Pre-U STPM Physics Term 1 CC039242a

Physics Term 1 STPM Chapter 6 Gravitation Speed of B, v2 = Gm1 (r – x) r 2 = (6.67 × 10–11)(8.0 × 1012)(5.0 × 108 ) (7.5 × 108 )2 = 6.89 × 10–4 m s–1 (ii) The period of orbit for A and B are the same because they are always colinear with the centre of mass G. Period of orbit, T = 2πr v = 2π × (2.5 × 108 ) 3.44 × 10–4 = 4.57 × 1012 s (d) If the mass of A >> mass of B, then the centre of mass of the system is almost at the centre of the star A. Hence, star A rotates about its axis and star B orbits about star A as centre. Energy of a Satellite 1. A satellite of mass m in circular orbit of radius r round the Earth of mass M has gravitational potential energy U = – GMm r 2. If the speed of the satellite is v, then mv 2 r = GmM r 2 mv 2 = GmM r Kinetic energy of satellite, K = 1 2 mv2 K = GMm 2r = – 1 2 U 3. Total energy of satellite, E = U + K = – GMm r + GMm 2r = – GMm 2r = 1 2 U 4. Figure 6.15 shows three types of energy which varies with the distance r from the centre of the Earth. Energy m 0 r Total energy, E = K + U Gravitational potential energy, U Kinetic energy, K Figure 6.15 Exam Tips If U = gravitational potential energy of satellite = – GMm r . Kinetic energy, K = – 1 2 U Total energy, E = K + U = 1 2 U Earth m v r M Satellite Figure 6.14 144 6

Physics Term 1 STPM Chapter 6 Gravitation 5. When a satellite in an orbit of radius r1 falls to another orbit of radius r2 < r1, (a) final potential energy U2 = – GMm r 2 < – GMm r 1 , initial potential energy. Hence, the potential energy decreases. (b) final kinetic energy K2 = GMm 2r 2 > GMm 2r 1 initial kinetic energy. (c) final total energy E2 = – GMm 2r 2 < – GMm 2r 1 , initial total energy. 6. There is a loss of energy because work is done to bring the satellite from a higher orbit to a lower orbit. Launching of Discovery on 8th November 1984. Astronaut Dale A. Gardner from Discovery trying to stabilise the Communication Satellite Westar VI. Example 10 (a) What is the relation between the gravitational pull on a mass and the gravitational potential energy of the mass in a gravitational field? (b) The figure on the right shows the variation of gravitational potential energy of a body of mass m in the gravitational field of the Earth with the distance r from the centre of the Earth, for values of r > RE, radius of the Earth. What is represented by the gradient of the tangent to the curve at r = RE? The body referred above is a rocket which is projected upwards from the Earth. At a certain distance R from the centre of the Earth, the total energy of the rocket (gravitational potential energy and kinetic energy) may be represented by a point on the line PQ. Five points A, B, C, D, E have been marked on this line. Which point (or points) could represent the total energy of the rocket A B C R D P r RE E Potential energy Tangent Energy Q 0 145 6

Physics Term 1 STPM Chapter 6 Gravitation (i) if it was momentarily at rest at the top of its trajectory? (ii) if it was falling towards the Earth? (iii) if it was moved away from the Earth with sufficient energy to reach an infinite distance? Explain your answers. Solution: (a) Gravitational pull, F = – dU dr = –(gravitational potential energy gradient) (b) Gradient of tangent at r = RE represents the magnitude of the gravitational pull of the Earth on the body. (i) Since the rocket stops momentarily at the top of its trajectory, its kinetic energy = 0. Total energy = Potential energy Hence, the total energy is represented by the point B. (ii) Since the rocket is falling when r = R, it has kinetic energy. Hence, total energy = Potential energy + Kinetic energy > Potential energy at r = R. Also the rocket does not escape from the gravitational attraction of the Earth. The total energy of the rocket is represented by the point C. (iii) Since the rocket has sufficient energy to reach an infinite distance, its kinetic energy  GMm R . Potential energy = – GMm R Total energy = Potential energy + Kinetic energy  – GMm R + GMm R  0 Hence, the total energy is represented by the point D or E. Quick Check 5 1. A planet of mass m moves in a circular orbit of radius R around the Sun of mass M with period T. Which of the following shows correctly how T depends on m, M and R? A T ∝ m C T ∝ R3 B T ∝ M D T ∝ R 3 2 2. A satellite P of mass m orbits round a planet with a period T. Another satellite Q of mass 2m orbits round the planet with a period 1 3 T. What is the value of the ratio: radius of orbit P radius of orbit Q ? A 0.667 C 1.50 B 1.48 D 2.08 3. A geosynchronous satellite of mass m appears stationary relative to the rotation of the Earth. If ω is the Earth’s angular velocity of rotation and M is the Earth’s mass, what is the radius of the satellite’s orbit? A ( GM ω2 ) 1 3 C ( GmM ω3 ) 1 3 B ( Gm ω2 ) 1 3 D ( GM ω3 ) 1 2 4. Star X of mass 2M and star Y of mass M perform circular motion about the common centre of mass under their gravitational attraction. What is the ratio of radius of X orbit radius of Y orbit ? A 1 4 C 1 B 1 2 D 2 146 6

Physics Term 1 STPM Chapter 6 Gravitation 5. Two stars of equal mass M move with constant speed v in a circular orbit of radius R about the common centre of mass as shown in the figure. What is the net force on each star? A zero C GM2 4R2 B GM2 R2 D Mv 2 2R 6. Why does the Earth stay in orbit at a constant distance from the Sun? A The gravitational pull of the Sun on the Earth is equal and opposite to the gravitational pull of the Earth on the Sun. B The gravitational pull of the Sun on the Earth is just sufficient for the centripetal force. C The gravitational pull of the Sun on the Earth is negligible. D The gravitational pull on the Earth by the Sun is balanced by the centripetal force on the Earth. 7. Which quantity is not necessarily the same for geosynchronous satellites around the Earth? A mass B period of orbit C direction of revolution D orbital plane 8. Which of the following graphs best represents the variation of the period T of a satellite with its radius of orbit r? A C B D G 9. A satellite P of mass m orbits round a planet of mass M in a circular orbit of radius r with angular velocity ω. (a) Write an expression for the centripetal force on the satellite in terms of m, ω and r. (b) What is the gravitational force on the satellite? (c) Use your answers from (a) and (b), deduce an expression for the period of orbit of the satellite. 10. A planet P of mass m orbits around the Sun of mass S1 in a circular orbit. In a different planetary system, a planet Q of mass m orbits about its Sun of mass S2. The radius of Q’s orbit is the same as the radius of P’s orbit but the period of planet P is half that of planet Q. Find the ratio of the mass S1 to S2. 11. The mass of Earth is 6.0 × 1024 kg and the mass of the Moon is 7.4 × 1022 kg. The distance between them is 3.8 × 108 m. (a) Find the position of the centre of mass of the system. (b) Explain why the Earth and the Moon must revolve round the common centre of mass, and not the Moon revolves round the centre of the Earth. 12. A satellite orbits the Earth in a circular orbit of radius r, and the period equal to the period of rotation of the Earth about its axis. If the radius of the Earth is 6.4 × 106 m, calculate the radius of the satellite’s orbit. 13.(a) Calculate the speed of a satellite orbiting close to the Moon’s surface. (b) What is the kinetic energy per unit mass of the satellite? (Radius of Moon = 1.74 × 106 m, mass of Moon = 7.35 × 1022 kg) 14.A satellite of mass m is in a circular orbit of radius r about the Earth. (a) Derive an expression for its kinetic energy T in terms of G, m, r and M, the mass of the P 147 6

Physics Term 1 STPM Chapter 6 Gravitation Earth. Hence, show that T = – 1 2 V, where V is the potential energy of the satellite. Write down the relation between T and the total energy E of the satellite. (b) Due to atmospheric friction, the total energy of the satellite decreases by ∆E. The change is so gradual that the orbit may be assumed to remain circular. Find in terms of ∆E, the corresponding changes in the satellite’s kinetic energy T and potential energy V. State clearly whether each change is an increase or a decrease. 15.(a) What is the velocity of projection of a satellite at a height of 160 km for it to move in a circular orbit around the Earth? (b) What is the period of the orbit? (Radius of Earth = 6.4 × 106 m, mass of Earth = 6.0 × 1024 kg) 6.5 Escape Velocity Students should be able to: • derive and use the equation for escape velocity ve = 2GM R and ve = 2gR Learning Outcome 1. The gravitational potential energy of a rocket of mass m on its launching pad is U0 = – GmM R where M = mass of Earth and R = radius of Earth 2. For the rocket to escape completely from the gravitational pull of the Earth, its distance from the Earth would be infinity, and its final gravitational potential energy U∞ = 0. 3. If sufficient energy in the form of kinetic energy is supplied to the rocket during launching, then Kinetic energy supplied = Gain in gravitational potential energy 1 2 mv 2 e = U∞ – U0 = 0 – ( – GmM R ) ve = 2GM R 4. The minimum velocity required by a body on the surface of the Earth to escape completely from the gravitational pull of the Earth is known as the escape velocity. Escape velocity ve = 2GM R in the direction perpendicular to the Earth’s surface. 5. To find the escape velocity from the surface of the Earth, ve = 2GM R and GM = gR2 ve = 2gR R = 6.4 × 106 m = 2 × 9.81 × 6.40 × 106 = 11.2 × 103 m s–1 = 11.2 km s–1 2017/P1/Q16 148 6

Physics Term 1 STPM Chapter 6 Gravitation 6. To reduce the energy required to launch a spacecraft to escape from the gravitational pull of the Earth, multistage rockets are used. After each stage, the empty fuel tank is ejected to reduce the load. The mass of the rocket decreases. As the spacecraft is further from the Earth, the value of the escape velocity decreases. Hence, less energy is required to achieve the escape velocity. Example 11 The mass of the Sun is 2.0 × 1030 kg and its radius is 7.0 × 108 m. If the mass of a helium atom is 6.6 × 10–27 kg, calculate (a) its gravitational potential energy on the surface of the Sun, (b) the minimum velocity of a helium atom for it to escape from the Sun. Solution: (a) Gravitational potential energy U = – GMm R = –(6.67 × 10–11)(2.0 × 1030)(6.6 × 10–27) 7.0 × 108 = –1.26 × 10–15 J (b) Kinetic energy required, 1 2 mv 2 e = GMm R ve = 2GM R = 2 × (6.67 × 10–11)(2.0 × 1030) 7.0 × 108 = 6.17 × 105 m s–1 Example 12 (a) Show that for a body to escape completely from the gravitational field of the Earth, it must be projected with a speed equal or greater than 2gR , where g is the acceleration due to gravity on the surface of the Earth and R is its radius. (b) The escape velocity from the Earth’s surface is 1.1 × 104 m s–1. What is its value at a height of 0.2R? (c) The root-mean-square (r.m.s.) speed of helium atoms at 290 K, the temperature of the lower atmosphere, is 1.3 × 103 m s–1. Calculate the r.m.s. speed of helium atoms at 1 500 K, the temperature of the atmosphere at a height of 0.2R. (d) The decay of radioactive mineral produces helium which escapes into the atmosphere. The total amount of helium produced by radioactive decay is much more than that found in the atmosphere. Suggest a reason to explain this observation. 149 6

