figure shown below, the force in the member BD is, 10 kN D A B 5 m 5 m 2.5 m C 5 kN (A) 5 kN (B) 10 kN (C) 15 kN (D) 20 kn Practice Problems 2 Direction for questions 1, 2 and 3: Two collinear forces of magnitudes 400 N and 200 N act along with a force 600 N, acting at an angle of 30° with the former forces, as shown below: 600 400 200 30 D A B C 1. The resultant force is (A) 1512 N (B) 1424 N (C) 1342 N (D) 1108 N 2. The angle made by the resultant force with the former forces is (A) 15° (B) 18° (C) 22° (D) 26° 3. If the forces 200 N and 600 N are collinear and acts at 30° with the force 400 N, then the resultant force shall be (A) 1812 N (B) 1526 N (C) 1423 N (D) 1164 N Direction for questions 4 to 6: Select the correct alternative from the given choices. 4. A force of 500 N is acting on a body. The magnitude of the force to be acted on the same body at 120° with the first force, so that the net force on the body is still 500 N acting along the bisector of 120°, is (A) 300 N (B) 500 N (C) 750 N (D) 100 N 5. P and Q are two forces acting at a point and their resultant force is R. When Q is doubled the resultant force is doubled. If again the resultant force is doubled when Q is reversed, then P : Q : R is: (A) 232 : : (B) 1 3 : : 2 (C) 326 : : (D) 3 5 : : 2 6. A force of 200 N is acting at a point making an angle of 30° with the horizontal. The components of this force along the x and y directions, respectively, are: (A) 173.2 N and 100 N (B) 200 N and 130 N (C) 250 N and 180.2 N (D) 135.5 N and 160 N Direction for questions 7 and 8: A small block of weight 200 N is placed on an inclined plane which makes an angle θ = 30° with the horizontal. 7. The component of the weight perpendicular to the inclined plane is (A) 160.2 N (B) 173.2 N (C) 140.2 N (D) 183.2 N 8. The component of the weight perpendicular to the inclined plane is (A) 70 N (B) 78 N (C) 98 N (D) 100 N 9. A circular disk of radius 20 mm rolls without slipping at a velocity v. The magnitude of the velocity at the point P is P O v v 30° 60°

3.28 |  Part III  •  Unit 1  •  Engineering Mechanics (A) 3v (B) 3 2 v (C) v 2 (D) 2 3 v 10. A stone with a mass of 0.2 kg is catapulted as shown in the figure below. The total force Fx (in N) exerted by the rubber band as a function of the distance x (in m) is given by Fx = 300 x2. If the stone is displaced by 0.2 m from the unstretched position (x = 0) of the rubber band, the energy stored in the rubber band is 0.2 Kg Stone of mass F f (A) 0.02 J (B) 0.3 J (C) 0.8 J (D) 10 J 1. The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of: [2004] K L M N O m (A) 0 N (B) 490 N in compression (C) 981 N in compression (D) 981 N in tension 2. Consider a truss PQR loaded at P with a force F as shown in the figure. The tension in the member QR is: [2008] F 45° 30° P Q R (A) 0.5F (B) 0.63F (C) 0.73F (D) 0.87F Direction for questions 3 and 4: Two steel truss members, AC and BC, each having cross sectional area of 100 mm2, are subjected to a horizontal force F as shown in figure. All the joints are hinged. 3. If F = 1 kN, the magnitude of the vertical reaction force developed at the point B in kN is: [2012] C F B A 45° 60° (A) 0.63 (B) 0.32 (C) 1.26 (D) 1.46 4. The maximum force F in kN that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 100 MPa is: [2012] (A) 8.17 (B) 11.15 (C) 14.14 (D) 22.30 5. A two member truss ABC is shown in the figure. The force (in kN) transmitted in member AB is [2014] A B C 10 kN 1 m 0.5 m 6. For the truss shown in the figure, the forces F1 and F2 are 9 kN and 3 kN, respectively. The force (in kN) in the member QS is (All dimensions are in m): [2014] P F1 F2 Q 3 1.5 3 3 2 R S T (A) 11.25 tension (B) 11.25 compression (C) 13.5 tension (D) 13.5 compression Previous Years’ Questions

Chapter 2  •  Free Body Diagrams – Trusses  | 3.29 7. Two identical trusses support a load of 100 N as shown in the figure. The length of each truss is 1.0 m; crosssectional area is 200 mm2, Young’s modulus E = 200 GPa. The force in the truss AB (in N) is ______ [2015] A C B 30° 100 N 30° 8. 4 m P Q 45° R 60° 100 kN For the truss shown in figure, the magnitude of the force in member PR and the support reaction at R are respectively [2015] (A) 122.47 kN and 50 kN (B) 70.71 kN and 100 kN (C) 70.71 kN and 50 kN (D) 81.65 kN and 100 kN 9. For the truss shown in the figure, the magnitude of the force (in kN) in the member SR is: [2015] 30 kN W V S R P U T Q 1 m 1 m 1 m 1 m (A) 10 (B) 14.14 (C) 20 (D) 28.28 10. A weight of 500 N is supported by two metallic ropes as shown in the figure. The values of tensions T1 and T2 are respectively. [2015] 30° 90° 120° 500 N T1 T2 (A) 433 N and 250 N (B) 250 N and 433 N (C) 353.5 N and 250 N (D) 250 N and 353.5 N 11. A two-member truss PQR is supporting a load W. The axial forces in members PQ and QR are respectively. [2016] (A) 2W tensile and 2W compressive (B) 3W tensile and 2W compressive L P R Q W 30º 60º (C) 3W compressive and 2W tensile (D) 2W compressive and 3W tensile 12. A force F is acting on a bent bar which is clamped at one end as shown in the figure. [2016] F 60º The CORRECT free body diagram is (A) F 60º Rx Ry M

3.30 |  Part III  •  Unit 1  •  Engineering Mechanics Answer Keys Exercises Practice Problems 1 1. B 2. A 3. A 4. B 5. C 6. A 7. A 8. D 9. A 10. C Practice Problems 2 1. C 2. A 3. D 4. B 5. A 6. A 7. B 8. D 9. A 10. C Previous Years’ Questions 1. A 2. B 3. A 4. B 5. 18 to 22 6. A 7. 100 8. C 9. C 10. A 11. B 12. A (B) F 60º Rx Ry M (C) F 60º Ry M (D) F 60º Rx Ry

Friction Definitions 1. Static friction: It is the friction between two bodies which is a tangential force and which opposes the sliding of one body relative to the other. 2. Limiting friction: It is the maximum value of the static friction that occurs when motion is impending. 3. Kinetic friction: It is the tangential force between two bodies after motion begins. Its value is less than the corresponding static friction. 4. Angle of friction: It is the angle between the action line of the total reaction of one body on another and the normal to the common tangent between the bodies when motion is impending. It is also defined as the angle made by the resultant (S) of the normal reaction (R) and the limiting force of friction (F) with the normal reaction R (see the figure given below). It is denoted by f. From the figure, we have: tan l m m = = = = F R R R coefficient of friction 5. Coefficient of static friction W P R S F f R S F F f It is defined as the ratio of the limiting force of friction (F) to the normal reaction (R) between two bodies (see above figure, where a solid body rests on a horizontal plane). It is denoted by m. m = = Limiting force of friction Normal reaction F R \ = F mR Friction, Centre of Gravity, Moment of Inertia ☞ Friction ☞ Laws of Friction ☞ Cone of Friction ☞ determination of the Center of Gravity of a Thin Irregular Lamina ☞ Integration Method for Centroid Determination in a Thin Lamina or Solid ☞ Area Moment of Inertia ☞ Polar Moment of Inertia ☞ Perpendicular Axis Theorem ☞ Centroid of Solids ☞ Mass Moment of Inertia ☞ Mass Moment of Inertia and Radius of Gyration CHAPTER HIGHLIGHTS Chapter 3

