Machines and Mechanisms ( PDFDrive )

342 CHAPTER THIRTEEN

3.5"
3

B
4 135°

C2 A
D

FIGURE 13.16 Kinematic diagram for Example Problem 13.5.

2. Sketch Free-Body Diagrams of the Mechanism Links

Link 2 is the piston/rod assembly/crosshead slot. Link 4 is the follower. Notice that link 3 is not a tangible link. It
is used in a kinematic simulation to separate the revolute joint on the follower and the sliding joint in the
crosshead slot. Thus, the mechanism is modeled with all lower-order joints. The kinematic diagram has four
links, two pin joints, and two sliding joints, and consequently one degree of freedom. The driver for this mecha-
nism is the movement of the fluid into the cylinder.

The free-body diagrams for links 2 and 4 are shown in Figure 13.17. Link 3 is not required for force
analysis. Notice that a friction force is shown opposing relative motion. The directions may seem confusing
and warrant further explanation.

Ff 24 Ff 42
45°
F21 C F24 3.5"
F42 B 135°

45° 4A
F4x1
300 lb 2

(b) Fy41 T41

(a)

F21 D

FIGURE 13.17 Free-body diagrams for Example Problem 13.5.

Consider link 4 (Figure 13.17b). The pin moves upward relative to the crosshead slot. Therefore, friction
will act to prevent this motion of the pin by acting downward. Similarly, consider link 2 (Figure 13.17a). The slot
moves downward relative to the pin (recall the definition of relative motion). Therefore, friction will act to
prevent this motion of the slot by acting upward.

3. Solve the Equilibrium Equations for Link 2

Link 2 (Figure 13.17a) is examined first because it contains the applied force. For this analysis, only the
x-equilibrium equation is required.
:+ ©F x = 0:

F24 = 300 lb ;‫ی‬

4. Solve the Equilibrium Equations for Link 4
The free-body diagram of link 4 (Figure 13.17b) will reveal the torque on the output shaft. Of course, Newton’s
first law dictates that F42 = F24.
Ff42 = m F42 = (0.15) (300 lb) = 45 lb
The torque can be determined by using the moment equilibrium equation.
+ ©MA = 0:

- (F42 cos 45°) (3.5 in.) + (mF42 cos 45°) (3.5 in.) + T21 = 0

Static Force Analysis 343

- [(300 lb) cos 45°] (3.5 in.) + [(45 lb) cos 45° (3.5 in.)] + T21 = 0
Finally, the torque exerted on the output shaft is

T21 = + 631 lb-in. = 631 lb-in., cw

PROBLEMS 13–7. A force that is applied to a control lever is shown in
Figure P13.7. Determine the moment, relative to the
Resultant Force pivot block, when b = 0°.
13–1. Determine the resultant for the forces shown in
Figure P13.1 when b = 25°. β
200 N 50 mm
100 N

140 N

β 160 N 80 mm

FIGURE P13.7 Problems 7–9.

FIGURE P13.1 Problems 1–3. 13–8. A force that is applied to a control lever is shown in
Figure P13.7. Determine the moment, relative to the
13–2. Determine the resultant for the forces shown in pivot block, when b = 60°.
Figure P13.1 when b = 65°.
13–9. A force that is applied to a control lever is shown in
13–3. Determine the resultant for the forces shown in Figure P13.7. Determine the moment, relative to the
Figure P13.1 when b = 105°. pivot block, when b = 130°.

Moment of a Force Static Machine Forces
13–4. A force is applied to a box wrench as shown in
Figure P13.4. Determine the moment, relative to the 13–10. Figure P13.10 shows an overhead lift device. If a
center of the nut, when b = 90°. 600-lb force is suspended from the top boom while
the mechanism is stationary, determine the force
25 lb β required in the cylinder. The top boom weighs 80 lb
and the weight of the cylinder is negligible.

1.5'
6'

2.5'

600 lb
8"

FIGURE P13.4 Problems 4–6. FIGURE P13.10 Problem 10.

13–5. A force is applied to a box wrench as shown in 13–11. Figure P13.11 shows a mechanism that raises pack-
Figure P13.4. Determine the moment, relative to the ages in a transfer mechanism. If a 100-N package sits
center of the nut, when b = 75°. on the horizontal link while the mechanism is
stationary, determine the torque required from the
13–6. A force is applied to a box wrench as shown in motor. The weights of the links are negligible.
Figure P13.4. Determine the moment, relative to the
center of the nut, when b = 110°.

344 CHAPTER THIRTEEN

2" 5"

100 mm 900 mm 100 N
400 mm

748 mm 700 mm 1500 lb 5"
3"

2"

FIGURE P13.11 Problem 11.

13–12. Figure P13.12 shows a mechanism that is used to 4"
shear thin-gauge sheet metal. If a 200-N force is
applied as shown, determine the force on the sheet FIGURE P13.14 Problem 14.
metal. The weights of the links are negligible.
13–15. A utility lift vehicle is shown in Figure P13.15.
200 N Determine the force required by the hydraulic cylin-
der to maintain the position of the bucket.
175 mm

100 mm 40° 5m
30 mm 2.4 m

60 mm 2.0 m

1200 N 0.9 m

75 mm 1.2 m

FIGURE P13.12 Problem 12. FIGURE P13.15 Problem 15.

13–13. Figure P13.13 shows an adjustable platform used to 13–16. A front loader is shown in Figure P13.16. Determine
load and unload freight trucks. Currently, a 1200-lb the force required from both hydraulic cylinders to
crate is located as shown. Draw a free-body diagram maintain the shovel position.
for each link. The platform weighs 400 lb, and the
weight of all other links is considered insignificant. 10" 30"

80" 12" 10"
16" 20"
44"
30" 1200 lb 15"
18" 40" 50"
6"
18" 60" 12"

24" 24" 24" FIGURE P13.16 Problem 16.

FIGURE P13.13 Problem 13. 13–17. A 500-lb crate is supported on a lift table, as shown
in Figure P13.17. Determine the force required in
13–14. The clamp shown in Figure P13.14 has a rated load the hydraulic cylinder to keep the platform in the
of 1500 lb. Determine the compressive force this position shown.
creates in the threaded rod, AB.

Static Force Analysis 345

20"

30" 3"
30" 3"
24"

35" 16" 8"
4"
16"
35" FIGURE P13.19 Problem 19.
Determine the instantaneous torque required to
20" operate this mechanism. It operates at a low speed,
45" 45" so inertial forces are negligible.

FIGURE P13.17 Problem 17.

13–18. Figure P13.18 illustrates a refuse truck capable of CASE STUDY
moving a dumpster from a lowered position, as
shown, to a raised and rotated position. Gravity 13–1. Figure C13.1 shows a mechanism that gives motion
removes the contents into the truck box. The dump- to plunger J. Carefully examine the components of
ster weighs 2400 lb and is shared equally by the two the mechanism, then answer the following leading
front forks. Determine the force in the two hydraulic
cylinders. J B
K A
32"
I HD C
Tube E

72"

60"
18"

5" 8" 18" FIGURE C13.1 (Courtesy, Industrial Press.)
10" 64"
120" questions to gain insight into its operation.

FIGURE P13.18 Problem 18. 1. As lever A is rotated, what type of motion does item
C exhibit?
13–19. Figure P13.19 shows a materials handling mecha-
nism that slides 8-lb packages along a counter. 2. What type of connection do items A and C have?
The coefficient of kinetic friction between the 3. What type of motion does ball H have?
package and counter is 0.25. The coefficient of 4. What type of motion does plunger J have?
kinetic friction between the collar and rod is 0.10. 5. What is the purpose of spring K?
6. What is the purpose of item E?
7. What is the purpose of this mechanism?
8. Compare this mechanism to another mechanical

concept that serves the same purpose.

CHAPTER

FOURTEEN

DYNAMIC FORCE ANALYSIS

OBJECTIVES chapter deals with force analysis in machines with significant
accelerations. This condition is termed dynamic equilibrium.
Upon completion of this chapter, the student will be The analysis of dynamic equilibrium uses many concepts
able to: of static equilibrium. Therefore, an understanding of the
topics presented in the previous chapter is necessary prior to
1. Understand the difference between mass and studying this chapter.
weight.
14.2 MASS AND WEIGHT
2. Calculate the mass moment of inertia of an object
either by assuming a similarity to a basic shape or Mass and weight are not identical. Mass, m, is a measure of
from the radius of gyration. the amount of material in an object. Mass can also be
described as the resistance of an object to acceleration. It is
3. Transfer the mass moment of inertia to an alternative more difficult to “speed up” an object with a large mass.
reference axis.
The weight, W, of an object is a measure of the pull of
4. Calculate inertial forces and torques. gravity on it. Thus, weight is a force directed toward the
center of the earth. The acceleration of gravity, g, varies
5. Determine the forces, including inertia, acting depending on the location relative to a gravitational pull.
throughout a mechanism. Thus, the weight of an object will vary. Mass, however, is
a quantity that does not change with gravitational pull.
14.1 INTRODUCTION As stated, it is used to describe the amount of material in
a part.
During the design of a machine, determining the operating
forces is critical. Consider the development of an automo- The magnitude of weight and mass can be related
tive windshield wiper system. A key task is the selection of through Newton’s gravitational law.
an electric motor that will drive the wipers. The torque
required to operate the system is the main attribute in this W = mg (14.1)
selection process. Different scenarios must be considered,
such as the fact that the car might be parked under a maple In most analyses on earth, the acceleration of gravity is
tree. Increased wiper friction due to the tree sap will assumed to be
demand greater motor torque. A common scenario occurs
during periods of heavy rain. The wipers will be operated g = 32.2 ft/s2 = 386.4 in./s2 = 9.81 m/s2 = 9810 mm/s2
on a high-speed setting. As the wipers oscillate at increased
speeds, large accelerations will result. Inertial forces will be This assumption is applicable to all machines and mecha-
created from the high accelerations. These forces may be nisms discussed in this book. Of course, in the case of
large enough to damage the components of the wiper designing machines for use in outer space, a different gravi-
system. In fact, the inertial forces created by the motion of tational pull would exist.
many high-speed machines exceed the forces required to
perform the intended task. In a reciprocating engine, such Mass and weight are often confused in the U.S.
as an automobile engine, the inertial forces can be greater Customary System; it is most convenient to use a derived
than the force produced by the gas pressure. In a gas unit for mass, which is the slug. This unit directly results
turbine, the inertial force on the bearings due to an unbal- from the use of equation (14.1)
anced rotor can be magnitudes greater than the weight of
the rotor. slug = [lb/ft/s2] = lb s2/ft