Physics Term 1 STPM Chapter 6 Gravitation Solution: (a) On the Earth’s surface, Gravitational potential energy = – GMm R When a body is projected with a velocity v  2gR Kinetic energy supplied  1 2 m ( 2gR )2  mgR Total energy of body  – GMm R + mgR g = GM R2  – GMm R + GMm R  0, the gravitational potential energy at infinity. Hence, the body will travel to infinity. (b) Escape velocity, ve = 2GM R on the surface of Earth. When r = R + 0.2R = (1.20)R Escape velocity, v1 = 2GM 1.20R = 1 1.20 × 1.1 × 104 = 1.00 × 104 m s–1 (c) vr.m.s. ∝ T 1.3 × 103 ∝ 290 At T = 1 500 K, v ∝ 1 500 v = 1 500 290 × 1.3 × 103 = 3.0 × 103 m s–1 (d) Helium atoms with speeds greater than the escape velocity would escape from the gravitational pull of the Earth. Helium atoms with speeds lower than the escape velocity remain in the Earth’s atmosphere. Quick Check 6 1. A body of mass m is projected from a planet of radius R. The acceleration of free fall is g. To escape from the gravitational field of the planet, the body must be projected with a speed of at least A gR C gR 2 B 2gR D mgR 2. The mass of a planet is 5.0 × 1024 kg and its radius is 6.1 × 106 m. The energy required for a body of mass 2.0 kg to escape completely from the planet is A 1.8 × 105 J B 5.5 × 107 J C 1.1 × 108 J D 2.2 × 108 J 3. The escape velocity at a height of 0.2R from a planet of radius R is 10 km s–1. What is the escape velocity from the surface of the planet? A 10 km s–1 B 11 km s–1 C 12 km s–1 D 13 km s–1 150 6

Physics Term 1 STPM Chapter 6 Gravitation 4. The mass of Jupiter is 320 times the mass of the Earth and its radius is 11 times the radius of the Earth. If the escape velocity from the surface of the Earth is 11 km s–1, what is the escape velocity from the surface of Jupiter? A 22 km s–1 C 320 km s–1 B 59 km s–1 D 653 km s–1 5. (a) (i) Show that a body projected from the Earth with a speed of 2gR will never return. Radius of the Earth is R. (ii) State two assumptions for this result to be valid. (iii) Calculate the escape speed from the Earth. (b) If mass of Earth mass of Moon = 81 and radius of Earth radius of Moon = 3.7, calculate the escape speed from the Moon. (c) If the mass of the Sun is 2.0 × 1030 kg, and its radius is 7.0 × 108 m, what is the escape speed from the Sun? 6. (a) A body of mass m is at a height h above the surface of the Earth. Write equations in terms of G, m, h and M mass of Earth and R radius of the Earth for (i) the gravitational potential at a height h. (ii) the gravitational potential energy of the body. (iii) the escape speed from a height h. (b) Show that the work done by the gravitational pull of the Earth when the body falls freely from a height h (h << R) to the surface of the Earth is GMmh R2 . (c) Deduce an expression for the velocity of projection of a body for it to reach a height equal to the radius R of the Earth. 7. (a) (i) Define gravitational field strength. (ii) Sketch a graph to show how the gravitational field strength varies with the distance r from the centre of the Earth. For value of r < R, radius of the Earth, state one basic assumption for the graph. (b) A stationary body is released from a height h from the surface of a planet of mass M and radius R. (i) Derive an expression for the velocity of the body when it hits the surface of the planet. (ii) If h << R, derive an expression for the gravitational potential energy U of the body at a height h. Neglecting terms greater than h R . (iii) If the gravitational potential energy mg0h = U0, g0 = acceleration of free fall on the planet surface, show how U is related to U0. Important Formulae 1. Newton’s law of gravitation F = G m1 m2 r 2 2. Gravitational field strength g = F m = GM r 2 3. Gravitational potential V = – GM r 4. Gravitational potential energy U = mV = – GMm r 5. Relation between F and U, g and V F = – dU dx , U = – ∫∞ r Fdx g = – dV dx , V = – ∫∞ r gdx 6. Speed, v of a satellite in circular orbit of radius r v = GM r 7. Energy of a satellite Gravitational potential energy, U = – GMm r , kinetic energy, K = GMm 2r Total energy, E = U + K = – GMm 2r 8. Relation between period T and radius of orbit r of a satellite: T2 ∝ r3 9. Escape velocity, ve = 2GM r 151 6

Physics Term 1 STPM Chapter 6 Gravitation STPM PRACTICE 6 1. The period of a satellite in circular orbit round the Earth depends on A the mass of the satellite B the radius of the orbit C the direction of motion of the satellite D the speed at which the satellite is launched from the Earth’s surface 2. A satellite of mass m is in an orbit of radius r above the Earth which has a mass M. The satellite then drops to an orbit of radius r 2 . The total energy of the satellite A decreases by GMm 2r B increases by GMm 2r C decreases by 3GMm 2r 2 D increases by 3GMm 2r 2 3. A satellite of mass m is in circular orbit of radius r about the Earth of mass M, and stays at a height h above the Earth’s surface. Which of the following represents the gravitational potential energy of the satellite? A mgh C – GMm 2r B – GMm r D GMm 2r 4. A satellite is moved from one circular orbit to another circular orbit further from the Earth. Which of the following quantities of the satellite increases? A Gravitational attraction B Gravitational potential energy C Linear speed D Centripetal acceleration 5. The figure shows two stars P and Q of masses 5M and M respectively separated by a distance r. 5M M P r Q The gravitational field strength at a distance x from P along the line joining P and Q is zero. What is the distance x? A 0.17r C 0.45r B 0.20r D 0.69r 6. Three point masses each of mass m are spaced uniformly along a straight line as shown in the figure. m x x m m What is the gravitational potential energy of the system? A – Gm2 x C – 5Gm2 2x B – 3Gm2 2x D – 3Gm2 x 7. A satellite is in circular orbit close to the surface of the Earth. The mass of the Earth is M and its radius is R. The speed of the satellite is A v = GM R B v = GM 2R C v = GMR D v = 2GMR 8. X and Y are points on the surface of the Earth and the Moon as shown in the figure below. Earth X Y Moon Which graph represents the variation of the gravitational potential V from X to Y? A V X Y 0 x C V X Y 0 x B V X Y 0 x D V X Y 0 x 152 6

Physics Term 1 STPM Chapter 6 Gravitation 9. A spacecraft moves away from the Earth. Which graph shows the variation of its gravitational potential energy U with the distance r from the centre of the Earth? [R = radius of the Earth] A U 0 r R C U 0 r R B U 0 r R D U 0 r R 10. A body of mass m is raised from the surface of the Earth to a height h = R, the radius of the Earth. If the mass of the Earth is M and G is the gravitational constant, the increase in the potential energy of the body is A GMm 2R C GMm R B 3GMm 4R D 2GMm R 11. What will happen if the Earth stops revolving about its axis? A The weight of objects on its surface increases. B Acceleration due to gravity at the poles decreases. C Objects on the Earth’s surface escape from the Earth. D Gravitational potential at the equator increases. 12. Two points P and Q at distances R and 2R respectively from the centre of the Earth are shown in the diagram below. Earth P Q R 2R The gravitational potential at P is –8 kJ kg–1. If an object of mass 2 kg is moved from P to Q, what is the change in gravitational potential energy of the object? A –8 kJ C +4 kJ B –4 kJ D +8 kJ 13. The mass of the Earth is 81 times the mass of the Moon. The distance of the Moon from the Earth is 3.8 ×108 m. The gravitational field strength at a point P along the line joining the Earth and the Moon is zero. What is the distance of the point P from the Earth? A 3.08 × 108 m C 3.69 × 108 m B 3.42 × 108 m D 3.75 × 108 m 14. Two stars P and Q which are of mass 3M and M respectively revolve around their centre of mass. What is value of the ratio of the speed of P to that of Q? A 1 5 C 1 3 B 1 4 D 1 15. The radius of the Moon’s orbit around the Earth is 60R, where R = radius of the Earth = 6 400 km. Calculate (a) acceleration of the Moon towards the Earth, (b) speed of the Moon round the Earth. (c) The period of the Moon’s orbit around the Earth is 27.3 days and the radius of its orbit is 3.85 × 105 km. Calculate the mass of the Earth. 16. The Earth has a mass of 6.0 × 1024 kg and a radius of 6.4 × 106 m. Calculate (a) the gravitational potential on the Earth’s surface. (b) the gravitational field strength at a distance of 600 km above the surface of the Earth. 17. The speed of a satellite at a height h above the Earth’s surface is 6500 m s–1. (a) Find the value of h. (b) Determine the period of orbit. (c) Explain why the satellite does not falls to the Earth although it has an acceleration towards the Earth. (Radius of the Earth = 6400 km). 153 6

Physics Term 1 STPM Chapter 6 Gravitation 18. (a) The radius of a planet is R and the acceleration of free fall on its surface is g. Derive expressions for (i) the mass M of the planet, (ii) the escape speed from the planet’s surface. (b) The speed of a satellite in circular orbit round the planet is v. Deduce an expression for the period of the satellite orbit. 19. The diameter of Mars is 6 900 km, the diameter of the Earth is 1.3 × 104 km. The mass of Mars is 0.11 times the mass of the Earth. (a) What is the ratio of the density of Mars to that of the Earth? (b) What is the value of the acceleration of free fall on the surface of Mars? (c) What is the escape velocity from the surface of Mars? 20. The acceleration of free fall on the surface of a planet of mass M and radius R is g. (a) Derive an expression relating M, R and g. Identify any other symbols in your expression. (b) Find in terms of R, the height h from the surface of the planet where the acceleration of free fall is 1 2 g. (c) Show that the acceleration of free fall, g on the Earth’s surface can be expressed as g = 4πρ Gr 3 where ρ = density and r = radius of the Earth. 21. A satellite of mass 1 200 kg is placed in a geostationary orbit at a distance of 4.23 × 107 m from the centre of the Earth. (a) Explain what is meant by the term geostationary orbit. (b) Calculate (i) the speed of the satellite. (ii) the angular velocity of the satellite. (iii) the acceleration of the satellite. (iv) the Earth’s gravitational pull on the satellite. (v) the mass of the Earth. (c) Explain why a geostationary satellite (i) must be placed vertically above the equator. (ii) must move from West to East. (iii) is often used for telecommunications. 22. (a) The diagram below shows how the gravitational potential V between the surface of the Moon and the surface of the Earth varies along the line between the centres of the Moon and Earth. At the point P, the gravitational potential is maximum. P Earth x y V / X 106 J kg–1 Moon 0 –1.3 –3.9 –62.3 (i) By considering the separate contributions of the Earth and the Moon to the gravitational potential, explain qualitatively why the graph has a maximum and why the curve is not symmetrical. (ii) State how the resultant gravitational force on the meteor at any point between the Moon and the Earth could be deduced from the above figure. (iii) Explain why a spacecraft needs more energy to travel from the Earth to the Moon than the energy needed for the return journey. (b) Certain meteorites found on the Earth are believed to be debris from volcanic eruptions on the Moon. When a meteor is at point P, the gravitational forces on it due to the Moon and Earth are FM and FE respectively. State the relation between FM and FE. Then, find the value of x y , where x and y are the distances of point P from the centre of the Moon and the centre of the Earth respectively. (c) For a meteor to reach the Earth, it must be projected from a volcano on the Moon with a certain minimum speed u. Use appropriate values from the above figure to find that speed. (Mass of Moon = 7.4 × 1022 kg, mass of Earth = 6.0 × 1024 kg) 154 6