3.32 |  Part III  •  Unit 1  •  Engineering Mechanics 7. Angle of repose R S F W a f The above figure shows a block of weight W on a rough and plane inclined at an angle a with the horizontal. Let R be the normal reaction and F be the force of friction. From applying the condition of equilibrium, algebraic sum of the forces resolved along the plane: = = W F sin a (1) Algebraic sum of the forces resolved perpendicular to the plane: = = W R cos a (2) From equations (1) and (2) tan a F R But tan f = F R \ = Angle of plane Angle of friction Suppose the angle of the plane a is increased to a value f, so that the block is at the point of sliding or the state of impending motion occurs, then at this angle, m = tanl = tana \ l = a Hence, the angle of repose is defined as the angle to which an inclined plane may be raised before an object resting on it will move under the action of the force of gravity and the reaction of the plane. Hence, angle of repose = angle of plane Laws of Friction First law: Friction always opposes motion and comes into play only when a body is urged to move. Frictional force will always act in a direction opposite to that in which the body tends to move. Second law: The magnitude of the frictional force is just sufficient to prevent the body from moving. That is, only as much resistance as required to prevent motion will be offered as friction. Third law: The limiting frictional force or resistance bears a constant ratio with the normal reaction. This ratio depends on the nature of the surfaces in contact. The limiting frictional resistance is independent of the area of contact. Fourth law: When motion takes place as one body slides over the other, the magnitude of the frictional force or resistance will be slightly less than the offered force at that condition of limiting equilibrium. The magnitude of the frictional force will depend only on the nature of the sliding surfaces and independent of the shape or extent of the contact surfaces. Force Determinations for Different Scenarios Least force required to drag a body on a rough horizontal plane: W P R F = μR q Force P is applied, at an angle q to the horizontal, on a block of weight W such that the motion impends or the block tends to move. Force, P W = - sin cos ( ) a q f Least force required, Pleast = W sinf Force acting on a block (weight = ) along a rough inclined plane: W P R F = μR a Motion direction For motion down the plane, P W= sin( - ) cos a f f For motion up the plane, P W= sin ( + ) cos a f f Force acting horizontally on a block (weight = ) resting on a rough inclined plane: For motion down the plane, P = W tan (a - f) For motion up the plane, P = W tan (a + f) W R P F = μR a Motion direction

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.33 Force acting, at an angle q to the plane, on a block (weight = ) resting on a rough inclined plane: P R W F = mR a q Motion direction sin( ) cos( ) For motion down the plane, P W= - + a f q f For motion up the plane, P W= + - sin( ) cos( ) a f q f Cone of Friction f f O R S B A Axis Direction of frictional force Point of contact Direction in which motion impends Let OR represent the normal reaction offered by a surface on a body and let the direction of impending motion be along OA while the direction in which the frictional force acts is in the opposite direction, i.e., along OB. Assuming that the body is in a state of limiting equilibrium, the resultant reaction S makes an angle of f with the normal OR. If the body slides in any other direction, the resultant reaction S will still make the same angle f with the normal. It is, therefore, seen that when limiting equilibrium is maintained, then the line of action of the resultant reaction should always lie on the surface of an inverted right circular cone; whose semi-vertical angle is f. This cone is known as the cone of friction. Solved Examples Example 1: Determine whether the 2 kN block, shown in the figure, will be held in equilibrium by a horizontal force of 3 kN? The coefficient of static friction is 0.3. 3 kN W = 2 kN 30° (A) 0.96 kN (B) 0.86 kN (C) 0.75 kN (D) 0.65 kN Solution: 3 kN 2 kN Applying the conditions of equilibrium and summing the force parallel and perpendicular to the plane, we have: Σ F(parallel to the plane) = 0 = -F - 2 sin30° + 3 cos 30° = 0 F = -2 × + × 1 2 3 0.866 = -1 + 2.598 = 1.598 kN Σ F(perpendicular to the plane) = 0 R - 2 cos 30° - 3 sin30° = 0 R = 2 × 0.866 + 3 × 0.5 = 1.732 + 1.5 = 3.232 kN This indicates that the value of F necessary to hold the block from moving up the plane is 1.598 kN. However, the maximum value obtainable as the frictional force, F = mR = 0.3 × 3.232 = 0.9696 kN This means that the block will move up the plane. Example 2: An effort of 2 kN is required just to move a certain body up an inclined plane of angle 15°, the force acting parallel to the plane. If the angle of inclination of the plane is made 20°, the effort required, again applied parallel to the plane, is found to be 2.3 kN. Find the weight of the body and the coefficient of friction. (A) 3.9 kN, 0.258 (B) 4.5 kN, 0.26 (C) 3.8 kN, 0.24 (D) 3.8 kN, 0.268 Solution: Let W be the weight of the body, m be the coefficient of friction and P be the effort when the inclination of the plane is a. Applying the conditions of equilibrium and summing the forces parallel and perpendicular to the plane, we have, Σ F(parallel to the plane) = 0 P - mR - Wsina = 0 (1) Σ F(perpendicular to the plane) = 0 R - W cosa = 0 (2) Eliminating R from equations (1) and (2) we have, P = mW cosa + W sina or