Thus, for machines with significant accelerations, Occasionally, the pound-mass (lbm) is also used as a measure
dynamic force analysis is necessary. The previous chapter dealt of mass. It is the mass that weighs 1 pound on the surface of
with force analysis in mechanisms without accelerations. This

346

Dynamic Force Analysis 347

the earth. Assuming that the standard value of gravity force of gravity, or weight. In dynamic force analysis, any
applies, the pound-mass can be converted to slugs by inertia effects due to the acceleration of the part will also
act at this point.
1 slug = 32.2 lbm
For complex parts, the location of the center of grav-
Generally stated, any calculation in the U.S. Customary ity is not obvious. A common method of determining the
System should use the unit of slug for mass. In the center of gravity is to divide the complex part into simple
International System, the common unit used for mass is the shapes, where the center of gravity of each is apparent.
kilogram (kg = N s2/m). The composite center of gravity can be determined from
a weighted average of the coordinates of the individual
14.3 CENTER OF GRAVITY cgs. For example, the x-component of the center of gravity
of a composite shape can be found from the following
The center of gravity, cg, of an object is the balance point of equation:
that object. That is, it is the single point at which the
object’s weight could be held and be in balance in all direc- xcg total = m 1 xcg 1 + m 2 xcg 2 + m 3 xcg 3 + Á (14.2)
tions. For parts made of homogeneous material, the cg is
the three-dimensional, geometric center of the object. For m1 + m2 + m3 + Á
many simple parts, such as a cylinder, the geometric center
is apparent. Locating the center of gravity becomes impor- Because the acceleration due to gravity will be the same for
tant in force analysis because this is the location of the the entire body, weight can be substituted for mass in
equation (14.2). Of course, similar equations can be written
for the y- and z-coordinates of the center of gravity.

EXAMPLE PROBLEM 14.1

The part shown in Figure 14.1 is made from steel (0.283 lb/in.3). Determine the coordinates of the center of
gravity.

2"
y

3" cg 0.5"
2" x

2" z 4"
10"

FIGURE 14.1 Part for Example Problem 14.1.

SOLUTION: 1. Divide the Link into Basic Shapes
This part can be readily divided into two components. The lower plate will be designated as component 1, and the
upper shaft will be designated as component 2.

2. Calculate the Weight of the Basic Shapes
The weight of parts is determined by calculating the volume of the parts and multiplying by the density of steel.

W1 = 10.283 lb/in.3 [(10 in.) (4 in.) (0.5 in.)] = 5.66 lb

W2 = 10.283 lb/in.32 c p (2 in.)2 (3 in.)d = 5.33 lb
4

These weights and cg coordinates are organized in Table 14.1.

348 CHAPTER FOURTEEN

TABLE 14.1 Basic Shapes Data for Example Problem 14.1

Component Weight (lb) xcg (in.) ycg (in.) zcg (in.)

1 5.66 (10/2 - 2) = 3 (0.5/2) = 0.25 0
2 5.33 0 (0.5/2) = 0.25 0

3. Use Equation (14.2) to Calculate the Center of Gravity
The coordinates of the center of gravity are found.

xcg total = Wpart 1 xcg part 1 + W part 2 xcg part 2

W part 1 + W part 2

(5.66 lb) (3 in.) + (5.33 lb) (0 in.)
= (5.66 + 5.33) lb = 1.545 in.

ycg total = Wpart 1 ycg part 1 + Wpart 2 ycg part 2

Wpart 1 + Wpart 2

(5.66 lb) (0.25 in.) + (5.33 lb) (2 in.)
= (5.66 + 5.33) lb = 1.099 in.

The center of gravity of both parts lies on the z-axis. Therefore, the center of gravity of the composite (total) part
will also lie on the z-axis. Therefore,

zcg total = 0

14.4 MASS MOMENT multiplying its mass, dm, by the square of the distance, r, to a
OF INERTIA reference axis, z. This distance is the perpendicular distance
from the axis to the arbitrary element dm.
The mass moment of inertia, I, of a part is a measure of the
resistance of that part to rotational acceleration. It is more The mass moment of inertia of the entire object is the
difficult to “speed up” a spinning object with a large mass sum of all particles that comprise the object. Mathematically,
moment of inertia. Mass moment of inertia, or simply the moment of inertia is expressed as
moment of inertia, is dependent on the mass of the object
along with the shape and size of that object. In addition, Iz = r2 dm (14.3)
inertia is a property that is stated relative to a reference point L
(or axis when three dimensions are considered). This refer-
ence point is commonly the center of gravity of the part. Because the definition involves r, the value of the mass

Figure 14.2 illustrates a general solid object. Notice that moment of inertia is different for each axis. For example,
a small element of the object has been highlighted. The mass
moment of inertia of this small element is determined by consider a slender rod. The mass moment of inertia relative

z to its longitudinal axis will be small because r is small

because r is small for each element of the rod. For an axis

that is perpendicular to the rod, the moment of inertia will

be large because r is large for the outermost elements.

Mass moment of inertia is expressed in the units of

mass times squared length. In the U.S. Customary System,
the common units are slug-squared feet (slug ft2), which
convert to pound-feet-squared seconds (lb ft s2). In the

International System, the common units used are kilogram-
squared meters (kg m2).

r 14.4.1 Mass Moment of Inertia of Basic
dm Shapes

z Equation (14.2) has been used to derive equations for
primary shapes. Table 14.2 gives these equations, which can
FIGURE 14.2 A general solid object. be used to compute the mass moment of inertia for
common solid shapes of uniform density.

Dynamic Force Analysis 349

TABLE 14.2 Mass Moments of Inertia

Shape Name Rendering Mass Moment of Inertia

Cylinder y Ix = 1 [mr 2]
2
r
x Iy = 1 [m(3r 2 + l 2)]
12
z
l Iz = 1 [m(3r 2 + l 2)]
12
y
Slender rod Ix = 0

Iy = 1 [ml 2]
12

zl x Iz = 1 [ml 2]
12

Thin disk y Ix = 1 [mr 2]
r 2

Iy = 1 [mr 2]
4

Iz = 1 [mr 2]
4
x
Rectangular block z x Ix = 1 [m (w 2 + h 2)]
12
y
w Iy = 1 [m (w 2 + l 2)]
12
h
Iz = 1 [m (h2 + l 2)]
zl 12

EXAMPLE PROBLEM 14.2

The part in Figure 14.3 weighs 3 lb. Determine the mass moment of inertia of the part, relative to an x-axis at the
center of the part.

3" 18" x

FIGURE 14.3 Part for Example Problem 14.2.

SOLUTION: 1. Determine the Mass of the Part

The part weighs 3 lb and it is assumed to be used on the earth’s surface. The mass can be calculated from
equation (14.1).

m= W = 3 lb = 0.093 slug
g 32.2 ft/s2

350 CHAPTER FOURTEEN

2. Calculate the Mass Moment of Inertia (Solid Cylinder)
In a true sense, this part is a solid cylinder with

r = 1.5 in. = 0.125 ft

l = 18 in. = 1.5 ft

The z-axis in Table 14.2 is equivalent to the x-axis in this analysis. The mass moment of inertia relative to this
axis at the center of the part is

lx = 1 [m(3r 2 + l 2)] = 1 [0.093 slug (3(0.125 ft)2 + (1.5 ft)2)]
12 12

= 0.0178 slug ft2 = 0.0178 lb ft s2

3. Calculate the Mass Moment of Inertia (Slender Rod)

This part may be approximated as a slender rod. Using this assumption, the mass moment of inertia is calculated
from Table 14.2 as

Ix = 1 [m (l )2] = 1 [0.093 slug (1.5 ft)2]
12 12

= 0.0174 slug ft2 = 0.0174 lb ft s2

The slender rod assumption underestimates the actual mass moment of inertia by only 1.15 percent. Apparently
this part could be approximated as a slender rod.

14.4.2 Radius of Gyration 14.4.3 Parallel Axis Theorem

Occasionally, the moment of inertia of a part about a spec- Mass moment of inertia is stated relative to an axis.
ified axis is reported in handbooks using the radius of Occasionally, the mass moment of inertia is desired relative
gyration, k. Conceptually, the radius of gyration is the to an alternate, parallel axis. A parallel axis transfer equation
distance from the center of gravity to a point where the has been derived [Ref. 11] to accomplish this task. To trans-
entire mass could be concentrated and have the same fer the mass moment of inertia from the x-axis to a parallel
moment of inertia. x′-axis, the transfer equation is

The radius of gyration can be used to compute the mass Ix œ = Ix ; m d2 (14.5)
moment of inertia by

I = m k2 (14.4) The value d in equation (14.5) is the perpendicular
distance between the two axes. Notice that the second term in
The radius of gyration is expressed in units of length. In equation (14.5) can be either added or subtracted. The term
the U.S. Customary System, the common units are feet (ft) is added when the reference axis is moved away from the
or inches (in.). In the International System, the common center of gravity of the basic shape. Conversely, the term is
units used are meters (m) or millimeters (mm). subtracted when the transfer is toward the center of gravity.