Physics Term 1 STPM Chapter 6 Gravitation 23. (a) Assuming that planets move in circular orbits about the Sun, show that the squares of their periods of revolution are proportional to the cubes of the radii of their orbits. What factors determine the value of the constant of proportionality? (b) The radius of the Earth’s orbit is 1.5 × 1011 m and the Earth’s period of revolution about the Sun is 3.2 × 107 s. The distance of the Moon from the Earth is 3.8 × 108 m and it makes one revolution about the Earth in 2.4 × 106 s. Find the ratio of the mass of the Sun to mass of the Earth. (c) A planet revolves round the Sun in an elliptic orbit. Sketch a diagram to show the orbit of the planet round the Sun. Mark on your diagram the position of the planet when its speed is maximum. Explain your answer. 24. (a) State Newton’s law of gravitation. (b) A communication satellite of mass 50 kg is put into an equatorial orbit with an angular velocity equal to that at which the Earth rotates about its axis, so that it is always above the same point on the Earth. (i) What is the angular velocity of the Earth’s rotation about its axis? Give your answer in rad s–1. (ii) Find the radius of the orbit. (iii) Calculate the change in (a) the potential energy, (b) the kinetic energy, of the satellite if it is moved from rest on the Earth’s surface to its orbit? [1 day = 8.6 × 104 s, radius of Earth = 6.4 × 106 m, mass of Earth = 6.0 × 1024 kg] ANSWERS 1 1. C 2. D Gmm‘ x2 = GMm‘ (d – x)2 3. 2.63 × 108 m 2 1. B 2. D 3. A 4. C 5. C 6. 4.74 × 1024 kg GMm R2 = mRω2 7. 5 × 1024 kg GM R2 = 10 m s–2 8. (a) a1 = GM2 R1 2 + Gm R2 2 (b) a2 = GM2 R1 2 – Gm R2 2 (c) a1 – a2 = 2Gm R2 2 3 1. D 2. C 3. B 4. C 5. C 6. D 7. C 8. D 9. 4.86 s T = 2π l g , gR2 = GM 10. 0.998 s g′ = (9.81 – Rω2 ) at the equator 11. (a) g = Gm r2 and r = (r + h) (b) (i) 1.66 N kg–1 gR2 = GM (ii) 3.46 × 108 m GME x2 = GMM (d – x)2 12. (a) (i) 2.71 × 10–3 m s–2 a = rω2 = r( 2p T )2 (ii) Gravitational force which causes the acceleration is the centripetal force. (b) On the Earth, g = 9.81 m s-2 = GM R2 , GM = gR2 Acceleration of Moon = GM r2 = gR2 r2 = (9.81)( 6.36 × 106 3.82 × 108 ) which equals answer to (a)(i). 4 1. B: Work done = m(VP – VQ) –6.0 × 109 = (500)[–5.0 × 107 – VQ] VQ = –3.8 × 107 J kg–1 2. C 3. B 4. A 5. A 6. B 7. C V = GM r ∝ 1 r , ΔK = m(VQ – VP) 8. v = ( 4 GM 23R ) 1 2 1 2 mv2 = m(GM R – GM 3R ) 2 m s-2 155 6

Physics Term 1 STPM Chapter 6 Gravitation 9. (a) g = 2.22 × 10–5 N kg–1 (b) –5.34 × 10–6 J kg–1 θ = tan–1 (0.750) = 37° A P B g θ 10. (a) 2 583 MJ (b) 9.0 N kg–1 g = – dV dx 11. (a) (i) Refer to page 133. (ii) Reason: At infinity, V = 0, or work is done by the Earth’s field. (b) 1.98 × 109 J 5 1. D 2. D 3. A 4. B 5. C 6. B 7. A 8. C 9. (a) F = mrω2 (b) GmM r 2 (c) mrω2 = mr( 2π T ) 2 = GmM r 2 T = 2π r3 GM 10. S1 S2 = 4 GS1m r 2 = mr( 2π T ) 2 , GS2m r 2 = mr( 2π 2T ) 2 11. (a) 4.63 × 106 m from the Earth –x = M(0) + m(d) M + m (b) Centre of mass not displaced since no external torque or force acts on the system. 12. 4.24 × 107 m mr ( 2π T ) 2 = GmM r 2 , GM = gR2 13. (a) 1.68 × 103 m s–1 (b) 1.41 × 106 J kg–1 14. (a) T = GMm 2R T = –E (b) ∆T = –∆E (increase) ∆V = 2∆E (decrease) 15. (a) 7.83 × 103 m s–1 v = GM r (b) 5.27 × 103 s T = 2πr v 6 1. B 2. C 3. B 4. B 5. (a) (i)(ii) Refer to page 148 (iii) 11 km s–1 (b) 2.4 × 103 m s–1 (c) 6.2 × 105 m s–1 6. (a) (i) Vh = – GM (R + h) (ii) U = – GMm (R + h) (iii) v = 2GM R + h (b) Use work done = UR – UR + h (c) 1 2 mv2 = U2R – UR v = GM R 7. (b) (i) 1 2 mv 2 = GMm R – GMm (R + h) v= 2GM ( 1 R – 1 R + h) (ii) Use U = – GMm (R + h) (iii) U = U0 – mg0 R STPM Practice 6 1. B : T = 4π2 r3 GM 2. A: ∆E = (– GMm 2(r 2 ) ) – (– GMm 2r ) = – GMm 2r 3. B : U = – GMm r 4. B : U = – GMm r . When r increases, U is less negative. 5. D : G(5M) x2 = GM (r – x) 2 x (r – x) = 5 x = 5 r – 5 x = 5 1 + 5 r = 0.69 r 6. C : U = – Gm2 x – Gm2 2x – Gm2 x = – 5Gm2 2x 7. A : mv2 R = GmM R2 v = GM R 8. B : Gravitational field strength = 0 at a point nearer to the Moon. Gravitational field strength = –gradient 9. D : U = – GMm r 10. A : U’ – U = – GMm 2R – (– GMm r ) = GMm 2R 11. A : Due to the rotation of the Earth, value of g deceases from the pole to the equator. At the equator, g’ = g – Rω2 , where g is gravitational field at the poles. 156 6

Physics Term 1 STPM Chapter 6 Gravitation If the Earth stops revolving (ω= 0), g is the same for all points on the Earth. Weight = mg > mg’ 12. D : VP = –8 kJ kg–1 = – GM R , VQ = – GM 2R = –4 kJ kg-1 ∆U = m(VQ – VP) = (2)[–4 – (–8)] kJ = 8 kJ 13. B: GME x2 = GMM (d – x)2 (ME = 81MM) x = 81 (d – x) x = 9.0(3.8 × 108 ) 1 + 9.0 m = 3.42 × 108 m 14. C: (3M)x = M(d – x) x = d 4 P and Q have the same ω ω = vP d/4 = vQ 3d/4 vp vQ = 1 3 15. (a) 2.73 × 10–3 m s–2 acceleration = ( R2 r 2 ) g (b) 1.02 km s–1 v2 r = acceleration (c) 6.07 × 1024 kg 16. (a) V = – GM R = – (6.67 × 10–11)(6.0 × 1024) 6.4 × 106 J kg–1 = –6.3 × 107 J kg–1 (b) g = (–) GM (R + h) 2 = (–) (6.67 × 10–11)(6.0 × 1024) (6.4 × 106 + 6.0 × 105 )2 N kg-1 = (–)8.2 N kg–1 17. (a) Total energy = – K.E. (– GMm 2(R + h) ) = 1 2 mv2 GM = gR2 R + h = gR2 v2 = 9.51 × 106 m h = (9.51 × 106 – 6.40 × 106 )m = 3.11 × 106 m (b) Period, T = 2π v = 2π(9.51 × 106 ) 6.5 × 103 s = 9.19 × 103 s (c) The acceleration is the centripetal acceleration. 18. (a) (i) mg = GmM R2 M = gR2 G (ii) 1 2 mv2 = GmM R and GM = gR2 v = 2gR (b) mv2 r = GmM r2 and GM = gR2 r = GM v2 = gR2 v2 T = 2πr v = 2πgR2 v3 C d x 3M M 19. (a) 0.7357 ρ = M V (b) 3.83 m s–2 g = GM R2 (c) 5.14 × 103 m s–1 v = 2GM r 20. (a) gR2 = GM (b) h = 0.414R (c) Use GM = gR2 M = 4 3 πR3 21. (a) Stationary relative to points on the Earth surface. (b) (i) 3.08 × 103 m s–1 (ii) ω = 2π rad/day = 7.27 × 10–5 rad s–1 (iii) 0.224 m s–2 a = rω2 (iv) 268.8 N F = ma (v) 6.01 × 1024 kg a = GM R2 22. (a) (i) At P, gravitational field strength = 0. Mass of Earth > mass of Moon. (ii) F = – dU dr = –mdU dr F = –(mass of meteor) × (gradient) (iii) P is closer to the Moon. Gravitational pull on the Moon < gravitational pull on the Earth. (b) FM = FE GMMm x2 = GMEm y2 x y = MM ME = 0.111 (c) FM = –FE x y = MMoon MEarth = 0.111 (d) u = 2.28 × 103 m s–1 23. (a) Use mrω2 = GMm r2 , ω = 2π T (b) 3.46 × 105 (c) Planet P At P, distance r of planet from the Sun is minimum. Potential energy U = – GMm r minimum. Kinetic energy, K = Total energy – U Kinetic energy is maximum 24. (a) Refer to page 126 (b) M = gR2 G GMm R2 = mg Assumption: Earth is a uniform sphere (c) (i) 7.27 × 10–5 rad s–1 (ii) 4.2 × 107 m mrω2 = GM R2 , GM = gR2 (iii) (a) ∆U = 2.7 × 109 J (b) ∆K = 2.4 × 109 J 157 6