3.34 |  Part III  •  Unit 1  •  Engineering Mechanics P = W (mcos a + sina) (3) When a = 15°, P = 2 kN and when a = 20°, P = 2.3 kN. Substituting in equation (3) we have, 2 = W(mcos a + sin a) 2 = W(mcos 15° + sin 15°) (4) 2.3 = W(mcos 20° + sin 20°) (5) Dividing equation (5) by (4), we have, 2 2 3 15 15 . 20 20 cos sin cos sin = ° + ° ° + ° m m 2 2 3 0 966 0 258 . 0 939 0 342 . . . . = × + × + m m or m[(2.3 × 0.969) - (2 × 0.939)] = × [(2 0. ) 342 - × ( . 2 3 0. ) 258 ] or 0.3507m = 0.0906 m = = 0 0906 0 3507 0 258 . . . From equation (5), 2.3 = W [0.258 × 0.939 + 0.342] = W (0.242 + 0.342) = 0.584W W = = 2 3 0 584 3 938 . . . kN Virtual Work Virtual displacement: Virtual displacement is defined as an infinitesimal (exceedingly small) and displacement, given hypothetically to a particle or to a body or a system of bodies in equilibrium consistent with the constraints. The displacement is only imagined and it does not have to take place for which it is called virtual displacement. Virtual work: Virtual work is defined as the work done, by a force on a body due to a small virtual (i.e., imaginary) displacement of the body. Principle of Virtual Work If a system of forces acting on a body or a system of bodies be in equilibrium and if the system be assumed to undergo a small displacement consistent with the geometrical conditions, then the algebraic sum of the virtual work done by the forces of the system is zero. y B A F k r r h C A′ y x x x q a To illustrate the principle of work, let us consider a body at equilibrium at a point A. A force F acts on the body and displaces it to the point A′, where the displacement consists of the following: 1. A very small rotation through the angle a about the origin of the rectangular 2-D coordinate system, say origin O in the xy plane. 2. A very small displacement h along the x-axis, and, 3. A very small displacement k along the y-axis. If the components of the force F along the x-axis and y-axis are Fx and F y respectively, then work done by the force F when its point of application is displaced from point A to A′ = + hFx kFy + - a( ) xFy yFx If a system of forces act on the body where h, k and a are the same for every force, then work done by all the forces: = h F ∑ x + k F ∑ y + - a∑ xFy yFx ( ), Where, ∑ FX and ∑ FY are the sums of the resolved parts of the forces along the x-axis and y-axis respectively, and ∑( ) xFy - yFx is the moments of the forces about the origin O. Since the system is in equilibrium, all the three terms in the above expression, for the work done by all the forces, is zero. Hence, the sum of the virtual works done by the forces is zero. Lifting Machine Lifting machines are defined as those appliances or machines which are used for lifting heavy loads. They are also called simple machines. Some commonly used machines are: 1. Lever 2. Inclined plane 3. Wedge 4. Wheel and axle 5. Winch crab 6. A pulley and system of pulleys 7. Screw jack Screw jack is the most important among all the above simple machines. Load or resistance: A machine has to either lift a load or overcome a resistance. It is usually denoted by W and its unit is N. Examples: A lifting device lifts a load or heavy weight whereas a bicycle overcomes the frictional resistance between the wheels and the road. Efforts: It is the force which is applied to a machine to lift a load or to overcome the resistance against a movement. It is usually denoted by P and its unit is N. Examples: Force applied on the pedals of a bicycle or on the handle of a screw jack.

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.35 Input of a machine: It is defined as the amount of total work done on the machine. This is measured by the product of the effort and the distance through which it moves. Input = Effort × Distance moved by the effort = P × y It has the unit of Nm. Output of a machine: It is defined as the amount of work got out of a machine or the actual work done by the machine Output of the machine: = Load × Distance through which load is lifted = W × x It has the unit of Nm. Velocity Ratio (VR): It is defined as the ratio of the distance moved by the effort and to the distance moved by the load during the same interval of time. VR = = Distance moved by the effort Distance moved by the load y x In all machines y > x. NOTE Mechanical advantage (): It is defined as the ratio of the load or weight lifted to the effort applied. MA W P = = Weight lifted Effort applied In all machines W > P. NOTE Ideal machine: It is defined as the machine which is absolutely free from frictional resistances. In such a machine, input = output. For an ideal machine, VR = MA Efficiency of a machine: It is the ratio of output of the machine to the input of the machine. h= output of the machine ×100 input of the machine = useful work done by the machine ×100 energy supplied to the machine = × × × W x P y 100 For an ideal machine, h = 100%. For an actual machine, h = = Ideal effort Actual effort Actual load Ideal load . Relation between MA, VR and h h= × × = = W x P y W P y x MA VR Frictional losses Output = Input - Losses due to friction Effort lost in friction = - P W VR Loss in load lifted due to friction = P × VR - W. Here P is the actual effort required to overcome resistance W or lift load W. Reversible and Irreversible Machine A machine is said to be reversible when the load W gets lowered on the removal of the effort. In such a case, work is done by the machine in reverse direction. A machine is said to be irreversible when the load W does not fall down on the removal of the effort. In such a case, work is not done in the reverse direction. The condition of irreversibility or self locking of a machine is that its efficiency should be less than 50%. Compound Efficiency It is defined as the overall efficiency of the combination of machines and it is the product of the efficiencies of the individual machines. The overall efficiency h of n machines coupled together is h h = = ∏ i i n 1 , where, ηi is the efficiency of the ith machine. Law of a Machine It is defined as the relationship which exists between the effort applied and the load lifted. P m= + W C P is the effort applied, W is the corresponding load, m and C are coefficients which are determined in any machine after conducting a series of tests and plotting the W versus P graph. The expression for maximum mechanical advantage is given by ( ) MA max m = 1 . The expression for maximum efficiency is given by hmax ( ) = . × 1 m VR Screw Jack It is a device for lifting heavy loads by applying comparatively a smaller effort at the end of the handle. The screw jack works on the principle of inclined plane. Nut l W d Screw head Handle

3.36 |  Part III  •  Unit 1  •  Engineering Mechanics It mainly consists of a nut which forms the body of the jack and a screw is fitted into it. The threads are generally square. The load W is placed on the head of the screw. By rotating the screw with a handle the load is lifted or lowered. Let W be the load lifted, a be the angle of helix of the screw and f be the angle of friction. Here, efficiency = + tan tan( ) a a φ , which shows that efficiency is independent of the load lifted or lowered. Assuming that the effort is applied at the end of the handle, let us consider the following two cases. Let the weight W be lifted: Let PE be the effort applied at the end of the handle. Let l be the length of the handle and let d be the mean diameter of the screw. Σ m about, the axis is zero. Let p be the pitch and m be the coefficient of friction, then: tan a p = p d tanf = m P Wd l p d d p E 2 ⋅ + - mp p m Let the weight W be lowered: Let Q be the effort applied at the circumference of the screw and let QE be the actual effort applied at the end of the handle. Q = W tan(f - a ) Q Wd l d p d p E = ⋅ - 2 + mp p m For an n-threaded screw, tana = np/pd. Differential Screw Jack Instead of only one threaded spindle as in the case of a simple screw jack it has two threaded spindles S1 and S2. The spindle S1 is screwed to the base which is fixed. S2 S1 l W This spindle carries both internal as well as external threads. The spindle S2 is engaged to spindle S1 by means of an internal thread. When spindle S1 ascends, the spindle S2 descends. This is also known as ‘Differential Screw’ jack. The principle of working of this jack is similar to the one as described in the above figure. Let ps1 = pitch of the threads on S1 ps2 = pitch of the threads on S2 Let the lever length be l and the effort be applied at the end of this lever. When the lever is moved by one revolution, the distance covered by the effort P is 2p l and correspondingly the load distance is equal to ps1 - ps2 . Then, velocity ratio ( ) VR . l p p S S = - 2 1 2 p ps1 is always greater than ps2 . Due to this difference, the mechanical advantage as well as the velocity ratio will be more. NOTE Direction for questions 3 and 4: A screw jack has a pitch of 12 mm with a mean radius of thread equal to 25 mm a lever 500 mm long is used to raise a load of 1500 kg. The coefficient of friction is 0.10. Example 3: Find the helix angle a and q (i.e., friction angle) (A) 6.2°, 4.5° (B) 4.85°, 5.7° (C) 4.85°, 5.7° (D) 4.36°, 5.7° Solution: Given P = 12 mm, d = 2r = 25 × 2 = 50 mm, l = 500 mm W = 1500 kg, m = 0.10, tanf = m = 0.10, f = 5.71° tan . a p p = = × = P d 12 50 0 076 a = 4.36° Example 4: What force is necessary when applied normal to the lever at its free end (A) 13.319 kg (B) 12.8 kg. (C) 14.5 kg (D) 18.3 kg. Solution: P wd l = + = × × × + 2 1500 50 2 500 tan(a φ) tan ( . 4 36 5. ) 71 P = 13.319 kg. Direction for questions 5, 6 and 7: A uniform ladder of weight 500 N and the length 8 m rests on a horizontal ground and leans against a smooth vertical wall. The angle made by the ladder with the horizontal is 60°. When a man of weight 500 N, stands on the ladder at a distance of 4 metre from the top of the ladder, the ladder is at the point of sliding. Example 5: Find the coefficient of friction in terms of RB. B 60° RB RA μRA A W + w