EXAMPLE PROBLEM 14.3

For the part shown in Figure 14.4, determine the mass moment of inertia of the part relative to an x-axis at the end of
the part.

x′

3" 18"

FIGURE 14.4 Part for Example Problem 14.3.

Dynamic Force Analysis 351

SOLUTION: The mass moment of inertia through the center of the part was determined in Example Problem 14.2 as

Ix = 0.0178 slug ft2
The distance of the transfer from the center to the end of the part is

d = 9 in. = 0.75 ft
Equation (14.5) can be used to transfer the reference axis to the end of the part. Notice that the second term is added
because the transfer is away from the center of gravity.

Ixœ = Ix + md 2 = 0.0178 slug ft 2 + (0.093 slug) (0.75 ft) 2
= 0.0701 slug ft2 = 0.0701 lb ft s2

14.4.4 Composite Bodies shapes from Table 14.2. The mass moment of inertia of each
basic shape is computed relative to an axis through the
In practice, parts cannot always be simply approximated by center of the entire part. Finally, the total mass moment
the basic shapes shown in Table 14.2. However, for more of inertia is determined by combining the values from the
complex parts, the determination of the moment of inertia individual shapes.
can be done by dividing the complex parts into several basic

EXAMPLE PROBLEM 14.4
The part in Figure 14.5 is made from steel. Determine the mass moment of inertia of the part, relative to a y-axis at the
center of the part.

2"
y

0.5"
3"

2" 4"
2"

10"

FIGURE 14.5 Part for Example Problem 14.4.

SOLUTION: 1. Identify the Basic Shapes and Determine Their Mass

The part can be divided into two component shapes, as in Example Problem 14.1. Using the weights determined
in that problem, the mass of the two parts is

m1 = W1 = 5.66 lb = 0.176 slug
g 32.2 ft/s2

m2 = W2 = 5.33 lb = 0.165 slug
g 32.2 ft/s2

2. Determine the Centroidal Mass Moment of Inertia of the Basic Shapes

Component 1 is a rectangular block and component 2 is a cylinder. Using Table 14.2, the mass moment of iner-
tia of each part is determined relative to their individual centers of gravity.
Component 1:

Iy = 1 [m (w 2 + l 2)] = 1 [0.176 slug [(4 in.)2 + (10 in.)2]
12 12

= 0.701 slug in.2 = 0.0118 slug ft2 = 0.0118 lb ft s2

352 CHAPTER FOURTEEN

Component 2:

Iy = 1 [m ( r 2)] (longitudinal axis) = 1 [0.165 slug (1 in.)2]
12 12

= 0.0138 slug in.2 = 0.0001 slug ft2 = 0.0001 lb ft s2

3. Utilize the Parallel Axis Theorem

The center of gravity information, determined in Example Problem 14.1, will be used to determine the mo-
ment of inertia for each component relative to the composite center of gravity. The parallel axis theorem is
used to accomplish this. Notice that the perpendicular distance between y-axes is along the x-direction.
Component 1:

d1 = (3.0 - 1.099) in. = 1.901 in. = 0.158 ft

Iyœ(component 1) = Iy (component 1) + m1d12 = 0.0118 slug ft2 + (0.176 slug) (0.158 ft)2

= 0.0162 slug ft2 = 0.0162 lb ft s2
Component 2:

d2 = 11.099 - 02 in. = 1.099 in. = 0.092 ft
Iy ¿(component2) = Iy(component2) + m2d22 = 0.0001slug ft2 + 10.165 slug2 10.0923 ft22

= 0.0015 slug ft2 = 0.0015 lb ft s2

4. Calculate the Composite Mass Moment of Inertia
Iyœ = Iy œ(component 1) + Iy œ(component 2)
= 10.0162 + 0.00152 slug ft2 = 0.0177 slug ft2 = 0.0177 lb ft s2

14.4.5 Mass Moment of Inertia— 14.5 INERTIAL FORCE
Experimental Determination
Section 13.4 listed Newton’s three principal laws of mechan-
One popular experimental method of determining the ics. The second law is critical for all parts that experience
mass moment of inertia of a part is to swing the part as a acceleration. It is stated as
pendulum. This method is illustrated in Figure 14.6.
SECOND LAW: A body that has an unbalanced force has
rcg
a. An acceleration that is proportional to the force,

b. An acceleration that is in the direction of the
force, and

c. An acceleration that is inversely proportional to
the mass of the object.

FIGURE 14.6 Mass moment of inertia experiment. For linear motion, this law can be stated in terms of the
acceleration of the link’s center of gravity, Ag; thus,

If the part is displaced a small angle and released, it will © F = mAg (14.7)
oscillate. The moment of inertia can be determined by
measuring the time to complete one oscillation, Δt. The Equation (14.7) can be rewritten as
mass moment of inertia of the part relative to an axis
through the center of gravity, has been derived [Ref. 11] as © F - 7 mAg = 0 (14.8)

Icg = mrcg c a ¢t b 2 - rcg d (14.6) Notice that the subtraction symbol (- 7 ) is used because
2p both force and acceleration are vectors.
g

Dynamic Force Analysis 353

The second term in equation (14.8) is referred to as the proportional to the magnitude of acceleration. Similarly,
inertia of a body. This term is defined as an inertial force, Fgi as the brakes in an automobile are slammed, decelerating
the vehicle, your head lurches forward, again opposing the
Fgi = - 7 mAg (14.9) acceleration of the automobile. This is Newton’s second
law in practice.
The negative sign indicates that the inertial force opposes
acceleration (it acts in the opposite direction of the accelera- Equation (14.8) can be rewritten as
tion). Inertia is a passive property and does not enable a
body to do anything except oppose acceleration. © F + 7 Fgi = 0 (14.10)

This notion is commonly observed. Imagine pound- This concept of rewriting equation (14.7) in the form of
ing on the gas pedal in an automobile, violently accelerat- equation (14.8) is known as d’Alembert’s principle. Using
ing the vehicle. Envision the tendency for your head to d’Alembert’s principle in force analysis is referred to as the
lurch backward during the acceleration. This is the inertial inertia–force method of dynamic equilibrium. It allows for
force, acting in an opposite direction to the acceleration analysis of accelerating links, using the same methods that
of the automobile. Further, the extent of the lurch is are used in a static analysis.

EXAMPLE PROBLEM 14.5

The compressor mechanism shown in Figure 14.7 is driven clockwise by a DC electric motor at a
constant rate of 600 rpm. In the position shown, the cylinder pressure is 45 psi. The piston weighs 0.5 lb, and
the coefficient of friction between the piston and the compressor cylinder is 0.1. The weight of all other
links is negligible. At the instant shown, determine the torque required from the motor to operate the
compressor.

8" 1.5"
45 psi
50°
2"

FIGURE 14.7 Mechanism for Example Problem 14.5.

SOLUTION: 1. Draw a Kinematic Diagram and Identify the Degrees of Freedom

This is a common in-line, slider-crank mechanism, having a single degree of freedom. A scaled kinematic
diagram is shown in Figure 14.8a.

B 3 8" C4
50° 2" 2 A (a)

1

FIGURE 14.8 Diagrams for Example Problem 14.5.

2. Decide on a Method to Achieve the Required Motor Torque

Because the piston is the only component without negligible weight, the inertial force, and the
acceleration, of this component must be determined. The acceleration of the piston (link 4) is strictly
translational and is identical to the motion of point C. Such acceleration analyses have been extensively
presented in Chapter 7.

Once the acceleration of the piston has been obtained, the subsequent inertial forces can be
calculated. Finally, free-body diagrams and the corresponding equations can be used to determine the required
torque.

354 CHAPTER FOURTEEN

3. Determine the Velocity of Points B and C

This type of analysis was extensively discussed in the earlier chapters of the book. The 2-in. crank is rotating at
600 rpm. The velocity of point B is

v2 = p
1600 rev/min2 = 62.8 rad/s, cw
30

VB = v2 rAB = 162.8 rad/s2 12 in.2 = 125.6 in./s Q40°

The direction of VB is perpendicular to link 2 and consistent with the direction of ω2, up and to the left. Using
CAD, a vector can be drawn to scale, from the velocity diagram origin, to represent this velocity. The relative

velocity equation for points B and C can be written as

VC = VB + 7 VC/B

A completed velocity diagram is shown in Figure 14.8b. Scaling the vector magnitudes from the diagram,
VC = 80.5 in./s :

VC/B = 82.2 in./s 79b°

(b)

(c)

FIGURE 14.8 Continued

Dynamic Force Analysis 355

4. Calculate Acceleration Components

The next step is to construct an acceleration diagram, which includes points B and C. Calculating the magni-
tudes of the known accelerations,

ABn = (VB)2 = 1125.6 in./s22 = 7888 in./s2 R50° (directed toward the
rAB 2.0 in. center of rotation, point A)

aBt = 0 (no angular acceleration of the 2-in. crank)

A n = 1VC/B22 = 182.2 in./s22 = 844 in./s2 11a° (directed from C toward B,
C/B rBC 8.0 in. measured from CAD)

Note that point A does not have a normal acceleration because the motion is strictly translational.
5. Construct an Acceleration Diagram

The relative acceleration equation for points B and C can be written as

ACn + 7 ACt = A n + 7 ABt + 7 ACn/B + 7 ACt /B
B

The completed acceleration diagram is shown in Figure 14.8c.
6. Measure the Piston Acceleration

Scaling the vector magnitudes from the diagram,
ACt /B = 5985 in./s2 Q79°

A t = 5378 in./s2 :
C

Because the tangential acceleration of point B is in the same direction as the velocity, the piston is accelerating
(speeding up), not decelerating.

7. Calculate the Inertial Force

Because the piston is the only link of considerable weight, its inertial force is computed by combining equations
(14.9) and (14.1).