7 Physics Term 1 STPM Chapter 7 Statics CHAPTER Concept Map STATICS 7 Bilingual Keywords Centre of gravity: Pusat graviti Particle: Zarah Equilibrium: Keseimbangan Torque: Tork Statics Equilibrium of a Particle Resultant force = 0 Triangle of forces Equilibrium of Rigid Bodies Torque τ = r × F Resultant force = 0 Resultant torque = 0 Centre of Gravity INTRODUCTION 1. When a particle is in equilibrium, it is either stationary or moving with constant velocity. 2. If the particle is stationary, then it is in static equilibrium. 158

159 7 Physics Term 1 STPM Chapter 7 Statics 7.1 Centre of Gravity Students should be able to: • define centre of gravity • state the condition in which the centre of mass is the centre of gravity Learning Outcomes 1. The centre of gravity, G of a rigid body is the point where the weight of the body acts. 2. When a rigid body is freely suspended in equilibrium, its centre of gravity is directly below the point of suspension. 3. This fact is used in experiment to locate the centre of gravity of a rigid body. 4. Figure 7.1(a) shows a lamina hanging freely from the point A. Its centre of gravity G is directly below A. With the aid of a plumb line, a vertical line through the point A is drawn on the lamina. 5. The lamina is then hung freely from another point B, and the vertical through B is drawn (Figure 7.1(b)). The position of the centre of gravity G is the point of intersection of the two lines. 6. Theoretically, the position of the centre of gravity G of a rigid body is determined by taking moments of the weight of particles in the body about certain axis. 7. Figure 7.2 shows a rigid lamina, which is made up of n particles of mass m1, m2, m3,..., mn with position coordinates (x1, y1), (x2, y2), (x3, y3),..., (xn, yn) respectively. 8. The mass of the lamina, M = m1 + m2 + m3 + ... + mn. 9. The weight Mg of the lamina acts through its centre of gravity G with coordinates (x –, y –). 10. Taking moments about the axis Oy: (Mg) x – = (m1g) x1 + (m2g) x2 +... + (mn g) xn = n ∑i = 1 (mi gxi) x – = ∑ (mi g xi) Mg = ∑ (mixi) M Similarly, taking moments about the axis Ox, yields y – = ∑ (mi yi) M Mg G A Lamina Pin Plumb line Mg G A B Plumb line Pin Figure 7.1 (a) (b) O x y m1 (x1 , y1) m1g m2g Mg mng m2 (x2 , y2) G ( , ) (xn , yn m ) n –x –y Figure 7.2 VIDEO Centre of Gravity

160 7 Physics Term 1 STPM Chapter 7 Statics 11. The coordinate (x –, y –) obtained above is based on the assumption that the value of g, the acceleration due to gravity is uniform throughout the whole rigid body. This is true for small bodies. 12. For a very large body, the value of g may not be uniform throughout the body. The coordinates of the centre of gravity is then given by x – = n ∑i = 1 (mi gi xi) ( n ∑i = 1 mi gi ) , y – = n ∑i = 1 (mi gi yi) ( n ∑i = 1 mi gi ) 13. The centre of gravity is at the same point as the centre of mass for a small body, or the body is in a uniform gravitational field where g is a constant. Example 1 A uniform square lamina of sides 8.0 cm has a section of it removed as shown in the figure. Where is the centre of gravity of the remaining section? Solution: The remaining lamina consists of a rectangle of area 32 cm2 with its C.G. at A(4,2) and a square of area 16 cm2 with its C.G. at B(2,6) 0 2 2 4 6 8 4 6 8 x / cm A(4,2) B(2,6) y / cm The weight of each section is proportional to its area. Coordinates of the centre of gravity of the remaining lamina are – x = (32)(4) + (16)(2) 32 + 16 cm and – y = (32)(2) + (16)(6) 32 + 16 cm = 3.3 cm = 3.3 cm

161 7 Physics Term 1 STPM Chapter 7 Statics Example 2 A uniform circular disc has a radius of 6.0 cm has a hole of radius 3.0 cm as shown in the figure. The disc is balanced horizontally at a point P. Find the point P. x / cm y / m –6 0 –6 6 6 Solution: x / cm y / cm –6 0 –6 6 6 P – x Q –3 The disc can be balanced horizontal if it is pivoted about its centre of gravity. Hence point P must be the centre of gravity of the disc. The weight of a circular disc is proportional to its area, A = πr2 ∝ r2 The original circular disc can be considered to consist of two sections. Section 1: Disc with size of the hole and its C.G. at Q(–3,0). Weight W1 ∝ 33 ∝ 9 Section 2: The remaining section which is shaded and its C.G. at P(– x,0) Weight W2 ∝ (62 – 32 ) ∝ 27 Weight of the whole disc (without the hole) W and its C.G. is at (0,0). Then for the whole disc, x-coordinate of C.G. = 0 = W1(–3) + W2(–x) W = (9)(–3) + (27)(–x) 36 – x = 1 cm Hence point P is 1 cm from the centre of the circular disc (see Figure).

162 7 Physics Term 1 STPM Chapter 7 Statics Quick Check 1 1. Three weights are fixed along a uniform rod which has a weight 5.0 N and length 0.80 m as shown in the figure. Where is the centre of gravity? 0.20 m 0.60 m 2.0 N 3.0 N 5.0 N 2. A rectangular piece of plywood measuring 2.0 m × 0.80 m has a section of it removed as shown in the figure. Where is the centre of gravity of the remaining section? x / m y / m –1.0 0 0.5 1.0 –0.4 –0.2 0.4 0.2 7.2 Equilibrium of Particles Students should be able to: • state the condition for the equilibrium of a particle • solve problems involving forces in equilibrium at a point Learning Outcomes 1. The condition for a particle to be in equilibrium is that the resultant force = 0. 2. Figure 7.3(a) shows three forces P, Q and R acting on a particle O. The particle is in equilibrium if the resultant force = 0 That is P + Q + R = 0 Figure 7.3(b) shows the vector addition of the three forces. 3. Triangle of forces If three coplanar forces P, Q and R on a particle O are in equilibrium, the three forces can be represented both in magnitude and direction by the sides AB, BC and CA of a triangle taken in order. 4. Conversely, if three coplanar forces acting on a particle O can be represented both in magnitude and direction by the sides of a triangle taken in order, then the three forces are in equilibrium. Figure 7.3 Triangle of forces P R O Q P Q R B C A (a) (b) Exam Tips Note that the direction of P, Q and R in Figure 7.3(b) all go round the triangle in the clockwise direction. 2017/P1/Q7

163 7 Physics Term 1 STPM Chapter 7 Statics Example 3 Forces of 5 N, 4 N and 3 N acting on a point are in equilibrium. If sin 37° = 0.6, find the angle between the 5 N and 3 N force. Solution: Since the forces are in equilibrium, they can be represented in magnitude and direction by the sides of a triangle. Also, 52 = 42 + 32 Hence, the triangle is a right-angled triangle. Angle between the 5 N and 3 N force α = 127° Exam Tips The angle α is shown in figure (b). The two forces must be from the same point. Example 4 Forces of 5 N, 7 N and 8 N are in equilibrium. What is the angle between the forces of 5 N and 8 N? Solution: The forces can be represented in magnitude and direction by the sides of a triangle. By using a suitable scale, construct a triangle ABC (Figure (a)) by using a pair of compass. Use a protractor to measure the angle θ. θ = 60° Angle between the 5 N and 8 N force α = (180° – 60°) = 120° Example 5 The figure shows a load S of weight W hanging vertically from a string tied to another string PQR at Q. The system is in equilibrium. Find the tension (a) T along QP, (b) F along QR of the string. 4 N 37 3 N 5 N α 3 N 4 N 5 N α (a) (b) θ α 7 N 5 N 8 N B A C 5 N 8 N 7 N α (a) (b) T P F R Q 30° S

164 7 Physics Term 1 STPM Chapter 7 Statics Solution: 30° W T F B A C T W P F R Q 30° S (a) (b) The forces W, T and F on the point Q are in equilibrium. Begin to draw the triangle of forces (Figure (b)) by drawing (i) the horizontal line CA to represent T, (ii) the vertical line AB to represent W, (iii) complete the triangle, BC represent F, (iv) mark the angle ACB = 30°. From figure (b), (a) tan 30° = W T T = W tan 30° (b) sin 30° = W F F = W sin 30° Alternative method (Resolution of forces) Since W, T and F are in equilibrium, the resultant force is zero. Hence, the algebraic sum of the components of the forces in any direction is zero. The horizontal and vertical directions are suitable directions to resolve forces. Resolving forces along the horizontal direction QX: F cos 30° – T = 0 T = F cos 30° .............................. a Resolving forces along the vertical direction QY, F sin 30° – W = 0 W = F sin 30° .............................. b (a) a b W T = sin 30° cos 30° = tan 30° T = W tan 30° (b) From b, F = W sin 30° T W X Y P F R Q 30° S

165 7 Physics Term 1 STPM Chapter 7 Statics Example 6 A load of mass 5 kg hangs from two strings of length 3.0 m and 4.0 m attached at points A and B respectively is separated horizontally by a distance of 5.0 m. Find the tension in each string. Solution: θ θ B A C Y X T T 2 1 3 m 4 m 5 m θ W Q R P T2 T1 W = (5 × 9.81) N = 49 N (a) (b) sin θ = 3 5 cos θ = 4 5 tan θ = 3 4 Method 1: Using triangle of forces Figure (a) shows the three forces W, T1, T2 acting on the point C which is in equilibrium. Figure (b) is the triangle of forces. sin θ = T1 W T1 = W sin θ = 49 × ( 3 5 ) = 29.4 N cos θ = T2 W T2 = 49 × ( 4 5 ) = 39.2 N Method 2: Using resolution of forces Resolving forces along the horizontal direction CX [Figure (a)]. T2 sin θ – T1 cos θ = 0 T2 = ( cos θ sin θ ) T1 .............................. a Resolving forces along the vertical direction CY. T1 sin θ + T2 cos θ = W ............................. b T1 sin θ + T1 ( cos θ sin θ ) cos θ = W T1 (sin2 θ + cos2 θ ) = W sin θ T1 = W sin θ = 49 × ( 3 5 ) = 29.4 N T2 = T1 tan θ From a = 29.4 × 3 4 = 39.2 N

166 7 Physics Term 1 STPM Chapter 7 Statics Example 7 (a) A force F1 acts on a body of weight W on a smooth plane inclined at 30° to the horizontal. The body is in equilibrium. Find F1 and R1, the normal reaction in terms of W and θ. θ W R1 F1 Solution: Method 1: Using triangle of forces (a) The triangle of forces is as shown. sin θ = F1 W F1 = W sin θ cos θ = R1 W R1 = W cos θ (b) From the triangle of forces tan θ = F2 W F2 = W tan θ cos θ = W R2 ∴ R2 = W cos θ Method 2: Using resolution of forces (a) Resolving forces parallel to the inclined plane, F1 – W sin θ = 0 F1 = W sin θ Resolving forces perpendicular to the inclined plane, R1 – W cos θ = 0 R1 = W cos θ (b) Resolving forces parallel to the inclined plane, F2 cos θ – W sin θ = 0 F2 = W ( sin θ cos θ ) = W tan θ Resolving forces perpendicular to the inclined plane, R2 – F2 sin θ – W cos θ = 0 R2 = F2 sin θ + W cos θ = W ( sin θ cos θ ) sin θ + W cos θ = sin2 θ + cos2 θ cos θ W = W cos θ (b) A horizontal force F2 prevents the body from moving as shown in the figure below. Find F2 and R2 in terms of W and θ. θ W R2 F2 θ B A W C F1 R1 θ X Y Z W F2 R2 θ θ W R1 F1 θ θ θ W R2 F2