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.37 (A) m = RB 1000 (B) m = 1400 RB (C) m = 500 RB (D) m = RB 500 Solution: Resolving all the forces RB = mRA RA = W + w = 500 + 500 = 1000 RB = m × RA = m × 1000 m = RB 1000 . Example 6: Find the reaction at B (i.e., RB) (A) 289 (B) 300 (C) 350 (D) 400 Solution: Taking moment at A, MA = 0 RB × ×8 = ××+ × × 3 2 500 8 2 1 2 500 4 1 2 RB = × + = 500 2 1000 6 92 289 . Example 7: Find the value of coefficient of friction (A) 0.370 (B) 0.486 (C) 0.289 (D) 0.355 Solution: From equation m = = = RB 1000 289 1000 0.289 Centre of Gravity The centre of gravity of a body is the point, through which the whole weight of the body acts, irrespective of the position in which body is placed. This can also be defined as the centre of the gravitational forces acting on the body. It is denoted by G or c.g. Centroid: It is defined as that point at which the total area of a plane figure (like rectangle, square, triangle, quadrilateral, circle etc.) is assumed to be concentrated. The centroid and the centre of gravity are one and the same point. It is also denoted by G or c.g.. Centroidal axis: It is defined as that axis which passes through the centre of gravity of a body or through the centroid of an area. Lamina: A very thin plate or sheet of any cross-section is known as lamina. Its thickness is so small that it can be considered as a plane figure or area having no mass. Determination of the Centre of Gravity of a Thin Irregular Lamina x1 x 2 a1 a2 y1 yG y2 xG G y o x The above figure shows an irregular lamina of total area A whose center of gravity is to be determined. Let the lamina be composed of small areas a1, a2… etc. Such that: A = a1 + a2 + … = Sai Let the distances of the centroids of the areas a1, a2, … etc. from the x-axis be y1, y2, … etc. respectively and from the y-axis be x1, x2, … etc. The sum of moments of all the small areas about the y-axis = a1 x1 + a2 x2 + … = S ai xi Let xG and yG be the coordinates of the centre of gravity G from the y-axis and x-axis respectively. From the principle of moments, it can be written that: AxG = Sai xi or x a x A a x a G i i i i i = ∑ = ∑ ∑ Similarly, it can be shown that: y a y a G i i i = ∑ ∑ 1. The axis of reference of a plane figure is generally taken as the bottom most line of the figure for determining yG and the left most line of the figure for calculating xG. 2. If the figure is symmetrical about the x-axis or y-axis, then the centre of gravity will lie on the axis of symmetry. 3. For solid bodies, elementary masses m1, m2, etc., are considered instead of the areas a1, a2, etc., and the coordination of centre of gravity are given as follows: x m x m y m y m G i i i G i i i = ∑ ∑ = ∑ ∑ , NOTES Example 8: Determine the position of the center of gravity for the following figure. 2 m 2 m 5 m 5 m 3 m 3 m 10 m 10 m

3.38 |  Part III  •  Unit 1  •  Engineering Mechanics Solution: y B C D E G′ G F C′ A H O x The x-axis and y-axis of reference are chosen as shown in the above figure such that the origin O coincides with the point A of the figure and the axes coincide with the left most and bottom most lines of the figure respectively. The position of the centre of gravity is determined with respect to the origin O. The figure is broken down into the three areas AHGG′, G′FC′B and CC′ED For rectangle AHGG′, Area A1 = 3 × 2 = 6 m2 c.g. coordinates, x1 2 2 = = 1 m y1 3 2 = = 1 5. m For rectangle G′FC′B, Area A2 = (2 + 10) × (5 - 3) = 24 m2 c.g. coordinates, x2 2 10 2 = 6 + = ( ) m y2 3 5 3 2 = + 4 - = ( ) m For rectangle CC′ED, Area A3 = 3 × 2 = 6 m2 c.g. coordinates, x3 10 2 2 = + = 11m y3 5 3 2 = = + 6 5. m c.g. of the figure coordinates, x A x A x A x A A A G = + + + + = 1 1 2 2 3 3 1 2 3 6 m y A y A y A y A A A G = + + + + = 1 1 2 2 3 3 1 2 3 4 m Integration Method for Centroid Determination in a Thin Lamina or Solid In this method, the given figure is not split into shapes of figures of known centroid as done in the previous section. The centroid is directly found out by determining Sai yi or Sai xi and Sai by direct integration. First moment of area: Consider a plane region of area A as shown in the following figure. y x G y x yG xG dA Plane region • Let dA be a differential (i.e., infinitesimal) area located at the point (x, y) in the plane region area A. Here, A dA A = ∫ First moments of the area about the x-axis and y-axis are respectively: M y X dA A = ∫ M x Y dA A = ∫ The coordinates (xG, yG) of the centre of gravity of the plane region is given by: x M A xdA dA G y A A = = ∫ ∫ y M A ydA dA G X A A = = ∫ ∫ 1. If the x-axis passes through the centre of gravity, then Mx = 0. Similarly, M y = 0, when the y-axis passes through the centre of gravity. 2. If the plane region is symmetric about the y-axis, then M y = 0 and xG = 0, i.e., the centre of gravity would lie somewhere on the y-axis. Similarly, Mx = 0 and yG = 0, if the plane region is symmetric about the x-axis, i.e., the centre of gravity would lie somewhere on the x-axis. NOTES If instead of a plane region, we have a plane curve of length L and on which a differential length dL is considered which is located at the point (x, y) on the curve, then the coordinates of the centre of gravity for the planar curve is given as follows: x M L xdL dL G y L L = = ∫ ∫ y M A ydL dL G X L L = = ∫ ∫

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.39 Example 9: The centre of gravity of the following shown area OBC, where the curve OC is given by the equation y = 0.625x2, with respect to the point O (0, 0) is: y C O x 4 B (A) (6, 5) (B) (6, 3) (C) (3, 5) (D) (3, 3) Solution: y y C x x dx O(0, 0) B(4, 0) Let us consider an elementary rectangular area of height y and width dx as shown in the above figure. Area of the elementary rectangle, dA = ydx = 0.625x2 dx Area of OBC, A d = = A x dx ∫ ∫ 0 625 2 0 4 0 4 . = × 0 625 4 3 3 . Moment of area about x-axis, M dA y x dx x X = = ∫ ∫ 0 4 2 0 4 2 2 0 625 0 625 2 . . = × 0 625 2 4 5 2 5 . Moment of area about y-axis, Mx = = dAx x dxx = × ∫ ∫ 0 4 2 0 4 4 0 625 0 625 4 4 . . Let xG and yG be the x and y coordinates of the centre of gravity of OBC with respect to the point O. Then, Mx = AyG and M y = AxG yG 0 625 2 4 5 3 0 625 4 3 2 5 3 . . × × × = xG = × × × 0 625 = 4 4 3 0 625 4 3 4 3 . . Example 10: The centre of gravity of the following hatched figure with respect to the point E is: y A 40 80 30 40 60 B C D F x E (A) (20, 30) (B) (37.84, 27.45) (C) (20, 27.45) (D) (37.84, 30) Solution: For Δ ABC, area A1 = 1 2 80 × (60 - 40) = 800 c.g. coordinates, x1 2 3 80 160 3 = × = y1 40 1 3 60 40 140 3 = + × - ( ) = For ACFE, area A2 = 40 × 80 = 3200 c.g. coordinates, x2 80 2 = = 40 y2 40 2 = = 20 For ΔCFD, area A3 = 1 2 30 × 40 = 600 c.g. coordinates, x3 50 2 3 = + × = 30 70 y3 1 3 40 40 3 = × = Since ΔCFD is cut out from the figure ABFE to obtain the hatched figure, the area of ΔCFD is assigned a negative sign. \ A3 = -600 Let xG and yG be the x and y coordinates of the centre of gravity of the hatched figure with respect to the point E, then: x A x A x A x A A A G = + + + + = 1 1 2 2 3 3 1 2 3 37.84 y A y A y A y A A A G = + + + + = 1 1 2 2 3 3 1 2 3 27.45 Theorems of Pappus–Guldinus A surface of revolution is a surface which can be generated by rotating a plane curve about a fixed axis.