F i 4 = - 7 m4 A g 4 = W4 (- 7 Ag4)
g g

= (0.5 lb) 15378 in./s22 = 6.96 lb ;
386 in./s2

Because the piston does not encounter rotational acceleration, rotational inertia is not observed.

8. Sketch Free-Body Diagrams of the Mechanism Links

Notice that link 3 (BC) is a simple link, containing only two pin joints. In addition, no other force is acting on this
link. Thus, it is a two-force member, and the forces acting on the link must be equal and along the line that con-
nects the two pins. The free-body diagram for link 3 is shown as Figure 14.8d. As before, the notation used is that
F32 is a force that is applied to link 3 as a result of contact from link 2.

Link 2 is also a simple link; it contains only two pin joints. However, a moment (torque) is also applied to this
crank. Thus, this link is not a simple, two-force member. Newton’s third law stipulates that a force that is acting at B
will be equal and opposite to F32. Thus, the direction of F23 is known as a result of Figure 14.8d. The angle between
links 2 and 3 was measured from the CAD model. A general pin joint at point A dictates that two reaction forces will
be present. The free-body diagram for link 2 is shown as Figure 14.8e.

Link 4 has sliding contact with link 1. This contact force will act perpendicular to the contact surface. The
force from the compressed gas will, similarly, act perpendicular to the piston surface. A friction force will oppose
the motion (velocity) of link 4. Also, Newton’s third law stipulates that a force that is acting at C will be equal and
opposite to F34. Thus, the direction of F43 is known as a result of Figure 14.8d. The free-body diagram for link 4
is shown as Figure 14.8f.

9. Solve the Dynamic Equilibrium Equations for Link 4

Link 4 is examined first because it contains the applied force. The gas force is calculated as

Fgas = pgas Apiston = pgas c p(dpiston)2 d = 45 lb/in.2 c p(1.5 in.)2 d = 79.5 lb ;
4 4

356 CHAPTER FOURTEEN
F32

3 F34
(d)

B 11° W = 0.5 lb
2" F23
F x F43 11° C4
39° 21
F41 Fgas
A2 F i (f) Ff 41
g4
T21

F y
21

(e)

The friction force is

Ff = mF41 = 0.1 F41

The two unknown forces on this link (Figure 14.8f) are solved by using the following equilibrium
equations:
:+ ©F x + 7 F i = 0:

F43 cos 11.0° - Fgas - Fgi 4 - Ff = 0

+ c ©F y = 0:

Solving these equations yields - F43 cos 11.0° + F41 - 0.5 lb = 0

F43 = + 89.8 lb = 89.8 lb R11°
F41 = + 16.6 lb = 16.6 lb c

10. Solve for Equilibrium of Link 3

Because link 3 is a two-force member (Figure 14.8d.), the equilibrium equations dictate that the forces have the
same magnitude, act along the same line, and are opposite in sense. Of course, Newton’s third law dictates that
F32 = F23. Thus, the forces acting on link 3 are

F34 = 89.8 lb 11a°

F32 = 89.8 lb R11°

11. Solve for Equilibrium of Link 2

The free-body diagram of link 2 (Figure 14.8e) will reveal the required motor torque. Of course, Newton’s third
law dictates that F32 = F23. The unknown forces and moment on this link are solved using the following
equilibrium equations:
:+ ©F x = 0:

F2x1 - F23 cos 11° = 0

+ c ©F y = 0:

F y + F23 sin 11° = 0
21

‫ی‬
+ ©MA = 0:

- T21 + (F23 sin 39°) (2 in.) = 0

Dynamic Force Analysis 357

Solving the three equations yields
F2x1 = +88.1 lb = 88.1 lb :
F2y1 = -17.1 lb = 17.1 lb T
T21 = + 113.0 lb in. = 113.0 lb in., cw

Because the torque is the desired value, solving only the moment equation was necessary.

14.6 INERTIAL TORQUE termed the angular inertia of a body. This term is used to
define an inertial torque, Tgi :
The concept of an inertial force, as described in equation
(14.7), is an extension of Newton’s second law for linear T i = -7 Ig a (14.13)
motion. For rotational motion, the second law can be g
summarized in terms of rotational acceleration and moment
of inertia, relative to an axis through the center of gravity. Again, the negative sign indicates that the inertial torque is
directed opposite to the angular acceleration.
© Mg = Ig a (14.11) Equation (14.12) can be rewritten as

Again, the subscript “g” refers to the reference point at the © M + 7 Tgi = 0 (14.14)
link’s center of gravity.
Similarly to linear motion, equation (14.11) can be rewritten as Equation (14.14) is termed the moment equation of dynamic
equilibrium. It is the rotational equivalent of d’Alembert’s
© Mg - 7 Tgi = 0 (14.12) principle described in Section 14.5. It allows for analysis of
accelerating links, using the same methods as are used in a
Notice that the subtraction symbol (- 7 ) is used because static analysis.
the directions of the moment and angular acceleration must
be accounted for. The second term in equation (14.12) is The following example problem will combine several of
the dynamic force analysis concepts presented in this chapter.

EXAMPLE PROBLEM 14.6

The mechanism shown in Figure 14.9 is used to lower and retract the landing gear on small airplanes. The wheel
assembly link weighs 100 lb, with a center of gravity as shown. The radius of gyration of the assembly, relative to
the center of gravity, has been experimentally determined as 1.2 ft. The motor link is rotating counterclockwise at
3 rad/s and accelerating at 10 rad/s2. For mass property estimation, the motor crank will weigh approximately 15 lb
and will be 2 ft long, 1 ft wide, and 0.25 ft thick. The connecting link is estimated to weigh 20 lb and can be
modeled as a 3.5-ft slender rod. Determine all forces acting on the joints of all links and the torque required to
drive the motor link.

0.75' 2.0'
1.77'

30°

3.0'

2.33'
3.0'

FIGURE 14.9 Landing gear for Example Problem 14.6.

358 CHAPTER FOURTEEN

SOLUTION: 1. Draw a Kinematic Diagram and Identify the Degrees of Freedom

This mechanism is the common four-bar linkage, having a single degree of freedom. A kinematic diagram is
given in Figure 14.10a.

2.0' 1.77'
.75'
1 .89'
D A ω, α

30°
2 g2

B

3

2.33' g3 1.5'
3'
4
C 3'
g4

(a)

2. Decide on a Method to Achieve the Required Motor Torque

Because all links have significant weight, the acceleration of the center of gravity of all links must be determined.
Such acceleration analysis has been extensively presented in Chapter 7. Once the accelerations have been established,
the subsequent inertial forces and torques can be calculated. Finally, free-body diagrams and the corresponding
equations can be used to determine the required torque.

3. Determine the Velocity of Points B and C

This type of analysis was extensively discussed in the earlier chapters of the book. The 1.77-ft crank is rotating at
3 rad/s. The velocity of point B is

VB = v2 rAB = 13 rad/s2 (1.77 ft) = 5.31 ft/s Q60°

The direction of VB is perpendicular to link 2 and consistent with the direction of ω2, up and to the left. Using CAD,
a vector can be drawn to scale, from the velocity diagram origin, to represent this velocity.

The relative velocity equation for points B and C can be written as

VC = VB + 7 VC/B

The vector diagram is constructed in Figure 14.10b. Scaling the vector magnitudes from the diagram,
VC = 5.00 ft/s Q30.7°

VC/B = 2.63 ft/s R51.4°

4. Calculate Acceleration Components
The next step is to construct an acceleration diagram, which includes points B and C. Calculating the magni-
tudes of the known accelerations,

ABn = (VB)2 = 15.31 ft/s22 = 15.93 ft/s2 30a° (directed toward the
rAB 1.77 ft center of rotation, point A)

ABt = a2 rAB = 110 rad/s22 (1.77 ft) = 17.70 ft/s2 Q60° (perpendicular to link 2,
consistent with α2)
ACn/B = (VC/B)2 = (2.63 ft/s)2 = 2.30 ft/s2 Q38.6°
rBC 3.0 ft (directed from C toward B,
measured from CAD)

Dynamic Force Analysis 359

(b)

FIGURE 14.10 Diagrams for Example Problem 14.6.

ACn = (VC)2 = (500 ft/s2) = 10.72 ft/s2 59.3a° (directed from C toward D,
rCD 2.33 ft measured from CAD)

5. Construct an Acceleration Diagram
The relative acceleration equation for points B and C can be written as

ACn + 7 ACt = ABn + 7 ABt + 7 ACn/B + 7 ACt /B

The acceleration polygon is constructed and shown in Figure 14.10c. Notice that the concept of the acceleration
image, as presented in Section 7.10, was used to determine the acceleration of the center of gravity of the three
moving links.

6. Measure the Acceleration of the Center of Gravity of All Links

Scaling the vector magnitudes from the diagram, ACt = 11.60 ft/s2 Q30.6°
ACt /B = 12.28 ft/s2 R51.4°

(c)

360 CHAPTER FOURTEEN

Ag2 = 11.91 ft/s2 78.0a° Ag3 = 19.21 ft/s2 89.4a°
Ag4 = 20.32 ft/s2 Q73.4°

The angular accelerations of the links can then be determined.

a3 = aCt /B = 12.28 ft/s2 = 4.1 rad/s2, counterclockwise
rBC 3.0 ft

a4 = aCt = 11.60 ft/s2 = 5.0 rad/s2, counterclockwise
rCD 2.33 ft

7. Calculate Mass Properties

The motor crank can be considered a rectangular block. From Table 14.2, the mass moment of inertia, at the cen-
ter of mass, relative to an axis normal to the broad side of the link is

Ig2 = 1 [m(W2 + l 2 )] = 1 a 15 lb b [(2 ft)2 + (1 ft)2] = 0.194 lb ft s2
12 12 32.2 ft/s2

The connecting arm can be considered a slender rod. From Table 14.2, the mass moment of inertia at the center
of mass relative to an axis normal to the length of the link is

Ig3 = 1 [ml 2] = 1 a 20 lb b (3.5 ft)2 = 0.634 lb ft s2
12 12 32.2 ft/s2

The radius of gyration of the wheel assembly has been experimentally determined. From equation (14.4),
the mass moment of inertia at the center of mass relative to an axis normal to the length of the
assembly is

Ig4 = ml2 = a 100 lb b (1.2 ft)2 = 4.472 lb ft s2
32.2 ft/s2

8. Calculate the Inertial Force
For the three moving links, the inertial force is computed by combining equations (14.9) and (14.1).