167 7 Physics Term 1 STPM Chapter 7 Statics 1. Three non-linear coplanar forces P, Q and R act on a particle at the origin of the x-y coordinate system. The resultant force on the particle is zero. Which of the following is incorrect? A P, Q and R forms a triangle. B P, Q and R are not mutually perpendicular. C The x-components and the y-components of P, Q and R are not all positive, or negative. D P and Q are of the same magnitude but in opposite directions. 2. Two forces P and Q act on a point X as shown in the figure below. P X Q Which of the following diagrams shows correctly the force F on X which would produce equilibrium? A Q F P C Q F P B Q F P D Q F P 3. A load is hung from a string as shown in the figure. Each part of the string would break if the tension exceeds 200 N. W 60° What is the maximum weight W that can be hung from the string? A 50 N C 115 N B 100 N D 173 N 4. A load hangs from two strings PQ and PR as shown in the figure. P Q R Load Which of the vector diagrams shows correctly forces at the point P? A C B D Quick Check 2

168 7 Physics Term 1 STPM Chapter 7 Statics 5. The diagrams show spring balances joined to demonstrate equilibrium of three coplanar forces. The readings give the magnitude of the forces. Which system is in equilibrium? A C 4 N 4 N 4 N 3 N 3 N 6 N B D 3 N 5 N 4 N 8 N 6 N 10 N 6. A ball of weight W hangs from a light string. When the wind blows, a constant horizontal force F acts on the ball, and the string makes an angle θ with the vertical. Which of the following is correct? A cos θ = F W C sin θ = W F B tan θ = F W D cos θ = W F 7. The forces 4 N, 4 N and 3 N acting on a point are in equilibrium. Find the angle between the two 4 N forces. 8. A load of 10 kg hangs from two strings of length 9 m and 12 m which are attached to two points at the same level and 15 cm apart. Calculate the tension in each string. 9. A wooden block of mass 2.0 kg is on a smooth plane inclined at an angle 30° to the horizontal. Calculate (a) the least force which is required to keep the block stationary on the inclined plane. (b) the normal reaction of the inclined plane. 10. A load of weight W hangs from a string as shown in the above figure. T X W F 60° Draw a triangle to show the equilibrium of the three forces W, F and T on the point X. Hence, find (a) F and (b) T in terms of W. Polygon of Forces 1. The principle used in the triangle of forces can be extended for a point under the action of a number of coplanar forces. 2. If a point O under the action of a number of forces are in equilibrium, then the forces can be represented by the sides of a polygon taken in order.

169 7 Physics Term 1 STPM Chapter 7 Statics O P T S R Q (a) C D E A P Q R S B T (b) Figure 7.4 3. Figure 7.4 shows a point O under the action of coplanar forces P, Q, R, S and T. The point O is in equilibrium. Then, P + Q + R + S + T = 0 ~ 4. The forces P, Q, R, S and T can be represented by the sides AB, BC, CD, DE and EA respectively of a closed polygon. 5. Figure 7.5(a) shows five coplanar forces A, B, C, D and E acting on a point. E C D B O A E D C B A R O (a) (b) Figure 7.5 6. Figure 7.5(a) is a vector diagram drawn to determine the resultant of A + B + C + D + E. Since A, B, C, D and E do not form a closed polygon, the forces are not in equilibrium. The resultant of the forces is represented in magnitude and direction by the vector R in Figure 7.5(b). That is, A + B + C + D + E = R 7. Alternatively, the resultant vector R can be determined by calculation using the method of resolving forces along two mutually perpendicular directions, say OX and OY. 8. Figure 7.6 shows three forces A, B and C acting on a point O, at angles of α, β and  to the OX direction. 9. To calculate the resultant of the three forces: Step 1 : Find the total components of the forces along OX. X = A cos α + B cos β + C cos  Step 2 : Find the total components of the forces along OY. Y = A sin α + B sin β + C sin  Step 3: Calculate the magnitude of the resultant force R using Pythagoras Theorem. R2 = X2 + Y2 R = X2 + Y2 Step 4 : Find the direction of the resultant force R from the OX direction. tan θ = Y X C B R A O X Y α θ β γ Figure 7.6

170 7 Physics Term 1 STPM Chapter 7 Statics Example 8 The diagram shows a point O in equilibrium under the action of five coplanar forces. Calculate the values of x and θ. O 2x N 4x N 3x N 2x N 13 N θ θ Solution: Since the point O is in equilibrium, total components of forces along OX direction = 0 3x sin θ + 2x sin θ – 4x = 0 5x sin θ = 4x sin θ = 4 5 ∴ θ = 53° 8’ Total components of forces along OY = 0 2x + 3x cos θ – 2x cos θ – 13 = 0 2x + x cos θ = 13 (sin = 4 5 2x + x ( 3 5 ) = 13 \ cos = 3 5 ) 13 5 x = 13 x = 5 Example 9 The figure shows a horizontal force F on a mass m on a smooth plane inclined at an angle θ to the horizontal. What is the magnitude of the resultant force on the mass? Solution: The forces on the mass are • mg, its weight acting vertically downwards. • R, the normal reaction of the inclined plane on the mass, perpendicular to the plane. • F Since the mass is on the plane, there is no motion in the direction perpendicular to the plane. Hence, the total components of forces perpendicular to the plane is zero. The mass can only move in the direction parallel to the inclined plane. The resultant force = the vector sum of the components of forces along the inclined plane = F cos θ – mg sin θ 4x N 2x N 2x N 3x N 13 N Y X O θ θ F θ F R mg θ θ

171 7 Physics Term 1 STPM Chapter 7 Statics 1. The figure shows a point O in equilibrium under the action of four forces. O θ 12 N 2 N 7 N p N Calculate the values of θ and p. Example 10 Three students A, B and C try to push a heavy object in the Ox direction. A applies a force of 200 N at an angle of 30° to Ox, and B pushes with a force of 400 N at an angle of 60° to Ox as shown in the figure. Find the smallest force and its direction that C must apply in order that the object O will move in the Ox direction. Solution: Method 1: Using vector diagram Begin by drawing the vector diagram according to scale. (i) Draw the horizontal line PS since resultant is along Ox. (ii) Draw FB = 400 N to scale. (iii) From the point Q, draw FA = 200 N to scale. (iv) The force which C applies Fc is the smallest when the line from R to the horizontal line is the shortest. (v) Hence, RS represents Fc in magnitude and direction Fc = 246 N. Method 2: Using resolving forces Suppose that the force C applied Fc makes an angle θ with Oy. Since the resultant of the 3 forces is along Ox, total components of forces along Oy = 0. 400 sin 60° – 200 sin 30° – Fc cos θ = 0 ∴ Fc cos θ = 246.4 N Fc = 246.4 cos θ Fc is minimum when cos θ = 1 θ = 0° Smallest value of Fc = 246.4 cos 0° = 246.4 N Direction of Fc is along negative Oy direction. Quick Check 3 O y x 60° 30° 200 N 400 N Q P S 60° FA = 200 N FB = 400 N FC Resultant force R x O y x A Fc B 60° 30° 200 N 400 N θ 2. The figure shows three forces of 10.0 N, 10.0 N acting and 8.7 N on a particle O. 10.0 N 10.0 N 8.7 N 90° 120° O

172 7 Physics Term 1 STPM Chapter 7 Statics Use a vector diagram to find the magnitude and direction of the fourth force that would keep the particle O in static equilibrium. 3. Three forces on 20 N, 40 N, and 50 N act on a point X. X 40 N 20 N 50 N Find the resultant force on X. 4. A force F acts on a body of mass m on a smooth plane inclined at an angle θ to the horizontal as shown in the figure. α m F θ What is the resultant force on the body? 7.3 Equilibrium of Rigid Bodies Students should be able to: • define torque as τ = r × F • state the conditions for the equilibrium of a rigid body • sketch and label the forces which act on a particle and a rigid body • use the triangle of forces to represent forces in equilibrium • solve problems involving forces in equilibrium Learning Outcomes Torque 1. Figure 7.7 shows a rigid body pivoted about a fixed axis. 2. When a force acts on the rigid body as shown in Figure 7.7, the body is rotated. 3. The turning effect of a force is known as the moment of the force. 4. A torque is produced by the force F on the rigid body. 5. The torque produced by a force is given by the vector product of the force F and r, the position vector of the point of application of the force from the axis of rotation, that is torque τ = r × F 6. Torque is a vector quantity, its direction is perpendicular to both r and F as shown in Figure 7.8. 7. The magnitude of the torque is τ = Fr sin θ where θ is the angle between the vectors r and F (Figure 7.8). (r sin θ) is the perpendicular distance from the axis to the line of action of the force F. Axis F Figure 7.7 Axis O r r sin θ F θ θ τ = r × F Figure 7.8 Exam Tips Torque = r × F not F × r 2008/P1/Q6, 2008/P1/Q9, 2010/P1/Q5, 2010/P2/Q1, 2013/P1/Q8, 2014/P1/Q9, Q19, 2015/P1/Q8, 2017/P1/Q18 INFO Torque

173 7 Physics Term 1 STPM Chapter 7 Statics 8. The magnitude of the torque is maximum when θ = 90°, that is when θ = 90°, torque, τ = Fr. 9. Based on this expression, the torque τ produced by a force may be defined as Torque, τ = Force × (Perpendicular distance of the force F from the axis of rotation) The unit of torque is N m. When θ = 0, torque = Fr sin 0 = 0 This happens when r and F are colinear (Figure 7.10). Example 11 A metre rule of weight 2.5 N is freely pivoted about a horizontal axis through a hole at the zero mark. What is the torque produced by the weight of the rule when the rule (a) is horizontal? (b) makes an angle of 30o with the horizontal? (c) is vertical? Solution: (a) The weight of the rule, F acts through the C.G. which is 0.50 m from the zero mark. When the rule is horizontal, F = 2.5 N is perpendicular to r = 0.50 m. Torque = Fr = (2.5)(0.50) N m = 1.25 N m (b) Perpendicular distance of force F from the axis through O is (0.50 cos 30o ) m. Torque = (2.5)(0.50 cos 30o ) N m = 1.08 N m (c) When the rule is vertical, the line of action of the force F passes through the zero mark. Torque = (2.5)(0) = 0 r F τ Axis Figure 7.9 Figure 7.10 r Axis F τ = 0 2.5 N 2.5 N 30°