3.40 |  Part III  •  Unit 1  •  Engineering Mechanics x L y y L r x G B r A r • For example, in the above figure, the curved surface of a cylinder is obtained by rotating the line AB about the x-axis. Theorem I The area of a surface of revolution is equal to the product of the length of the generating curve and the distance traveled by the centroid of the curve while the surface is being generated. The generating curve must not cross the axis about which it is rotated. NOTE In the above figure, length of the generating curve = L. Distance traveled by the centroid while the surface is being generated = 2pr (circumference of a circle of radius r) \ Area of the surface of the cylinder generated = L × 2pr = 2prL A body of revolution is a body which can be generated by rotating a plane area about a fixed axis. y A B x r r O y x r r (a) (b) For example, in the above figure, the volume of a sphere is obtained by rotating the semi-circle OAB about the x-axis. Theorem II The volume of a body of revolution is equal to the product of the generating area and the distance traveled by the centroid of the area while the body is being generated. The theorem does not apply if the axis of rotation intersects the generating area. NOTE In the figure, generating area = 1 2 pr2. Distance traveled by the centroid of the area while the body is being generated = 2 4 3 p p × r (circumference of a circle of radius 4r 3p ) \ Volume of the sphere generated = × × = 1 2 2 2 3 4 3 4 3 p p p p r r r Example 11: A quartered circular arc AB when rotated about the y-axis generates a surface of area Ay. The same r r r A B x y arc when rotated about the x-axis generates a surface of area Ax . If the ratio A y : Ax is related to the length r by the equation Ay Ax krn = , where k, n are constants, then the value of k and n respectively are (A) 0.27 and 0 (B) 0.27 and 1 (C) 3.75 and 0 (D) 3.75 and 1 Solution: Length of the arc = 1 2 p r; x coordinate of the centroid of the arc = - 2 2 r r p . Distance travelled by the centroid when the arc is rotated about the y-axis = 2 2 p p × -1 p r( ) Using Pappus–Guldinus theorem I, A r r r r y = - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × - 2 = - 2 2 2 1 2 1 2 p p p p p p ( ) ( ) y coordinate of the centroid of the arc = -r 2r p . Distance travelled by the centroid when the arc is rotated about the x-axis = 2 2 p p × - p r( ) . Using Pappus–Guldinus theorem I, A r r r r x = - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × - = - 2 2 2 2 2 p p p p p p ( ) ( ) \ = = - - Ay Ax krn 2 1 2 ( ) p p ⇒ = = - - n k 0 2 1 2 and ( ) . p p

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.41 Example 12: A solid ring (torus) of circular cross-section is obtained by rotating a circle of radius 25 mm about the x-axis as shown in the following figure. 100 mm x y If the density of the material making up the circular crosssection is 7800 kg/m3, the weight of the ring generated is: (A) 82.6 N (B) 94.4 N (C) 123.4 N (D) 90.6 N Solution: y coordinate of the centroid of the circle = 100 mm = 0.1 m Area of the circle = p × (0.025)2 Distance traveled by the centroid of the circle while generating the ring = 2p × (0.1) (circumference of a circle of radius 0.1 m) Using Pappus–Guldinus theorem II, Volume of the ring generated = p × (0.025)2 × 2p × (0.1) = 0.001233 m3 Weight of the generated ring = 7800 × 0.001233 × 9.81 = 94.4 N. Area Moment of Inertia In a plane region of area A, a differential area dA located at the point (x, y) is considered as shown in the below figure. r x O x y Y dA Plane region The moment of inertia of the area about the x-axis and y-axis respectively are: I y x dA I x dA A y A = = ∫ ∫ 2 2 and I x and I y are also called as the second moments of the area. Polar Moment of Inertia In the above figure, the polar moment of inertia of the area about the point O (actually, about an axis through the point O, perpendicular to the plane of the area) is J r dA J I I A x y 0 2 0 = = + ∫ The above equation states that the polar moment of inertia of an area about a point O is the sum of the moments of inertia of the area about two perpendicular axes that intersect at O. Radius of Gyration In the above figure, the radii of gyration of an area about the x-axis, y-axis and the origin O are: k I A k I A and k J A x x y y o o = = , = Parallel Axis Theorem The moment of inertia of a plane region area about an axis, say AB, in the plane of area through the centre of gravity of the plane region area be represented by I G, then the moment of inertia of the given plane region area about a parallel axis, say OX, in the plane of the area at a distance d from the centre of gravity of the area is I X = IG + Ad 2, Plane region, area = A X A B O G d Where, I X = moment of inertia of the given area about the OX axis IG = moment of inertia of the given area about AB axis A = area of the plane region d = perpendicular distance between the parallel axes AB and OX G = centre of gravity of the plane region Perpendicular Axis Theorem If IOX and IOY are the moments of inertia of a plane region area about two mutually perpendicular axes OX and OY in the plane of the area, then the moment of inertia of the plane region area IOZ about the axis OZ, perpendicular to the plane and passing through the intersection of the axes OX and OY is: I I OZ = + OX IOY