Fgi 2 - 7 m 2Ag 2 = W2 (- 7 Ag 2)
g

= (15 lb) 111.91 ft/s22 = 5.55 lb R78.0°
32.2 ft/s2

Fgi 3 = - 7 m3Ag 3 = W3 (- 7 Ag 3)
g

= (20 lb) 119.21 ft/s22 = 11.93 lb R89.4°
32.2 ft/s2

Fgi 4 = - 7 m4Ag4 = W4 (- 7 Ag4 )
g

= (100 lb) 120.32 ft/s22 = 63.11 lb 73.4b°
32.2 ft/s2

9. Calculate the Inertial Torque
For the three moving links, the inertial torque is computed with equation (14.13).
Tgi2 = - 7 Ig 2a 2 = (0.194 lb ft s2) 110 rad/s22

= 1.94 ft lb, cw

Tgi3 = - 7 Ig 3a3 = (0.634 lb ft s2) 14.1 rad/s22

= 2.60 ft lb, cw

Dynamic Force Analysis 361

Tgi4 = - 7 Ig 4a 4 = (4.472 lb ft s2) 15 rad/s22

= 22.36 ft lb, cw

10. Sketch Free-Body Diagrams of the Mechanism Links
Because the weight of all links is to be included in the analysis, there are no two-force members. Thus,
all contact forces at the joints are general and are represented by their orthogonal components. The
free-body diagram of link 4 is shown in Figure 14.10d. The free-body diagram of link 3 is shown in Figure
14.10e. Of course, Newton’s third law, declaring that F34 and F43 have the same magnitude and opposing
directions, still applies. Finally, the free-body diagram of link 2 is shown in Figure 14.10f. Because each
link has more than three unknown forces, the equilibrium equations from all links will need to be solved
simultaneously.

11. Generate Equilibrium Equations for Link 4
The following dynamic equilibrium equations are generated from the free-body diagram of link 4
(Figure 14.10d).

:+ ©F x + 7 Fgi = 0:

F4x1 - F4x3 - Fgi 4 cos 73.4° = 0

F4x1 - F4x3 - 18.03 lb = 0

+c ©Fy + 7 F i = 0:
g

F y - F4y3 - W4 - Fgi4 sin 73.4° = 0
41

F y - F y - 160.48 lb = 0
41 43

‫ی‬+ ©MD + 7 Tgi = 0:

- F4x3[2.33 ft (sin 59.4°)] - F4y3 [2.33 ft (cos 59.4°)] - W4[3.0 ft(cos 59.4°)]

- Fgi4 [cos (73.4° - 30.6°)] [3.0 ft] - Tgi4 = 0

y y

F 41 F 21

x 1.77'

F 41 T21 x

59.4° F 21

30°

i 78° x

T g2 F 23

2.33' y .89' i y
3.0'
F 43 (f) F g2 F 23
W2
i
y y
F g4
F 43 F 32
73.4°
i 1.5'
30.6°
T g4 W3 38.6°
W4
x

F 32

(d) y i

F 34 T g3

x 89.4°

F 34 i

F g3
3.0'

(e)

FIGURE 14.10 Continued

362 CHAPTER FOURTEEN

Substituting values gives
- 2.000 F4x3 - 1.186 F4y3 - 313.98 ft lb = 0

12. Generate Equilibrium Equations for Link 3

The following dynamic equilibrium equations are generated from the free-body diagram of link 3
(Figure 14.10e).
:+ ©F x + 7 Fgi = 0:

F x - F3x2 - Fgi 3 cos 89.4° = 0
34

F3x4 - F3x2 - 0.13 lb = 0

‫یی‬+ c ©F y + 7 Fgi = 0:
F3y4 - F3y2 - W3 - Fgi3 sin 89.4° = 0
F3y4 - F3y2 - 31.93 lb = 0

+ ©MB + 7 Tgi = 0:
F3x4 [3.0 ft (cos 38.6°)] - F3y4 [3.0 ft (sin 38.6°)] (2.33 ft) + W3 [1.5 ft (cos 38.6°)]
+ Fgi 3 [cos (38.6° - 0.6°)] [1.5 ft] - Tgi 3 = 0

Substituting values gives

2.344 F3x4 - 1.872 F y + 34.95 ft lb = 0
34

13. Generate Equilibrium Equations for Link 2

The following dynamic equilibrium equations are generated from the free-body diagram of link 2
(Figure 14.10f).
:+ ©Fx + 7 Fgi = 0:

F2x3 + F2x1 + Fgi 2 cos 78° = 0

F2x3 + F y + 1.15 lb = 0
21

+ c ©Fy + 7 Fgi = 0:

F2y3 + F2y1 - W2 - Fgi 2 sin 78° = 0
F2y3 + F2y1 - 20.43 lb = 0

+ ©MA + 7 Tgi = 0:

T21 + F2x3 [1.77 ft (sin 30°)] + F2y3 [1.77 ft (cos 30°)] - W2 [0.89 ft (cos 30°)]

Fgi 2 [sin (78° - 30°)] [0.89 ft] - Tgi2 = 0

Substituting values gives

T21 + 0.885 F2x3 + 1.533 F y - 17.17 ft lb = 0
23

14. Solve the Equilibrium Equations

A total of nine equilibrium equations have been generated. As previously stated, Newton’s third law stipulates
that the following magnitudes are equal.

F4x3 = F3x4 F y = F y
43 34

F2x3 = F3x2 F y = F2y3
23

Dynamic Force Analysis 363

Therefore, nine unknown quantities remain. Solving the nine equilibrium equations, simultaneously, gives the
following results:

F4x1 = - 78.41 lb = 78.41 lb ;
F4y1 = + 58.38 lb = 58.38 lb c

F4x3 = - 96.44 lb = 96.44 lb : and F x = 96.44 lb ;
34

F4y3 = - 102.09 lb = 102.09 lb c and F3y4 = 102.09 lb T

F3x2 = - 96.32 lb = 96.32 lb : and F2x3 = 96.32 lb ;

F y = - 134.03 lb = 134.03 lbc and F y = 134.03 lbT
32 23

F2x1 = + 95.17 lb = 95.17 lb :
F2y1 = + 154.46 lb = 154.46 lb c

T21 = + 307.88 ft lb = 307.88 ft lb, cw

PROBLEMS the mass moment of inertia about its centroidal axial
axis.
Mass and Mass Moment of Inertia
14–8. The part shown in Figure P14.5 is a solid cylinder 2 ft
14–1. The mass of a connecting rod from an internal in diameter, 3 ft long, and weighing 48 lb. Determine
combustion engine has been determined to be the mass moment of inertia about a centroidal axis,
2.3 kg. Compute the weight of the rod. perpendicular to its length.

14–2. A robotic gripper was weighed at 4.5 lb. Determine 14–9. The part shown in Figure P14.9 is a slender rod, 14
the mass of the gripper. in. long, rotating about an axis perpendicular to its
length and 3 in. from its center of gravity. Knowing
14–3. A robotic gripper was weighed at 4.5 lb and has a that the rod weighs 2 lb and has a diameter of
radius of gyration relative to a certain axis at the center 1.25 in., determine its mass moment of inertia
of gravity of 5 in. Determine the mass moment of iner- about that axis.
tia of the part relative to this axis.
L
14–4. A 6-kg mechanism link has a radius of gyration d
relative to a certain axis at the center of gravity of
150 mm. Determine the mass moment of inertia of δ
the part relative to this axis.
FIGURE P14.9 Problems 9 and 10.
14–5. For the part shown in Figure P14.5, calculate the
mass moment of inertia and the radius of gyration
about a centroidal longitudinal axis of a 14-in.-long
shaft that weighs 5 lb and has a diameter of 0.625 in.

L 14–10. The part in Figure P14.9 is a slender rod, 0.4 m long,
rotating about an axis perpendicular to its length
Longitudinal axis and 0.12 m from its center of gravity. Knowing that
d the rod has a mass of 6 kg, determine its mass
moment of inertia about that axis.
Axis perpendicular to length
14–11. Determine the moment of inertia of the steel link
FIGURE P14.5 Problems 5–8. (r = 0.183 lb/in.3) shown in Figure P14.11 with
respect to the y-axis.

Ø 0.375"

0.75" y

14–6. For the part shown in Figure P14.5, calculate the mass Ø 0.375"
moment of inertia and the radius of gyration about a
centroidal longitudinal axis of a 1200-mm-long shaft 0.75"
that has a mass of 100 kg and a diameter of 50 mm.
1.75" 0.125"
14–7. The part shown in Figure P14.5 is a solid cylinder 2 ft 3.25"
in diameter, 3 ft long, and weighing 48 lb. Determine
FIGURE P14.11 Problem 11.

364 CHAPTER FOURTEEN

14–12. Determine the moment of inertia of the steel link 0.65 m
(r = 0.183 lb/in.3) shown in Figure P14.12 with
respect to the y-axis. 0.32 m
0.3 m
0.375"

0.75" y Ø 0.375" 0.4 m 0.1 m 0.4 m
0.2 m
0.125" 30Њ
0.6 m
0.75"

1.75" 0.75"
3.25"

FIGURE P14.12 Problem 12. 0.37 m

Inertial Forces FIGURE P14.16 Problem 16.18.