174 7 Physics Term 1 STPM Chapter 7 Statics Quick Check 4 1. A uniform metre rule is pivoted about 50 cm mark. Forces of 7 N, 4 N and 9 N act on the rule is as shown in the figure. 0 20 50 100 cm 50° 7 N 4 N 9 N (a) What is the torque produced by each of the forces? (b) What is the resultant torque on the rule? 2. Forces act along the side of a rectangular lamina measuring 0.40 m × 0.60 m is as shown in the figure. O F3 = 2 N F1 = 3 N F2 = 6 N F4 = 5 N Find the resultant torque on the lamina. Conditions for the Equilibrium of a Rigid Body 2016/P1/Q16 1. A rigid body might not be in equilibrium when the resultant force on it is zero. 2. Figure 7.11 shows a rigid body, freely pivoted at O, under the action of two equal and opposite forces F where the lines of action are separated by a distance d. 3. The resultant force on the rigid body is zero, but the body is not in equilibrium. 4. The force F, acting at A produces a torque in the anticlockwise direction about O. The force F at B also produces a torque in the anticlockwise direction. Hence, total torque in the anticlockwise direction about O = Fx + F (d – x) = Fd 5. Since torque = Iα I = moment of inertia α = angular acceleration the rigid body is not in equilibrium. It rotates with an angular acceleration α. 6. From the above discussion, two conditions are necessary for a rigid body to be in equilibrium. (a) Resultant force = 0 (b) Algebraic sum of moments of forces about any axis = 0 7. If a rigid body under the action of a number of coplanar forces is not in equilibrium, (a) there is a resultant force, the rigid body has a linear acceleration, a or (b) there is a resultant torque, the rigid body has an angular acceleration, α or both (a) and (b). d B F A F O x d – x Figure 7.11

175 7 Physics Term 1 STPM Chapter 7 Statics 8. Figure 7.12 shows a rigid body subjected to three coplanar forces in equilibrium. 9. The centre of gravity G is directly below the point of suspension O. 10. The weight W of the body acts vertically downwards. 11. When produced upwards, the line of action of W passes through the point O. 12. The tensions in the string T1 and T2 also pass O. 13. Hence, the lines of action of the three forces acting on a rigid body pass through a common point. 14. The three forces may be represented by the sides of a triangle taken in order (Figure 7.12(b)). Example 12 A uniform rod PQ of length 10 m and mass 2 kg rests on two supports A and B, 2.0 m from the end P and 4.0 m from the end Q respectively. A load of 3 kg is placed at the end of P and a load of 2 kg at the end of Q. Calculate the reactions at the supports A and B. Solution: 3g 2g 2g P R A B S Q 2 m 3 m 1 m 4 m Suppose R and S are the reactions at the supports. Since the rod is in equilibrium, the resultant force is zero. Total upwards force = total downwards force. R + S = 3g + 2g + 2g g = acceleration due to gravity = 7g Taking moments about A, clockwise moments = anticlockwise moments 2g × 3 + 2g × 8 = 4 × S + 3g × 2 4 S = 16g S = 4g = 4 × 9.81 N = 39.2 N From R + S = 7g R = 7g – 4g = 3g = 29.4 N T2 T1 O θ G W String T2 T1 W θ Figure 7.12 (a) (b)

176 7 Physics Term 1 STPM Chapter 7 Statics Example 13 A uniform ladder of length 12 m and weight 200 N leans against a smooth wall such that it is at an angle of 60° to the floor. A boy of weight 400 N stands 1 3 of the way from the lower end. Find (a) the normal reaction at the wall, (b) the magnitude and direction of the resultant force acting on the lower end of the ladder. Solution: Suppose R = normal reaction at the wall F = resultant force at the end A Taking moments about the end A, clockwise moments = anticlockwise moments 200 (6 cos 60°) + 400 (4 cos 60°) = R (12 sin 60°) 600 + 800 = 12 R sin 60° R = 1 400 12 sin 60° = 135 N Resolve forces horizontally, F sin θ = R = 135 N Resolve forces vertically, F cos θ = (400 + 200) N = 600 N F2 sin2 θ + F2 cos2 θ = 1352 + 6002 F = 615 N F sin θ F cos θ = 135 600 tan θ = 0.225 θ = 12.7° Example 14 _ L 2 3 60 N Back bone Pivot Pelvis Muscle Pivot 12° T H V G _ L 1 2 L 60 N 250 N (a) (b) Figure (a) shows a man bending to lift a load of 60 N. His back bone may be considered as a rod pivoted at the end as shown in figure (b). The weight of the upper torso of the man is 250 N and acts through the point G. 400 N 200 N R F G C A B 60° θ AB = 12 m AG = 6 m AC = 1 3 × 12 m = 4 m

177 7 Physics Term 1 STPM Chapter 7 Statics Calculate (a) the tension T in the back muscle. (b) the compression H on the back bone. (c) the vertical reaction V on the pivot. Solution: (a) Taking moments about the pivot, Clockwise moments = Anticlockwise moments 60 L + 250 × L 2 = T × ( 2 3 L sin 12°) T = 185 × 3 2 sin 12° = 1 335 N (b) Resolving forces horizontally, T cos 12° = H H = 1 335 cos 12° = 1 306 N (c) Resolving forces vertically, T sin 12° + V = 250 + 60 V = 310 – 1 335 sin 12° = 32.5 N From the above example, you will notice that a very large force T acts along the back muscle and the back bone experiences a great compression when a man bends his back to lift a heavy load. This is bad for the back. We should bend our knees not our back when lifting a heavy load. Quick Check 5 Bend your knees not your back 1. A uniform rod of length 4 m is pivoted at A. Loads of 400 N and 100 N hang from the ends when the rod is in equilibrium. 1 m 3 m 400 N 100 N A What is the weight of the rod? A 500 N C 200 N B 300 N D 100 N 2. A uniform metre rule of weight 2.0 N is pivoted at the 70 cm mark. A weight of 5 N hangs from the 100 cm mark. 100 cm 70 cm Pivot 5 N When the metre rule is horizontal, what is the resultant moment about the pivot? A Zero B 0.15 N m C 1.1 N m D 1.5 N m 3. A uniform beam of length L is supported by two vertical forces F1 and F2 as shown in the figure. F1 F2 L _L 4 _L 8 What is the value of the ratio F1: F2? A 2 : 1 B 2 : 3 C 3 : 2 D 3 : 5

178 7 Physics Term 1 STPM Chapter 7 Statics 4. A uniform rod pivoted at its centre of gravity, O is subjected to the forces as shown in the figure. If the rod is in equilibrium, what is the force F? 30° 4 m 4 m 100 N OPivot F A 200 N B 100 N C 87 N D 50 N 5. The figure below shows a heavy piece of wood freely hinged at P. The end Q of the wood is supported by a wire tied at R on the wall. The weight of the wood is W and the tension in the wire is T. X P R W S T Q Wire Wood Wall The direction of the force on the wood at P is along A PQ B PS C SP D PX 6. A ladder PQ stands on a rough floor and the upper end leans against a rough wall. The weight of the ladder is W and the forces on the ladder at the ends P and Q are Y and X respectively. Which of the diagrams show correctly the forces on the ladder? A X P Y W Q B X P Y W Q C X P Y W Q D X P Y W Q 7. Forces of equal magnitude F act on identical square laminas. Which lamina is in static equilibrium? A F O F C F O F B F F O D F F O 8. A rod PQ of negligible weight is freely hinged at P. Forces Fx and Fy at the end Q keep the rod at an angle of 30° to the horizontal. 30º P Q Fy Fx If Fx = 6.0 N, what is the force Fy? A 3.0 N B 3.5 N C 5.2 N D 10.4 N

179 7 Physics Term 1 STPM Chapter 7 Statics Important Formulae 1. Condition of equilibrium of a particle Resultant force = 0 2. When three forces, F1 , F2 and F3 on a particle are in equilibrium, the forces can be represented by the sides of a triangle (figure below). F1 F2 F3 F1 F2 F3 F1 + F2 + F3 = 0 3. Torque τ = r × F Magnitude of torque = Fr sin θ 4. Conditions for equilibrium of a rigid body • Resultant force = 0 • Resultant torque = 0 STPM PRACTICE 7 1. An object is acted on by two forces X and Y. A force F is applied to hold the object in equilibrium. Which triangle of forces shows correctly the relation between the forces X, Y and F? A F X Y C F X Y B F X Y D F X Y 2. A half-metre rule of a smooth horizontal surface is free to rotate about a vertical axis through a hole at the 20.0 cm mark. Forces of equal magnitude 4.0 N act at the ends of the rule in opposite directions as shown in the figure below. 4.0 N 50 cm 0 20.0 cm 4.0 N 30° 30° What is the resultant torque on the half-metre rule? A 0 C 2.0 N m B 1.0 N m D 4.0 N m 3. A heavy beam of weight W rests on a smooth sphere with one end on a rough floor as shown in the figure. Rough floor Which figure correctly shows the forces on the beam?

180 7 Physics Term 1 STPM Chapter 7 Statics A W C W B W D W 4. A metre rule is free hinged at the 0 cm mark and is free to rotate about a horizontal axis about the hinge. The weight of the rule is 4.00 N. A load of weight 2.00 N is fixed at the 80.0 cm mark as shown in the figure. Hinge 0 cm 30° 4.00 N 50.0 cm 80.0 cm 2.00 N Metre rule What is the torque on the metre rule when it makes an angle of 30° to the vertical? A 1.80 N m C 2.40 N m B 2.00 N m D 3.60 N m 5. A heavy uniform beam of length l is supported at P and Q as shown in the diagram. _l _ F2 F1 2 3 l 3 P Q What is the ratio F1 F2 of the reactions at the supports? A 1 3 C 2 3 B 1 2 D 2 6. The figure shows a uniform diving-board of weight 300 N held horizontally by two supports at P and Q. A diver of weight 600 N stands at one end. Q Y 4 m 12 m P X What are the reactions X and Y at the supports P and Q respectively? X Y A 1 050 N 1 350 N B 1 050 N 2 250 N C 1 350 N 1 350 N D 1 350 N 2 250 N 7. A metre rule where centre of gravity is not at the 50 cm mark is hung horizontally from two spring balances at the 0 cm and 100 cm mark. The readings of the spring balances are 72 g and 48 g respectively. The centre of gravity of the metre rule is at A 20 cm mark B 30 cm mark C 40 cm mark D 70 cm mark 8. A picture weighing 10.0 N hangs freely from a string XYZ as shown below. What is the tension T in the string? A 2.9 N C 5.8 N B 5.0 N D 8.7 N 9. A uniform rod of weight W is freely hinged at P. It is held horizontally by a force F acting at an angle 60° to the rod as shown in the figure. F P l 3l 60° What is the value of F? A 0.200 W C 0.500 W B 0.385 W D 0.577 W T TT T X Z 60° 60° Y 10.0 N