3.42 |  Part III  •  Unit 1  •  Engineering Mechanics Y Plane region X Z O I OZ is also called as the polar moment of inertia and the axis OZ is called as the polar axis. NOTE Example 13: In the below figure, the axes AB and OX are parallel to each other. If the moments of inertia of the rectangle PQRS along the axis AB, which passes through the centroid of the rectangle, and the axis OX are I G and I X respectively, then the value of IX/IG is X A Q R G P S B O • (A) 4 (B) 12 (C) 3 (D) 0.25 Solution: From parallel axis theorem, we have IX = IG + A (perpendicular distance between axes)2, Let PQ = d and QR = b, then the perpendicular distance between the axes = d 2 \ = I I + = A + d I bd d X G G 2 2 4 4 So, I I bd I X G G = +1 4 3 To determine I G, let us consider a rectangular strip of thickness dy at a distance y from the axis AB as shown below: A B P S Q R y dy d 2 d 2 −d b Area of the rectangular strip = bdy Moment of inertia of the strip about the axis AB = (bdy) y2 Moment of inertia of the rectangle PQRS about the axis AB, I by dy bd G d d = = - ∫ 2 2 2 3 12 \ = I I X G 4. Example 14: The moment of inertia for the following hatched figure about the axis AB (which passes through the centroid of the figure), where AB = DC = 30 m, PQ = SR = 20 m, BC = AD = 20 m and QR = PS = 10 m, is: A B C A B D S R P Q (A) 6.78 × 104 m4 (B) 5.41 × 103 m4 (C) 1.83 × 104 m4 (D) 2.6 × 105 m4 Solution: Moment of inertia of the hatched figure = moment of inertia of  ABCD - Moment of inertia of  PQRS = × × - × = × × - × = 1 12 1 12 30 20 20 10 18333 33 3 3 3 3 4 ( ) ( ) . . DC AD SR QR m Example 15: A circular section of diameter d is lying on the xy–plane where the centre of the circular section coincides with the origin O as shown in the following figure. y x z O If the moments of inertia of the circular section along the x, y and z axes are IX, IY and IZ respectively, then which of the following statements is NOT correct? (A) I d X = p 4 32 (B) IX = IY (C) I d Z = p 4 32 (D) I d Y = p 4 64

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.43 Solution: y x z dr • O r Let us consider an elementary ring of thickness dr and located at a distance r from the origin O. Area of the elementary ring = 2prdr. Moment of inertia of the elementary ring about the z-axis = 2prdr × r2 = 2pr3dr. Moment of inertia of the whole circular section about the z-axis = = ∫ 2 32 3 0 2 4 p p r dr d D/ . From the symmetry of the circular section, it can be written that I X = I Y. From the perpendicular axis theorem,we have, IZ = I X + IY i.e., I Z = 2IX \ = I d I X Y p 4 64 Description Shape L xc yc Horizontal line y x a a 2 a 0 Vertical line y a x a 0 2 a Inclined line with q y x a q a cos 2 a θ       sin 2 a θ       Semicircular arc y r r p r 0 2r π Quarter circular arc y y • x x CG πr 2 2r π 2r π Circular arc y x a/2 a/2 ar 2 sr inα/2 α 0 (Continued)

3.44 |  Part III  •  Unit 1  •  Engineering Mechanics Description Shape L xc yc Rectangle y c b x b/2 h/2 bh 2 b 2 h Square y a c a x a2 2 a 2 a Parallelogram a b c y x a ab sina cos 2 b a + α sin 2 a α Triangle a h b y x 2 bh 3 a b + 3 h Semi circle c o y x 3p 4R R • πR2 2 0 4R 3π Quarter circle c y yc xc R x • πr2 2 4R 3π 4R 3π Sector of a circle y xc • x a a R2a 2 sin 3 R α α 0 Quarter ellipse y xc yc a x b • πab 3 4a 3π 4b 3π

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.45 Quarter parabola y y 2 = kx xc yc a x b • πab 3 3 5 a 3 5 b General spandrel y y = kxn xc yc o a x c b • 3 ab 3 4 a 3 4 b Description Figure Xl Y l Rectangle b/2 b/2 a/2 a/2 x y 3 12 ab 3 12 ba Circle x r y πr4 4 πr4 4 Ellipse b a π ab3 4 π ba3 4 Triangle y h x b C h/3 3 36 bh 3 36 hb Quadrant Circle C r x 4r 3p 4r 3π 0.0549 r 4 0.0549 r4

3.46 |  Part III  •  Unit 1  •  Engineering Mechanics Centroid of Solids If dm is an elemental mass in a body of mass M and xG, yG are the coordinates of the center of gravity of the body from the reference axes y-axis and x-axis respectively, then: X xdm dm xdm M y ydm dm ydm M G G = = = = ∫ ∫ ∫ ∫ ∫ ∫ , Let us consider a right circular solid cone whose centre of gravity is to be determined. Let the diameter of the base of the right circular solid cone be 2R and its height H as shown in the following figure. Since the cone is symmetric about the VX axis, its centre of gravity will lie on this axis. The cone can be imagined to be consisting of an infinite number of circular discs with different radii, parallel to the base. y X x H B 2R A C V F D E dy Consider one such disc of radius x, thickness dy and at a depth y from the vertex of the cone, i.e., from V. From the geometry of the above figure, x R y H or X yR H = = Volume of disc = = p p X dy y R H dy 2 2 2 2 If r is the density of the material making up the cone, then dm = r p y R H dy 2 2 2 \ = = [ ] = ∫ ∫ ∫ y ydm dm y R H dy y R H dy G y H H H H r p r p 3 2 2 0 2 2 2 0 0 3 4 3 4 \ Centroid or centre of gravity of a right circular cone is situated at a distance of 3 4 H from its vertex V and lies on its axis VX. Example 16: In the homogenous hollow hemisphere, shown in the following figure, OP = 10 cm = the radius of the hemisphere. The points P, G and O lie on a straight line that is perpendicular to the base CD. If G is the centroid of the hollow hemisphere, then which one of the following statements is not correct? P G C O D • (A) OG = 5 cm (B) OG = OP 3 8 (C) CO = 10 cm (D) OD = 2 × OG Solution: The centre of gravity of a hollow hemisphere with respect to the x-axis would lie on an perpendicular axis along which the homogeneous hemisphere is symmetrical. Since G is the centre of gravity, then the hemisphere should be symmetrical along OP, i.e., CO = OD. It can also be deciphered that CO = OD = radius of the hemisphere = OP = 10 cm. Now OG will be equal to R/2, where R is the radius of the hollow hemisphere. \ OG = 0.5 OP = 5 cm It can be written OP = CO = OD = 2 OG, hence the option (B) is NOT correct. Option B would be right if the hemisphere had been a homogeneous solid hemisphere. NOTE Mass Moment of Inertia The Moment of Inertia of an element of mass is the product of the mass of the element and the square of the distance of the element from the axis. The mass moment of inertia of the body with respect to Cartesian frame xyz is given by: I y xx z dm y z dv v = + = + ∫ ∫ ( ) ( ) 2 2 2 2 r I x YY z dm x z dv v = + = + ∫ ∫ ( ) ( ) 2 2 2 2 r I x zz y dm x y dv v = + = + ∫ ∫ ( ) ( ) , 2 2 2 r where, IXX, IYY and IZZ are the axial moments of inertia of mass with respect to the x-, y- and z-axes respectively. For thin plates essentially in the x-y plane, the following relations hold. I y dm I x dm I z dm x y dm I I I xx YY zz zz XX YY = = = = + = + ∫ ∫ ∫ ∫ 2 2 2 2 2 ( ) Izz is also called the polar moment of inertia.