14–13. The compressor mechanism shown in Figure P14.13 Inertial Torques
is driven clockwise by a DC electric motor at a Figure P14.19 shows a link that weighs 4 lb and is rotating clock-
constant rate of 800 rpm. In the position shown, the wise at 20 rad/s. For Problems 14–19 and 14–20, determine the
cylinder pressure is 70 psi, and the piston weighs magnitude of the inertial force and the inertial torque at the center
0.75 lb. The coefficient of friction between the of gravity if:
piston and the compressor cylinder is 0.1. The
weight of all other links is negligible. At the instant 0.75 in. dia.
shown, determine the torque required from the
motor to operate the compressor.

7" 1.5"
70 psi
65°
1.75"

FIGURE P14.13 Problems 13–15. 16"

14–14. For the compressor mechanism described in 1"
Problem 14–13, determine the torque required from 60°
the motor if the motor is rotating at 800 rpm and
accelerating at a rate of 5000 rad/s2. FIGURE P14.19 Problems 19 and 20.

14–15. For the compressor mechanism described in 14–19. The link accelerates at 600 rad/s2.
Problem 14–13, determine the torque required from 14–20. The link decelerates at 600 rad/s2.
the motor if the motor is rotating at 800 rpm and
decelerating at a rate of 5000 rad/s2. Figure P14.21 shows a 10-kg link that rotates counter-
clockwise at 15 rad/s. Determine the magnitude of the iner-
14–16. The materials handling mechanism, shown in Figure tial force and the inertial torque at the center of gravity if:
P14.16, slides 4-kg packages along a counter. The 14–21. The link accelerates at 400 rad/s2.
machine operates with the crank rotating counter- 14–22. The link decelerates at 400 rad/s2.
clockwise at a constant rate of 120 rpm. The coeffi- 14–23. Figure P14.23 shows a slider-crank mechanism.
cient of kinetic friction between the package and
counter is 0.15. The weight of all the mechanism links Link 2 rotates clockwise at a constant 200 rad/s. The
is negligible. Determine the instantaneous torque weight of link 2 is negligible, link 3 is 3 lb, and link 4
required from the motor to operate this mechanism. is 2 lb. The radius of gyration of link 3 relative to the

14–17. For the materials handling mechanism described in
Problem 14–16, determine the torque required from
the motor if the motor is rotating at 120 rpm and
accelerating at a rate of 100 rad/s2.

14–18. For the materials handling mechanism described in
Problem 14–16, determine the torque required from
the motor if the motor is rotating at 120 rpm and
decelerating at a rate of 100 rad/s2.

Dynamic Force Analysis 365

75 mm 3. The inertial force and torque of link 4,
20°
4. The pin forces at B and C, and
ω
5. The torque to drive the mechanism in this position.
750 mm
14–26. Repeat Problem 14–25 with b = 90°.

14–27. Figure P14.27 shows a small hydraulic jack. At this
instant, a 10-lb force is applied to the handle. This
causes the 3.5-in. link to rotate clockwise at a
constant rate of 5 rad/s. The weight of links 2 and
3 is negligible, and link 4 is 1.5 lb. Determine the
following:

150 mm

15 mm 5" Lift
60°
FIGURE P14.21 Problems 21 and 22. 5"

10 lb
3.5"

B 2" 4" C4 FIGURE P14.27 Problem 27.
1.5" 2 3 20 lb
Aβ 1. The linear acceleration of the piston,

1 2. The inertial force of link 4,

FIGURE P14.23 Problems 23 and 24. 3. The pin forces, and

center of gravity is 3 in. For b = 45°, determine the 4. The force developed on the piston due to the
following: hydraulic fluid.

1. The linear acceleration of link 4 and the center of 14–28. Figure P14.28 shows a mechanism for a transfer
gravity of link 3, conveyor. The driving link rotates counterclockwise
at a constant rate of 25 rpm. The box weighs 50 lb as
2. The angular acceleration of link 3, shown. The weight of the driving link and the
coupler are negligible. The weight of the conveyor
3. The inertial force and torque of the coupler link, link is 28 lb and the center of gravity is at its
midspan. The radius of gyration of the conveyor
4. The pin forces at B and C, and link relative to the center of gravity is 26 in. For
b = 30°, graphically determine the following:
5. The torque to drive the mechanism in this position.
8" 10" 50 lb
14–24. Repeat Problem 14–23 with b = 120°. 18" 16"

14–25. Figure P14.25 shows a four-bar mechanism. Link 2 β
rotates counterclockwise at a constant 10 rad/s. The 4"
weight of links 2 and 3 is negligible, and link 4 is 17 kg.
The radius of gyration of link 4 relative to the center FIGURE P14.28 Problems 28 and 29.
of gravity is 45 mm. For b = 45°, determine the
following: 1. The linear acceleration of the center of gravity of the
conveyor link,
1. The linear acceleration of the center of gravity of
link 4, 2. The rotational acceleration of the conveyor link,
3. The inertial force and torque of the conveyor link,
2. The angular acceleration of link 4, 4. The pin forces, and
5. The torque required to drive the mechanism.
A β C 40 mm 14–29. Repeat Problem 14–28 with b = 100°.
25 mm
2 110 mm D
B 4 50 Nmm

25 mm
3

120 mm

FIGURE P14.25 Problems 25 and 26.

366 CHAPTER FOURTEEN mechanism, then answer the following leading
questions to gain insight into its operation.
CASE STUDY 1. As gear A rotates clockwise, describe the motion of
gear B.
C14–1 Figure C14.1 shows a mechanism that gives motion
to slides C and D and is used in a wire-stripping 2. As gear A rotates clockwise, what is the immediate
machine. Carefully examine the components of the motion of slide C?

CE D 3. Discuss the action that takes place as pin E reaches
the end of the slot.
B
4. Discuss precisely the continual motion of slides C
F and E.

G 5. Discuss how this motion could possibly be used in a
A wire-stripping machine.

FIGURE C14.1 (Courtesy, Industrial Press.) 6. What is the purpose of spring G?

7. How would the mechanism change if a “stiffer”
spring were installed?

ANSWERS TO SELECTED EVEN-NUMBERED
PROBLEMS

Chapter 1 3–58. C = 19.22 E = 17.52
3–60. B = 8.81 C = 117.7
1–26. n = 4, jp = 4, jh = 0, M = 1 3–62. D = 38.12 F = 238.9
1–28. n = 4, jp = 4, jh = 0, M = 1
1–30. n = 4, jp = 4, jh = 0, M = 1 Chapter 4
1–32. n = 6, jp = 7, jh = 0, M = 1
1–34. n = 4, jp = 4, jh = 0, M = 1 4–2. ¢x = 2.189 in. :
1–36. n = 4, jp = 4, jh = 0, M = 1
1–38. n = 6, jp = 7, jh = 0, M = 1 4–4. ¢RP = 8.420 in. R27.5°
1–40. n = 6, jp = 7, jh = 0, M = 1
1–42. n = 6, jp = 7, jh = 0, M = 1 4–6. ¢Rpiston = 47.10 mm ;
1–44. n = 9, jp = 11, jh = 0, M = 2
1–46. n = 4, jp = 4, jh = 0, M = 1 4–8. ¢ucrank = 23.0°, ccw
1–48. n = 8, jp = 10, jh = 0, M = 1 4–10. ¢Rend = 2.029 in. 55.1°a
1–50. n = 6, jp = 7, jh = 0, M = 1
1–52. Crank-rocker 4–12. ¢uhandle = 22.2°, ccw

1–54. Crank-rocker 4–14. ¢uram = 17.6°, ccw
4–16. ¢Rend = 22.644 in. 44.9°a

4–18. ¢uhandle = 34.4°, ccw

4–20. ¢uwheel = 16.3°, cw

4–22. ¢Rend = 203.4 73.9b° Q45.5°
4–24. ¢Rcarrier = 249.7 mm

Chapter 3 4–26. ¢Rbox = 0.579 m T

3–2. A = 17.3 in. 4–28. ¢Lcylinder = 1.566 in.
4–30. ¢Rclaw = 29.62 mm 85.2°a
3–4. R = 12 in.

3–6. s = 156.6 mm 4–32. ¢Lspring = 1.118 in., shorter

3–8. x = 11.5 in. y = 16.4 in. 4–34. ¢uram = 3.03°, cw

3–10. s = 175 mm 4–36. ¢ubed = 14.0°, ccw

3–12. L = 8 ft, 8 in. 4–38. ¢RP = 7.247 10.0b°
4–40. ¢Rpiston = 66.82 mm ;
3–14. h = 11.3 ft

3–16. y = 11.7 ft 4–42. ¢Rstamp = 1.570 in. T

3–18. h = 83.1 in. 4–44. ¢uram = 14.4°, cw
3–20. R = 24.18 Q18.1°
4–46. ¢utop handle = 16.8°, cw

3–22. R = 212.13 25.0b° 4–48. ¢L cylinder = 68.1 mm, shorter

3–24. R = 221.20 1138.5.b1°a° 4–50. ¢Rbox = 0.362 m T
3–26. R = 24.18 4–52. ¢Rclaw = 30.87 mm Q86.52°

3–28. R = 212.13 25.0b° 4–54. ¢uram = 5.5°, cw

3–30. R = 221.2 13.4b° 4–56. (¢Rpiston)max = 90.0 mm

3–32. J = 8.074 Q6183..73b°6° 4–58. (¢uram)max = 46.3°
3–34. J = 5.587
4–60. (¢uwheel assy)max = 57.6°

3–36. J = 212.13 R65.0° 4–62. (¢uram)max = 29.5°

3–38. J = 8.074 Q6183.3b.8° ° 4–64. (¢Rblade)max = 1.513 in.
3–40. J = 5.587
4–66. (¢uwiper arm)max = 72.8°