181 7 Physics Term 1 STPM Chapter 7 Statics 10. (a) Three coplanar forces F1, F2 and F3 act on a point A as shown in Figure. State or show on a vector diagram, the condition for the three forces to be in equilibrium. (b) Four coplanar forces F4, F5, F6 and F7 which do not pass through a common point of a rigid body B are as shown in Figure (b). State or show on a vector diagram, the conditions for the rigid body to be in equilibrium. 11. (a) State the conditions for equilibrium for a rigid body. (b) A uniform rod PQ of mass m and length L is held horizontal with the end P against a wall and a light cord attached to the end Q. The cord makes an angle q with the rod as shown in the figure. The coefficient of static friction between the end P and the wall is 0.40. P Rod Cord Wall θ Q Determine the maximum value of the angle q for the rod to remain in equilibrium in the horizontal position. 12. A sign board of weight 240 N and length 1.20 m is hinged at the point B and supported by a wire AC as shown in the figure. The tension in the wire is T and the reaction at the hinge is R. (a) Show on the figure the forces on the sign board. (b) Calculate the tension T in the wire and the reaction R. 13. A person supports a load of 20 N in his hand as shown in Figure (a). The system of the hand and load is equivalent to the system shown in Figure (b). The rod represents the fore arm which exerts a force of 65 N at G, T is the tension in the biceps. 20 N Biceps Elbow (a) 20° 3.5 cm 6.5 cm 25 cm 65 N 20 N T G (b) Calculate (a) the tension T in the biceps. (b) the horizontal and vertical components of the force at the elbow. 14. (a) State the conditions for equilibrium for a rigid body. (b) A load of 0.50 kg is suspended from a string OP, and pulled by a horizontal force F using another string as shown in the diagram. The string makes an angle of 30° with the vertical. 30º O T P F (i) Find the tension T in the string, and the force F. F3 F2 F1 A (a) B F5 F4 F7 F6 (b) 1.20 m C A B 60° AWAS

182 7 Physics Term 1 STPM Chapter 7 Statics 1 1. Distance of C.G. from the weight of 2.0 N –x = (2.0)(0) + (3.0)(0.20) + (5.0)(0.80) + (5.0)(0.40)m (2.0 + 3.0 + 5.0 + 5.0) = 0.44 m 2. Complete rectangle with C.G. at (0,0) = section removed with C.G. at(0.5,0) + section remaining with C.G. at (–x,0) 0 = (0.4 × 0.5) × (0.5) + [2.0 × 0.8 – 0.4 × 0.5]–x (2.0 × 0.8) –x = (0.071) m 2 1. D 2. A 3. B 4. D 5. B 6. B 7. 136° 8. T1 = 0.8 W = 78.4 N T2 = 0.6 W = 58.8 N 9. (a) mg sin 30° = 9.81 N (b) R = mg cos θ = 16.97 N 10. (a) F = W sin 60° (b) T = W tan 60° 3 1. 67.38°, 13 N p sin q = 12 p cos q + 2 = 7 2. 10.0 N 8.7 N F4 = 5.0 N 10.0 N 60° 3. 50 N, 53.13° from the horizontal 4. F cos α – mg sin θ Resultant force acts parallel to the incline. 4 1. (a) Taking torque in the anticlockwise direction as positive, Torque = (7 N)(0.50 m) = 3.5 N m Torque = (–4 N)(0.30 m) = –1.2 N m Torque = (9 sin 50° N)(0.50 m) = 3.4 N m (b) Resultant torque = 5.7 N m 2. Taking anticlockwise torque as positive. Resultant torque = F1(0.30 m) – F2(0.20 m) + F3(0.30 m) – F4(0.20 m) = –0.70 N m Or, taking clockwise torque as positive, resultant torque = 0.70 N m 5 1. D 2. C 3. B 4. D 5. B 6. B 7. C 8. B (ii) The strings will break if the tension exceeds 6.00 N. The angle that the string OP makes with the vertical is increased by increasing the force F. What is the maximum angle that the string OP can make with the vertical? (c) One end of a uniform rod of weight 28 N is on a rough floor and the other end is against a smooth wall as shown in the figure. The coefficient of static friction between the floor and the rod is 0.25. θ (i) Draw a diagram to show the forces on the rod. (ii) Determine the maximum value of the angle θ before the rod slides. ANSWERS

183 7 Physics Term 1 STPM Chapter 7 Statics STPM Practice 7 1. D : For equilibrium, arrows of forces point in the anticlockwise (or clockwise) direction. 2. B : Resultant torque = 4.0 sin 30o (0.20) + 4.0 sin 30o (0.30) N m = 1.0 N m 3. B : Reaction on beam by smooth surface is normal to the surface. The three forces should pass through a common point. 4. A : Torque = (4.00)(0.50 sin 30o ) + (2.00)(0.80 sin 30o ) N m = 1.80 N m 5. A : Take moments about the CG, F1(l/2) = F2( l/6), F1/F2 = 1/3 6. D : Take moments about P, Y(4) = (300)(6) + (600)(12), Y = 2250 N X + 300 + 600 = Y = 2250 N, X = 1350 N 7. C : Take moments about the CG, (72 g)x = (48g)(100 – x), hence x = 40 cm 8. C : 2T sin 60o = 10.0 N, T = 5.8 N 9. B : Take moments about P, F (3l sin 60o ) = W(l), F = 0.385 W 10. (a) F1 + F2 + F3 = 0 ~ F3 F2 F1 (b) • F4 + F5 + F6 + F7 = 0 • Algebraic sum of moments about any points = 0 11. (a) Resultant force = 0 Resultant torque = 0 (b) N T W θ F = µN T(L sin q) = W( L 2 ) T = W 2 sin q N = T cos q, F = µN = µT cos q W = F + T sin q = µT cos q + T sin q = ( W 2 sin q )(0.40 cos q + sin q) Hence, q = 22° 12. (a) R T 1.20 m 240 N C A B 60° 30° AWAS (b) Horizontal forces, T cos 30o = R cos 30o hence T = R Vertical forces, T sin 30° + R sin 30° = 240 N T = R = 240 N 13. (a) T = 410 N (b) Horizontal component = 140 N Vertical component = 300 N 14. (a) Refer to page 174 (b) (i) T cos 30° = (0.50)(9.81) T = 5.66 N F = T sin 30° = 2.83 N (ii) T > F, Tmax = 6.00 N (6.00) cos θ = (0.50)(9.81) θ = 35° (c) (i) N R F W θ N: reaction R: normal reaction F: static friction W: weight R = W, F = µW = N Take moments about the lower end, N(l cos θ) = W(l/2) sin θ Tan θ = (0.25)(28) θ = 82°

CHAPTER Concept Map DEFORMATION OF 8 SOLIDS Bilingual Keywords Brittle: Rapuh Dislocation: Perkehelan Ductile: Mulur Elastic limit: Had kenyal Limit of proportionality: Had kekadaran Plastic deformation: Canggaan plastik Strain: Terikan Stress: Tegasan Yield point: Takat alah Young’s modulus: Modulus Young Deformation of Solids Stress and Strain Stress = Longitudinal force, F Cross-sectional area, A Strain = Extension, e Original length, l Extension-force Graph Stress-strain Graph e 0 Glass F Copper Rubber Strain 0 Glass Stress Rubber Copper Young’s modulus = Stress Strain Strain energy 184