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.47 Mass Moment of Inertia and Radius of Gyration I K m I K m I K m K I m K I m K I m xx x yy y zz z x xx y YY z zz = = = = = = 2 2 2 The parallel-axis theorem for the mass moment of inertia states that the mass moment of inertia with respect to any axis is equal to the moment of inertia of the mass with respect to a parallel axis through the centre of mass plus the product of the mass and the square of the perpendicular distance between the axes. Mathematically IAB = IG + md2 For a thin plate, I t I I t I I t I xx xx YY YY zz zz ( ) mass area mass area mass a = = = ( ) ( ) ( ) ( ) r r r ( rea) Where t is the uniform thickness and r is the mass of the thin plate. I I I zz = + xx YY The mass moment of inertia about a centroidal axis perpendicular to a uniform thin rod of length , mass m and small cross section is given by I m YY = 1 12 2  Radius of gyration about a centroidal axis perpendicular to a uniform thin rod of the length , mass m and a small cross section is given by Ky =  12 The mass moment of inertia about the longitudinal and transverse axes passing through the centre of mass of a rectangular prism (block) of cross section (axb), uniform density r and length  is given by I m xx = + b 1 12 2 2 ( )  . I I YY zz m a b m a = + = + 1 12 1 12 2 2 2 2 ( ) ( )  In the above case, if the three axes were chosen through a corner instead of centre of mass, the results are: I I I xx YY zz m b m a b m a = + = + = + 1 3 1 3 1 3 2 2 2 2 2 2 ( ) ( ) ( )   For a right circular cylinder of radius R, length or height  and mass m, the mass moment of inertia about the centroidal x-axis is given by I m R xx = + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 4 2 4 12  Solid Body Centroid Mass moment of inertia Solid hemisphere z G R O x y xG = yG = 0 3 8 Z R G = 2 2 5 XX YY ZZ I I = = I m = R Solid sphere z G R O x y xG = yG = zG = 0 2 2 5 2 5 y XX YY ZZ I I I mR K R = = = = (Continued)

3.48 |  Part III  •  Unit 1  •  Engineering Mechanics Solid Body Centroid Mass moment of inertia Solid cylinder R G O z y L x xG = yG = 0 2 G L z = 2 2 2 1 1 4 3 1 2 XX YY ZZ I I mR mL I mR = = + = Rectangular block (cuboid) z b G O a x y L xG = yG = 0 2 G L z = 2 2 2 2 2 2 1 1 12 3 1 1 12 3 1 ( ) 12 xx yy zz I ma mL I mb mL I m a b = + = + = + Slender rod (thin cylinder) L G x y z O 2 G L z = yG = zG = 0 I xx = 0 2 3 YY ZZ mL I I = = Solid disk z R O x y xG = yG = zG = 0 2 2 4 2 2 XX YY zz z mR I I mR I r K = = = =

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.49 Exercises Practice Problems 1 Select the correct alternative from the given choices. 1. A belt supports two weights W1 and W2 over a pulley as shown in the figure. If W1 = 2000 N, the minimum weight W2 to keep W1 in equilibrium (assume that the pulley is locked and m = 0.25) is: • O A B T1 T2 W1 W2 2000 N b (A) 911.9 N (B) 812.8 N (C) 913 N (D) 715.5 N 2. B A C • • r = 0.25 m 25 cm 50 cm 45° A rotating wheel is braked by a belt AB attached to the lever ABC hinged at B. The coefficient of friction between the belt and the wheel is 0.5. The braking moment exerted by the vertical weight W = 200 N is: (A) 98.23 Nm (B) 95.96 Nm (C) 95.00 Nm (D) 93.24 Nm 3. A screw jack has square threaded screw of 5 cm diameter and 1 cm pitch. The coefficient of friction at the screw thread is 0.15. The force required at the end of a 70 cm long handle to raise a load of 1000 N and the force required, at the end of the same handle to raise the same load, if the screw jack is considered to be an ideal machine, respectively, are: (A) 7.702 N and 2.123 N (B) 7.702 N and 2.273 N (C) 8.162 N and 1.850 N (D) 8.162 N and 1.798 N Direction for question 4: A locomotive of weight W is at rest. 4. The reactions at A and B are B W RA RB A a a b C P • • (A) W N 2 (B) 2WN (C) 2 3 WN (D) 3WN Direction for question 5: When it is pulling a wagon, the draw bar pull P is just equal to the total friction at the points of contact, A and B. 5. The new magnitudes of the vertical reactions at A and B respectively are: (A) Wa Pb a Wa Pb a - + 2 2 , (B) W W 2 2a , (C) W W 2 3 , (D) W W 2 2 3 , 6. A four wheel vehicle with passengers has a mass of 2000 kg passengers. The road, on which the vehicle is moving, is inclined at an angle q with the horizontal. If the coefficient of static friction between tyres and the road is 0.3, the maximum inclination q at which the vehicle can still climb is: • • 1 m 0.25 m mg 0.5 q q (A) 18° (B) 16.7° (C) 15° (D) 17.2° 7. A weight W of 2000 N is to be raised by a system of pulleys as shown in the following figure. W P • • 2000 N + d v 2 y −d

3.50 |  Part III  •  Unit 1  •  Engineering Mechanics The value of the force P which can hold the system in equilibrium is: (A) 5000 N (B) 1000 N (C) 2000 N (D) 1500 N Direction for questions 8, 9 and 10: A weight of 600 N just starts moving down a rough inclined plane supported by a force of 200 N acting parallel to the plane and it is at the point of moving up the plane when pulled by a force of 300 N parallel to the plane. F R q q P = 200 N • 8. The values of the normal reaction R and the limiting friction F, respectively are: (A) 500 cosq and 500mcosq (B) 400 cosq and 400mcosq (C) 600 cosq and 600mcosq (D) 600 cosq and 500mcosq 9. The inclination of the plane q is: (A) 30° (B) 25.6° (C) 24.6° (D) 32.1° 10. The coefficient of friction is: (A) 0.092 (B) 0.1124 (C) 0.1510 (D) 0.2130 Practice Problems 2 Direction for questions 1 to 10: Select the correct alternative from the given choices. 1. The block shown in figure below is kept in equilibrium and prevented from sliding down by applying a force of 600 N. The co-efficient of friction is 3 5 . The weight of the block would be: 600 N 30° (A) 4000 N (B) 2500 N (C) 3000 N (D) 5000 N 2. Mention the statements which are governing the laws of friction between dry surfaces. (i) The friction force is independent on the velocity of sliding. (ii) The friction force is proportional to the normal force across surface of contact (iii) The friction force is dependent on the materials of the contacting surfaces. (iv) The friction force is independent of the area of contact. (A) 2, 3, 4 (B) 1 and 3 (C) 2 and 4 (D) 1, 2, 3 and 4 3. The limiting friction between two bodies in contact is independent of (A) Nature of surfaces in contact; (B) The area of surfaces in contact; (C) Normal reaction between the surfaces. (D) All of the above. 4. A body of weight 50 N is kept on a plane inclined at an angle of 30° to the horizontal. It is in limiting equilibrium. The co-efficient friction is the equal to: (A) 1 3 (B) 3 (C) 1 50 3 (D) 3 5 5. A man of weight 60 N stands on the middle rung of a ladder of weight 15 N. The co-efficient of friction between contacting surfaces is 0.25. The reaction at the floor is: (A) 80 N (B) 73.25 N (C) 85.6 N (D) 72.75 N 6. Determine the effort required at the end of an arm 50 cm long to lift a load of 5 kN by means of a simple screw jack with screw threads of pitch 1 cm if the efficiency at this load is 45%. (A) 40.8 N (B) 43.6 N (C) 44.8 N (D) 35.36 N 7. Determine the effort needed if the jack in above question is converted into a differential screw jack with internal threads of pitch 7 mm and efficiency of operation is 30%. (A) 15.9 N (B) 19.8 N (C) 17.2 N (D) 18 N 8. A wooden block is being split by a 20° wedge with a force of 70 N applied horizontally as shown. Taking the coefficient of friction between wood and the wedge as 0.4 estimate the vertical force tending to split the wood apart. 70 N 20° Wedge