3–42. J = 212.13 R65.0° 4–68. (¢Rslide pin)max = 44.50 mm

3–44. J = 26.094 R11.4° Chapter 5

3–46. J = 109.76 R24.1° 5–2. b = 49.1°, v = 109 rpm
5–4. t1 = 0.188s, t2 = 0.142 s
3–48. J = 101.68 18.3b° 5–6. t1 = 0.067 s, t2 = 0.53 s
3–52. J = 26.10 R2141..14°°a 5-8. Q = 1.714, v = 63.2 rpm
3–54. J = 109.8 5-10. Q = 2.083, v = 162.2 rpm

3–56. J = 101.68 18.3b°

367

368 Answers to Selected Even-Numbered Problems

5–12. L2 = 4 mm, v = 750 rpm 7–8. ashaft = 7.85 rad/s2, cw
5–14. b = 20°, v = 100 rpm
5–16. b = 12.6°, v = 4286 rpm 7–10. ¢Ractuator = 85 in.
5–18. b = 8.6°, v = 1818 rpm
5–20. b = 0°, v = 17.14 rpm 7–12. ¢Rlinear motor = 15 in. T
5–22. b = 19.3°, v = 33.3 rpm
5–24. b = 16.36°, v = 200 rpm 7–14. ¢Ractuator = 10 in.
5–26. b = 8.57°, v = 300 rpm
5–28. b = 6.92°, v = 40 rpm 7–16. AnB = 17,770 in/s2 70b°
5–30. b = 49.09°, v = 17.6 rpm AB = 22,872 in./s2 R71.0°
5–32. b = 51.43°, v = 240 rpm 7–18. AA = 5158 in./s2 6.9°a
7–20. AA/B = 25.46 mm/s2 11.3°a
7–22.

7–24. AC/B = 1.35 ft/s2 42.3°a

7–26. Apiston = 31,341 in./s2 :

7–28. Apiston = 37,194 in./s2 :

Chapter 6 7–30. Aneedle = 29,271 mm/s2 c

7–32. Ablade = 58.97 in./s2 ;

6–2. ¢t = 37.5 s 7–34. Ablade = 103.73 in./s2 ;

6–4. v = 22.37 min/hr 7–36. ahorse = 5.22 rad/s2, cw

6–6. vmin = .0167 rpm 7–38. anozzle = 9.80 rad/s2, ccw
6–8. ¢Rtotal = 72 in.
6–10. ¢Rtotal = 7.5 in. 7–40. anozzle = 78.55 rad/s2, cw
6–12. V = 125.66 ft/min
7–42. awheel assy = 2.08 rad/s2, cw

7–44. Apiston = 93,195 in./s2 :
6–14. VB = 90 ft/s R 20° 7–60. AcB3/B2 = 900 mm/s2 Q30°
6–16. VB/A = 5.94 ft/s Q59.4°
6–18. VA/B = 15.72 ft/s 25.9°a 7–62. AcB3/B2 = 900 mm/s2 60b°

6–20. Vpiston = 272.55 in./s ; Chapter 8

6–22. Vblade = 59.63 in./s ; (spreadsheet/program results at shown at 120° crank angle)

6–24. Vblade = 5.94 in./s T 8–2. ¢R4 = 123.9 mm at u2 = 120°
8–4. u4 = 16.6° at u2 = 120°
6–26. vwiper blade = 2.50 rad/s, ccw 8–6. V4 = - 2577 mm/s at u2 = 120°
8–8. v4 = 3.27 rad/s at u2 = 120°
6–28. vbath = 0.85 rad/s, cw 8–10. A4 = 2349 mm/s2 at u2 = 120°
8–12. v4 = 204.4 rad/s at u2 = 120°
6–30. vsegment gear = 2.232 rad/s, ccw

6–32. Vblade = 112.91 mm/s c

6–34. Vcylinder = 4.39 in./s, compressing

6–36. Vright piston = 150.68 in./s 45°b Chapter 9
6–38. Vpackage = 775 mm/s : Q12.9°
6–40. Vplatform = 15.99 ft/min 9–14. vcam = 10.9 rpm, Vmax = 0.5 in./s

6–42. Vblade = 112.64 in./s : 9–16. vcam = 42.9 rpm, Vmax = 4.0 in./s

6–44. vwiper arm = 1.88 rad/s, cw 9–18. vcam = 13.3 rpm, Vmax = 2.5 in./s

6–46. vsegment gear = 3.827 rad/s, cw 9–20. vcam = 17.1 rpm, Vmax = 3.1 in./s

6–48. Vcylinder = 4.39 in./s, compressing 9–22. vcam = 20 rpm, Vmax = 2.0 in./s

6–50. Vpackage = 953 mm/s : 9–28. vcam = 9.4 rpm, Vmax = 0.94 in./s
9–30. Vmax = 1.25 in./s, Amax = 3.13 in./s2
6–74. Vpiston = 230.3 in./s ; 9–32. Vmax = 8.0 mm/s, Amax = 8.0 mm/s2
9–34. Vmax = 189 mm/s, Amax = 2961 mm/s2
6–76. Vblade = 4.82 in./s T 9–36. Vmax = 1.43 in./s, Amax = 6.41 in./s2
9–62. vout = 7.46 rad/s, aout = 123 rad/s2
6–78. vbath = 1.245 rad/s, cw

6–80. vblade = 0.071 rad/s, ccw

6–82. Vright piston = 288.3 in./s 45°b
6–84. Vplatform = 12.79 ft/min Q12.9°

6–86. Vblade = 91.11 in./s :

6–88. vwiper arm = 31.46 rad/s, cw Chapter 10

6–90. vsegment gear = 5.421 rad/s, cw 10–2. p = 0.393 in.
10–4. D = 84 mm
6–92. VCylinder = 2.64 in./s, compressing 10–6. mp = 1.53
6–94. V - 2. Avehicle = 10.60 ft/s2 10–8. mp = 1.47
10–10. C = 1.125 in.
Chapter 7 10–12. D1 = 2 in., D2 = 4.5 in.
10–14. C = 3.5 in.
7–4. ¢RA = 40 mm c
7–6. acam = 9.82 rad/s2, cw

Answers to Selected Even-Numbered Problems 369

10–16. Vt = 67.7 in./s Chapter 12
10–18. Vt = 90.3 in./s
10–20. N1 = 24 teeth, N2 = 96 teeth 12–2. p = .0357 in., l = 2.87°
10–22. Pd = 8, N1 = 16, N2 = 64
10–24. Pd = 4, N1 = 20, N2 = 60 12–4. p = .020 in., l = 3.65°
10–26. Pd = 12
10–28. D1 = 3.0 in., D2 = 13.33 in. 12–6. ¢Rram = 2.5 in. c
10–30. D1 = 2.0 in., D2 = 10.25 in.
10–32. Pd = 12, D1 = 4.0 in., D2 = 24.0 in. 12–8. ¢Rtable = 0.154 in. c
10–34. ¢u = 2.12 rev
12–10. ¢Rplate = 2.756 in. T
10–36. ¢s = 42.4 in.
12–12. ¢Rplatform = 2.564 in.Q28.32°9.7b°
10–38. v = 10.6 rpm 12–14. ¢Rend = 6.445 in.
10–40. pn = 0.28 in.
12–16. ¢Rfront end = 0.921 in. Q86.7°
10–42. Pd = 7.6, N1 = 19, N2 = 38
10–44. gp = 14.9°, gg = 75.1° 12–18. Vnut = 0.167 in./s T
10–46. gp = 10.4°, gg = 64.6°
10–48. Pd = 12, Nw = 2, Ng = 50, l = 5° 12–20. Vtable = 0.154 in./s c
10–50. v5 = 100 rpm, cw; c = 8.5 in. 12–22. Vplatform = 7.236 in./s Q32.8°
10–52. v8 = 30 rpm, cw; C = 17.97 in.
10–54. s4 = 0.74 in., c = 4.625 in. 12–34. e = 26%
10–56. Pd = 8
10–58. v1 = 3576 rpm 12–36. e = 47.2%
10–60. v1 = 2520 rpm, C = 9 in.
10–62. N = 17-68, 17-68, 18 -45, 18-45, 17 -34 12–38. e = 24.5%

10–64. ¢uwindow = 8°ccw 12–40. M10 * 1.50 and M8 * 1.25
10–66. ¢uwindow = 10.8°ccw
10–68. v4 = 2160 rpm, cw Chapter 13
10–70. v6 = 536 rpm, cw 13–2. R = 248 Q66°
10–72. v8 = 378 rpm, ccw
13–4. M = 200 in. lbs, cw
13–6. M = 188 in. lbs, cw
13–8. M = 18.9 Nm, cw
13–10. Fcyl = 3733 lbs(C)
13–12. Fmetal = 868 N T
13–14. Fscrew = 1200 lb(C)
13–16. Front cyl = 2137 lbs(C)

Rear cyl = 7182 lbs(C)
13–18. Front cyl = 5000 lbs(T)

Rear cyl = 11,110 lbs(C)

Chapter 11 Chapter 14

11–2. vout = 479 rpm, cw 14–2. m = 0.14 slugs
11–4. vout = 313 rpm, ccw
11–6. vin = 2700 rpm, ccw 14–4. I = 0.135 kg m2
11–8. c = 22.375 in., T = 162°
14–6. Ix = 31.25 kg m2
11–10. c = 25.618 in., u = 144° 14–8. Iz = 1.49lb ft s2
14–10. Iz = 0.166 kg m2
11–12. 5V belt 14–12. Iy cg = 0.00626 lb in. s2

11–14. 3V belt 14–14. Tmotor = 199.49 in. lbs, cw

11–16. 3V belt 14–16. Tmotor = 14.04 Nm, ccw

11–18. vout = 84 rpm, cw 14–18. Tmotor = 10.22 Nm, ccw
11–20. vin. = 760 rpm, cw 14–20. Fcig = 52.25 lbs Q3.7°
11–22. c = 48.724 in., u = 125°
T i = 132.5 in. lbs, cw
11–24. c = 52.424 in., u = 96° cg

11–26. No. 80 Chain 14–22. Fcig = 1377 N R49.4°

11–28. No. 100 Chain Tcig = 19.5 Nm, ccw

REFERENCES

1. Barton, Lyndon, Mechanism Analysis: Simplified Graphical 9. Mabie, Hamilton and Reinholtz, Charles, Mechanisms and
and Analytical Techniques, 2nd ed., Marcel Dekker Inc., Dynamics of Machinery, 4th ed., John Wiley and Sons Inc.,
New York, 1993. New York, 1987.