Physics Term 1 STPM Chapter 8 Deformation of Solids INTRODUCTION 1. When a ball is being kicked, its shape changes instantaneously, but the ball is able to regain its original shape, as it moves through the air. 2. Elasticity is the property of a body which enables the body to regain its original shape with the removal of the deforming force. 3. Not all materials are able to regain the original shape after the removal of the deforming force. Materials which are unable to regain the original shape are known as inelastic. Examples of inelastic materials are clay and plasticine. 4. Most materials are elastic up to a certain limit, known as the elastic limit. Beyond the elastic limit, a material suffers permanent deformation. 8.1 Stress and Strain Students should be able to: • define stress and strain for a stretched wire or elastic string Learning Outcome 1. Figure 8.1 shows a wire of natural length l, fixed at one end and stretched by a longitudinal force F. The extension produced is e, and A is the cross-sectional area of the wire. F A l e Figure 8.1 2. Stress = Longitudinal force, F Cross-sectional area, A Stress = F A Strain = Extension, e Original length, l Strain = e l 3. Below the elastic limit, the wire returns to its original length when the stress is removed. Beyond the elastic limit, the wire no longer returns to its original length when the stress is removed. Example 1 A steel wire of length 2.5 m and cross-sectional area of 1.5 × 10–6 m2 is fixed at one end. A load of 10 kg is hung from the free end so that the wire hangs vertically. The extension in the wire is 0.82 mm. Calculate (a) the stress applied (b) the strain Info Physics Sungai Johor Bridge The bridge is 1.708 meters (1.7 km) long, with a 500-metre (0.500 km) main span across the Johor River. The two main pylons of the bridge rise to a height of 143 meters. Cables are used to support the main span. 185 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Solution: (a) Force on the wire F = mg Stress = F A = 10 × 9.81 1.5 × 10–6 = 6.54 × 107 N m–2 Example 2 The figure shows a suspension bridge which uses a system of cables. All the cables are made of same materials and have the same diameter. The cables are attached to points evenly spaced along the bridge. It may be assumed that the weight of the bridge is evenly distributed along the points where the cables are attached. A B T1 T2 Bridge Cable ff fi Compare (a) the tension T1 and T2 in the cables attached to A and B, (b) the stresses in the two cables. Solution: (a) Suppose the load of the bridge acting at each point A, and B is mg. Resolving forces vertically, T1 cos α = mg T1 = mg cos α T2 cos β = mg T2 = mg cos β α < β cos α > cos β Hence, T1< T2 (b) (Stress)1 = T1 A (Stress)2 = T2 A Since T1 < T2 (Stress)1 < (Stress)2 (b) Strain = Extension Original length = 0.82 × 10–3 2.5 = 3.28 × 10–4 β mg T2 α mg T1 186 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Quick Check 1 1. A vertical copper wire of length 2.00 m and cross-sectional area 1.5 mm2 supports a load of 2.5 kg at its lower end. The extension of the wire is 2.48 mm. Calculate (a) the stress (b) the strain 2. A load hangs from a vertical steel wire with diameter d. The stress produced is 5.0 × 106 N m–2. The same load hangs from another steel wire of the same length but of diameter 2d. What is the strain in this wire? 3. When a load hangs from the lower end of a steel wire W, the stress in the wire is s, and the strain is n. Two steel wires that are identical to wire W are joined together (i) in series and (ii) in parallel. The same load is hung from the lower end of the combined wires. Load (i) (ii) Load (i) (ii) Find in terms of s and n, the stress and strain in each of the wires that are joined (i) in series, and (ii) in parallel. 8.2 Force-extension Graph and Stress-strain Graph Students should be able to: • sketch force-extension graph and stress-strain graph for a ductile material • identify and explain proportional limit, elastic limit, yield point and tensile strength • define the Young’s modulus • solve problems involving Young’s modulus • distinguish between elastic deformation and plastic deformation • distinguish the shapes of force-extension graphs for ductile, brittle and polymeric materials Learning Outcomes (a) (b) Load, m s Reference wire Test wire Fixed mass V, Vernier scale W (fixed) Load, m Reference wire Spirit level Test wire Spherometer Figure 8.2 Searle’s apparatus 2009/P1/Q17, 2010/P1/Q18, 2010/P1/Q15, 2014/P1/Q10, 2015/P1/Q9, Q19, 2016/P1/Q19 VIDEO Ductile and Brittle Material 187 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 1. Figure 8.2(a) is the simplified version of Searle’s apparatus and Figure 8.2(b) shows the actual Searle’s apparatus used to measure the extension in wire when stretched by a load hanging from its lower end. 2. Two identical long thin wires of the same material hang from the ceiling. • the purpose of using long thin wires is that the extension produced would be bigger and can be measured more accurately • two wires are used to eliminate errors due to (i) thermal expansion when temperature rises during the experiment, (ii) the supports yielding under heavy load. 3. A constant mass hangs from the reference wire to straighten it and to prevent kinks along the wire. 4. A metre rule is used to measure the initial length l of the test wire. Its diameter d is measured using a micrometer screw gauge. 5. The force stretching the wire F = mg, where m is the mass of the load and g is the acceleration due to gravity. The mass of the load is increased in steps of 0.5 kg and the corresponding extension e is measured. 6. A vernier scale (Figure 8.2(a)) and a spherometer with a spirit level (Figure 8.2(b)) are used to measure the extension. (a) The initial reading of the vernier scale is noted, or the spirit level is first levelled and the spherometer reading is noted. (b) An extra load is added to the test wire, and the new vernier reading is noted, or the spirit level relevelled, and its reading is noted. (c) The extension produced is equal to the difference in the scale readings. 7. The force-extension graph and the stress-strain graph for a steel wire are as shown in Figure 8.3. Force, F Y A L B C D 0 Extension, e Stress = F / A Y A L B C D 0 Strain = e / l (a) Force-extension graph (b) Stress-strain graph A: Limit of proportionality L: Elastic limit B: Yield point C: Maximum force (or stress) D: Wire breaks Figure 8.3 8. In both graphs: (a) From O to A: A straight line through the origin implies that • the extension e is directly proportional to the force F, Hooke’s Law is true, F = ke • stress is directly proportional to the strain The point A is the limit of proportionality. Below the limit of proportionality, Extension ∝ Force and Stress ∝ Strain (b) Just beyond the point A is the elastic limit L. Below the elastic limit, the wire experiences elastic deformation. It returns to its original length with the removal of the force, along the path LAO. The work done in stretching the wire is completely recoverable. 188 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Exam Tips The difference between elastic limit and limit of proportionality. Elastic limit Limit of proportionality Below elastic limit, wire returns completely to its original length. Within limit of proportionality, extension is directly proportional to the force. 9. (a) When the wire is stretched beyond the elastic limit, it does not recover completely to its original length. The path taken is along BY which is almost parallel to AO. (b) Permanent deformation or permanent extension produced is represented by OY. Thus, plastic deformation occurred. 10. During plastic deformation, the work done is partially and not completely recoverable. The energy dissipated from the wire during the process is represented by the area OBYO in Figure 8.3(a). 11. (a) The point B is known as the yield point. At the yield point, plastic deformation starts. Atomic planes slide over each other, due to the movement of dislocations in the atomic plane. (b) The dislocations are due to the present of impurity atoms. In steel, the impurity atoms are carbon atoms. (c) The wire thins uniformly. 12. When the force is increased further, the extension (or strain) produced is bigger. 13. At C, the force or stress is maximum. The wire thins non-uniformly and necks are formed. Any effort to save the wire from breaking by reducing the force or stress fails. The wire breaks at D. 14. The tensile strength of the wire is the maximum stress that the wire can withstand before breaking. Tensile strength = maximum force cross-sectional area 15. Materials which undergo plastic deformation before breaking are ductile. Substance like glass which only undergoes elastic deformation before breaking is said to be brittle. 16. (a) Figure 8.4 shows the stress-strain graph for steel, glass, copper and rubber. Glass does not undergo plastic deformation. (b) A small stress produces a large strain for rubber. When rubber stretched by a small force, a large extension is produced. This is because rubber is a polymer. When stretched, the rubber molecules straighten. When the force is released, the rubber regained its original length, unless it is stretched beyond the elastic limit. Stress Steel Breaks Breaks Breaks Copper Rubber Strain Glass 0 Figure 8.4 189 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Example 3 (a) What is the essential difference between elastic deformation and plastic deformation of a solid? (b) Name a solid which shows both elastic and plastic behaviour, and state the conditions for each type of behaviour to occur. (c) Name a solid which shows plastic behaviour. Solution: (a) Elastic deformation: Object returns completely to its original dimensions when the force is removed. Plastic deformation: Object does not return completely to its original dimensions when the force is removed. Permanent deformation is produced. (b) Steel undergoes • elastic deformation below its elastic limit. • plastic deformation beyond its elastic limit. (c) A strip of plastic shows plastic behaviour. Example 4 The tensile strength of a material is the ratio of the tension which would break a specimen of the material to the cross-sectional area of the specimen. (a) A glass rod of radius 2.5 mm is clamped vertically. Increasing load is attached to the lower end of the rod. The rod breaks when the load reaches 300 kg. What is the tensile strength of the glass rod? (b) Estimate the load that will break a glass fibre of radius 1.0 μm. (c) In an experiment, it is found that the glass fibre in (b) does not break until the load is 0.01 N. Explain the difference in the value of the load compared to that in (b). Solution: (a) Tensile strength = Breaking tension Cross-sectional area = 300 × 9.81 π(2.5 × 10–3)2 = 1.50 × 108 N m–2 (b) Breaking tension = Tensile strength × Cross-sectional area = (1.50 × 108) × (π × (1.0 × 10–6)2 ) = 4.7 × 10–4 N (c) Experimentally, the glass fibre does not break until the force reaches 0.01 N which is much greater than the calculated value of 4.7 × 10–4 N because the glass fibre has different properties compared to the glass rod. 190 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Young’s Modulus 2013/P1/Q9 1. The Young’s modulus of the material of a wire is defined as follows. Young’s modulus, E = Stress, F/A Strain, e/l 2. The values of the Young’s modulus of some common materials are given below. Material Young’s modulus/N m–2 Aluminium 7.0 × 1010 Brass 9.0 × 1010 Copper 1.1 × 1011 Lead 1.5 × 1010 Platinum 1.7 × 1011 Steel 2.0 × 1011 Tungsten 3.5 × 1011 Nylon 2.0 × 109 3. Hooke’s law states that within the elastic limit, the extension e of a wire is directly proportional to the force applied. Hence, force, F = ke where k is the force constant of the wire. 4. From the definition of Young’s modulus, E = Stress, F/A Strain, e/l Force, F = ( AE l )e = ke (from Hooke’s law) Force constant of a wire, k = AE l which depends on • E, Young’s modulus of the material of the wire • l, length of the wire • A, cross sectional area of the wire 5. The Young’s modulus of the material of the wire is determined using Searle’s apparatus which was discussed earlier (Refer Figure 8.2). The extension e for different mass m is measured, and a graph of extension e against mass m is plotted (Figure 8.5). The gradient of the straight section of the graph equals to e m. 6. The value of Young’s modulus is calculated using the expression Young’s modulus, E = Stress, F/A Strain, e/l (F = mg, A = πd2 4 ) = mgl ( πd2 4 )e = ( 4gl πd2)( m e ) = ( 4gl πd2)( 1 gradient of graph) Mass, m Extension, e Gradient = –– e m 0 Figure 8.5 INFO Young’s Modulus 191 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Example 5 The following results are obtained in an experiment to determine the Young’s modulus of a metal. Length of wire = (3.255 ± 0.005)m Diameter of wire = (0.64 ± 0.01) mm Force applied on wire = (26.8 ± 0.1) N Extension of wire = (1.45 ± 0.05) mm (i) Calculate the value of the Young’s modulus of the metal and the uncertainty in the value. (ii) Suggest two modifications to the experiment which would improve the accuracy of the results. Solution: (i) Using Young’s modulus, E = F A e l A = πd2 4 = 4Fl πd2 e = 4 × (26.8) × (3.255) π (0.64 × 10–3)2 (1.45 × 10–3) = 1.87 × 1011 N m–2 ΔE E = ( ΔF F + 2Δd d + Δl l + Δe e ) ∆E = ( 0.1 26.8 + 2(0.01) 0.64 + 0.005 3.255 + 0.05 1.45 ) (1.87 × 1011) N m–2 = 0.13 × 1011 N m–2 Hence, E = (1.9 ± 0.1) × 1011 N m–2 (ii) Modifications • use a longer wire, or • use a thinner wire, or • increase the force on the wire so that the extension produced is larger. Example 6 A wire of length 0.50 m is fixed horizontally between two supports 0.50 m apart. When a mass of 7.5 kg is hung from the mid-point of the wire, the middle of the wire sags a distance of 1.0 cm. If the diameter of the wire is 2.8 mm, calculate the Young’s modulus of the material of the wire. Solution: BC2 = 252 + 12 BC = 25.02 cm Total extension, e = 2 (50.02 – 50.00) cm = 0.04 cm A 25.0 cm 25.0 cm 1.0 cm B θ T T T T mg C 192 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Resolving forces vertically, 2T cos θ = mg T = mg 2 cos θ E = F A e l = Fl eA E = mgl 2eA cos θ cos θ = 1.0 25.02 = 7.5 × 9.81 × 0.50 × 25.02 2(0.04 × 10–2) π(1.4 × 10–3)2 × 1.0 = 1.87 × 1011 N m–2 Example 7 Two wires P and Q have the same original length and cross-sectional area of 2.00 mm2 but are made of different materials. The wires are stretched by a force F. The figure below shows how the extension e produced varies with the force F. (a) Use the graphs shown to deduce the ratio of the Young’s modulus EP of the material of wire P to EQ the Young’s modulus of the material of wire Q. (b) The table below shows the Young’s modulus and the tensile strength of five different metals. The tensile strength of a material is the maximum stress on the material before it breaks. Two of the five metals are the material of wires P and Q. Tin Silver Copper Platinum Steel Young’s modulus/× 1010 Pa 5 8 13 17 21 Tensile strength/× 107 Pa 2 29 12 33 40 (i) Assuming that the percentage error in the value of EP EQ in (a) above is less than 5%, identify the possible pairs of metals for wires P and Q. (ii) Use the above graphs and data in the above table to identify the actual metals for wire P and wire Q. Wire Q Wire P 250 200 F / N e / mm 150 100 50 1.0 0 2.0 3.0 4.0 193 8