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.51 (A) -54.24 N, 54 N (B) -65 N, 64 N (C) -48 N, 46 N (D) -56 N, 54 N 9. A screw thread of screw jack has a mean diameter of 10 cm and a pitch of 1.25 cm. The co-efficient of friction between the screw and its nut housing is 0.25. The force F that must be applied at the end of a 50 cm lever arm to raise a mass 6000 kg, is: (A) 1985 N (B) 1723 N (C) 1630 N (D) 1874 N 10. Efficiency of the screw jack in problem above is: (A) 12% (B) 13.7% (C) 15% (D) 16.4% Previous Years’ Questions 1. An elevator (lift) consists of the elevator cage and a counter weight, of mass m each. The cage and the counterweight are connected by a chain that passes over a pulley. The pulley is coupled to a motor. It is desired that the elevator should have a maximum stopping time of t seconds from a peak speed v. If the inertia of the pulley and the chain are neglected, the minimum power that the motor must have is: [2005] Pulley Cage v m Counter weight Chain v m (A) 1 2 mv2 (B) mv t 2 2 (C) mv t 2 (D) 2mv2 t 2. If a system is in equilibrium and the position of the system depends upon many independent variables, the principle of virtual work states that the partial derivatives of its potential energy with respect to each of the independent variable must be: [2006] (A) -1.0 (B) 0 (C) 1.0 (D) ∞ 3. A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is m = 0.2. A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right? [2009] T G 100 N μ = 0⋅2 (A) 176.2 (B) 196.0 (C) 481.0 (D) 981.0 4. A 1 kg block is resting on a surface with coefficient of friction m = 0.1. A force of 0.8 N is applied to the block as shown in the figure. The friction force is: [2011] 0⋅8 N 1 kg (A) 0 (B) 0.8 N (C) 0.98 N (D) 1.2 N 5. A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure. Block R is tied to the wall by a massless and inextensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F (in kN) needed to move the block S is: [2014] P Q R S F (A) 0.69 (B) 0.88 (C) 0.98 (D) 1.37 6. A block weighing 200 N is in contact with a level plane whose coefficients of static and kinetic friction are 0.4 and 0.2 respectively. The block is acted upon by a horizontal force (in newton) P = 10t, where t denotes the time in seconds. The velocity (in m/s) of the block attained after 10 seconds is: ____ [2014]

3.52 |  Part III  •  Unit 1  •  Engineering Mechanics 7. A body of mass (M) 10 kg is initially stationary on a 45° inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.5.? The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is _______ [2014] M 45° 8. A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-Y axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip it about point Q without slipping. What are the minimum values of the force (in newton) and the static coefficient of friction m between the floor and the wardrobe, respectively? [2014] Y P Q Y 2 m 4 m (A) 490.5 and 0.5 (B) 981 and 0.5 (C) 1000.5 and 0.15 (D) 1000.5 and 0.25 9. For the same material and the mass, which of the following configurations of flywheel will have maximum mass moment of inertia about the axis of rotation OO′ passing through the center of gravity. [2015] O O′ O O′ (A) Solid Cylinder (B) Rimmed wheel (C) Solid sphere (D) Solid cube O O′ O O′ 10. The value of moment of inertia of the section shown in the figure about the axis-XX is: [2015] 120 45 45 60 All dimensions are in mm X X 15 15 30 30 (A) 8.5050 × 106 mm4 (B) 6.8850 × 106 mm4 (C) 7.7625 × 106 mm4 (D) 8.5725 × 106 mm4 11. A block of mass m rests on an inclined plane and is attached by a string to the wall as shown in the figure. The coefficient of static friction between the plane and the block is 0.25. The string can withstand a maximum force of 20 N. The maximum value of the mass (m) for which the string will not break and the block will be in static equilibrium is _______ kg. [2016] Take cos q = 0.8 and sin q = 0.6 Acceleration due to gravity g = 10 m/s2 θ 12. The figure shows cross-section of a beam subjected to bending. The area moment of inertia (in mm4) of this cross-section about its base is ________. [2016]

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  | 3.53 10 8 10 10 R4 R4 All dimensions are in mm 13. A system of particles in motion has mass center G as shown in the figure. The particle i has mass mi and its position with respect to a fixed point O is given by the position vector ri . The position of the particle with respect to G is given by the vector ρi . The time rate of change of the angular momentum of the system of particles about G is (The quantity ri indicates second derivative of ρi with respect to time and likewise for ri ). [2016] O ri r ρ G m System boundary i (A) Σi i i i r m× r (B) Σi i r × m ri i (C) Σi i i i r m× r (D) Σi i r r × mi i  Answer Keys Exercises Practice Problems 1 1. A 2. B 3. B 4. A 5. A 6. B 7. B 8. C 9. C 10. A Practice Problems 2 1. C 2. A 3. B 4. A 5. D 6. D 7. A 8. A 9. B 10. B Previous Years’ Questions 1. C 2. B 3. C 4. B 5. D 6. 4.8 to 5 7. 56 to 59 8. A 9. B 10. B 11. 5 12. 1873 to 1879 13. B

☞ Dynamics ☞ Rectilinear Motion: Displacement, Distance, velocity and Acceleration ☞ Motion at a Uniform Acceleration ☞ Vertical Motion Under Gravity ☞ Kinetics of a Particle ☞ Differential Equation of Rectilinear Motion ☞ Motion of a Particle Acted Upon by a Constant Force ☞ Freely Falling Body ☞ Dynamics of a Particle ☞ Work and Energy ☞ Law of Conservation of Energy ☞ Impact ☞ Elastic Impact ☞ Plastic or Inelastic Impact ☞ Coefficient of Restitution CHAPTER HIGHLIGHTS Dynamics Dynamics is the branch of mechanics dealing with the motion of a particle or a system of particles under the action of a force. Dynamics is broadly divided into two categories: 1. Kinematics 2. Kinetics Kinematics is the study of motion of a body without any reference to the forces or other factors which causes the motion. Kinematics relates displacement, velocity and acceleration of a particle of system of particles. Kinetics studies the force which causes the motion. It relates the force and the mass of a body and hence the motion of the body. So in fact, the motion of a particle or body is largely covered and interpreted by Kinematics and Kinetics. Types of Motion The rate of change of position is motion. The type of motion is explained by the type of path traced by it. If the path traced is a straight line, the motion is said to be rectilinear motion or translation. If the path traced by the motion (or path traversed by the particle) is a curve, it is known as curvilinear motion. When the curve becomes a circle, then it is known as circular motion. The two types of motion, i.e., rectilinear and curvilinear motions, explained above can be together termed as the general plane motion. Rectilinear Motion: Displacement, Distance, Velocity and Acceleration 1. Displacement and distance: A B x x x Let the particle be at the position A at any point of time t. Let the position of the particle be at B at time t + dt (dt > 0). Then the particle is said to move from A to B. The change in position is the displacement x. It is the shortest distance between A and B. Distance is the length of the path described by the particle from point A to point B. y z x R P Q Let a body start from a point P and move towards a point Q and then turn and reach a point R. During this course of motion, the total displacement is denoted by x. The distance traversed is given by y + z. When the motion of a particle is considered along a line segment, both distance and displacement are the same in magnitudes. NOTE Rectilinear Motion Chapter 4