2. Baumeister, Theodore III, Avallone, Eugene, and Sadegh, 10. Martin, George, Kinematics and Dynamics of Machines,
Ali, Mark’s Standard Handbook for Mechanical Engineers, 2nd ed., Waveland Press Inc., Long Groove, IL, 2002.
11th ed., McGraw-Hill Book Company, New York, 2006.
11. Norton, Robert, Design of Machinery, 4th ed., McGraw-Hill
3. Chironis, Nicholas and Sclater, Neil, Mechanisms and Book Company, New York, 2008.
Mechanical Drives Sourcebook, 4th ed., McGraw-Hill Book
Company, New York, 2007. 12. Uicker, John, Pennock, Gordon, and Shigley, Joseph, Theory
of Machines and Mechanisms, 4th ed., Oxford University
4. Erdman, Aurthur, Sandor, George, and Kota, Sridhar, Press, New York, 2010.
Mechanism Design, Vol 1: Analysis and Synthesis, 4th ed.,
Prentice Hall, Upper Saddle River, NJ, 2001. 13. Townsend, Dennis and Dudley, Darle, Dudley’s Gear Hand-
book, 2nd ed., McGraw-Hill Book Company, New York,
5. Kepler, Harold, Basic Graphical Kinematics, 2nd ed., 1991.
McGraw-Hill Book Company, New York, 1973.
14. Waldron, Kenneth and Kinzel, Gary, Kinematics, Dynamics,
6. Jensen, Preben, Cam Design and Manufacture, 2nd ed., and Design of Machinery, 2nd ed., John Wiley and Sons Inc.,
Marcel Dekker, New York, 1987. Hoboken, NJ, 2004.

7. Jensen, Preben, Classical Modern Mechanisms for Engineers 15. Wilson, Charles and Sadler, Peter, Kinematics and Dynamics
and Inventors, Marcel Dekker, Inc., New York, 1991. of Machinery, 3rd ed., Pearson Education, Upper Saddle
River, NJ, 2003.
8. Jones, Franklin, Holbrook, Horton, and Newell, John,
Ingenious Mechanisms for Designers and Inventors, Vols. I–IV, 16. Working Model Demonstration Guide, Knowledge Revolution
Industrial Press Inc, New York, 1930. Inc., San Mateo, CA, 1995.

370

INDEX

Note: The letter ‘i’ and ‘t’ followed by locators refers to illustrations and tables cited in the text

A relative acceleration summary problems, C
208i–209i, 210
Absolute motion, 128 Cam follower displacement diagrams
Acceleration. See also Coriolis acceleration, relative velocity method, 137–141, 138i, analytical summary problems, 253–254
140i, 140i description of, 224–225, 238i
Normal acceleration, Relative graphical summary problems, 252–253
acceleration, Tangential acceleration, relative velocity method summary problems,
Total acceleration 162i–164i, 164–165 Cam followers, 223
analysis of, 175 motion scheme nomenclature, 227–228
general summary problems, 206 vector component addition, 53–55, 54i, 54t prescribed motion, 225–226
kinematic analysis, 2 vector component subtraction, 55–60, types of, 224–225, 224i
paths of, 179–180, 180i
point of interest, 4 56i–57i, 59t Cam followers motion scheme
vector, 43 vector magnitude determinations, 66–71, constant acceleration, 228, 228t, 229t, 230i
velocity profile, 178–179, 179t, 206–207, 207i constant velocity, 228, 228t
Working Model, 208i–209i, 213 67i–70i, 66t cycloidal motion, 230–236, 233i, 234i,
Acceleration curves vector triangle addition, 50–51, 51i, 53–54, 235i, 236i
description of, 202 harmonic motion, 228–230, 228t, 230i
graphical differentiation, 202–203, 203i 59–60, 59i, 759, 69, 69i
numerical differentiation, 204–205, 205 vector triangle subtraction, 57i, 55–57 Cam joint
summary problems, 206i–209i, 212–213 velocity curves summary problems, definition of, 3, 3i
Acceleration images, description of, 196–197 kinematic diagram symbol, 5t–6t
ACME thread types, 314, 315i, 317t 162i–164i, 165
Actuators Angle method, 50–51 Cams
definition of, 4 Angle of contact description of, 223, 224i, 226–227, 226i,
role of, 12–13 types of, 223–224, 224i
types of, 12–13 belt drive, 303–304, 303t
Addendum, gear terminology, 260, 260i, 263i chain drive, 308–309 Case studies
Air/hydraulic motors, actuator type, 12 Angular acceleration, 173, 174i acceleration analysis, 213–214, 213i–214i
Algebraic solution, common mechanism, Angular displacement, 73, 73i belt/chain drives, 312–313, 312i–313i
142, 190–191 Angular inertia of a body, 355 cam design/analysis, 255–257, 256i–257i
American Gear Manufacturer’s Association Angular position vector, 72–73 displacement/position analysis, 108, 108i
(AGMA) Angular velocity dynamic force analysis, 364, 364i
diametrical pitch, 261 and linear velocity, 126–127, 127i gears, 297–299, 298i–299i
gear quality, 273, 273t links, 125–126, 126i kinematic motion/classification, 29–30, 30i
gear standardization, 264–266, 264t and relative velocity, 128–129 machine analysis, 42, 42i
American National Standards Institute (ANSI), Annular gears, 259 machine design, 121–122, 121i–122i
gear standardization, 265 “Archimedes Screw,” See Auger screws screw mechanisms, 326–327, 326i–327i
Analytical method Assembly circuit spread sheets, 222, 222i
cam follower displacement diagrams branch defect, 120 static force analysis, 343, 343i
summary problems, 253–254 circuit defect, 119 vectors, 71, 71i
cycle position analysis, 96–98, 97i, 98i Auger screws, 323, 323i velocity, 168–169, 168i–169i
cylindrical cam profile design, 250, 255 Automatic Dynamic Analysis of Mechanical
disk cam profile design, 242–249, 243i–245i, Center distance
247i, 249i Systems (ADAMS®), 24, 31 belt drive, 302–303
disk cam profile design summary problems, Avoirdupois pound, 328 chain drive, 308–309, 308i
254i–255i, 255 gear mesh relationship, 266i, 267
displacement analysis, 79, 79i, 80–81 B
displacement analysis summary problems, Center of gravity, 345–346, 345i, 346t
101i–105i, 106 Backlash Chain drive
instant center location summary problems, gear mesh relationship, 270–271
162i–164i, 164–165 gear terminology, 260 description of, 300, 306–309
instant center method, 123, 152, 153i selection summary problems, 312
instant center method summary problems, Ball screws, 315, 317i, 321 Chain drive geometry, 308i, 308–309, 312
162i–164i, 165 Base circle Chain drive kinematics, 309–310, 309
limiting positions analysis, 91–93, 92i Chain length, 308–309, 308i
limiting positions summary problems, cam profile design, 237, 237i Chain pitch, 307, 307t
101i–105i, 105–106 gear terminology, 260i, 262i–263i Chain speed, 309
mechanism analysis technique, 24 Base diameter, 260 Chains, types of, 306i, 306–307
motion curves summary problems, 254 Bellcrank, 3i, 4 Change point, 19t, 20, 21
relative acceleration analysis, 188i, Belt drive geometry, 302–303, 302i, 311 Circular pitch, 260, 260i
188–190, 190t Belt drive kinematics, 303–306, 303i, 305i, 311 Clearance, 260
Belt drives Clockwise, 125
description of, 300–301 Closed-form position analysis
selection summary problems, 310 four-bar linkage, 87
types of, 300–301 in-line slider-crank, 81–83, 81i–82i
Belt length, 302, 303t offset slider-crank, 84–86, 84i–85i
Belt speed, 303–304 Coarse pitch, 261t, 264t
Belts, 300–302, 301i–302i, 301t Cog belt, 301, 301i
Bevel gear kinematics, 283–284, 283i
Bevel gears, 259–260, 259i, 283 371
Bull gear, 266

372 Index Cylindrical cams Drivers, 12–13, 15
analytical profile design, 250 Drum cam, 224, 224i
Coincident joint, 16–18 description of, 224, 224i, 362 Dynamic Analysis of Dynamic Systems (DADS®),
Combined motion schemes graphical profile design, 249–250, 250ii
dynamic analysis program, 31
comparisons, 237t D Dynamic equilibrium, 328
computer software for, 236 Dynamic force analysis
description, 236 D’Alembert’s principle, 351
goals of, 236 Dedendum, 260i, 260 design questions, 2
jerk, 236 Degree of freedom purpose of, 344
modified sinusoidal acceleration, 237 static force analysis, 343, 343i
modified trapezoidal acceleration, 236 beer crusher computation, 10i, 10–11 and static equilibrium, 328
polynomial displacement, 236 description of, 8 strategy for, 31
trapezoidal acceleration, 236 equation for, 74 Dynamic force analysis programs, 24
Common mechanisms, algebraic solutions, four-bar mechanism calculation, 19, 19i
lift table computation, 15–16, 15i–16i E
142, 190–191 mechanism types, 8, 8i
Common units, 173, 329, 346, 348 outrigger computation, 13–14, 13i–14i Eccentric crank, 14, 14i
Commutative law of addition, 49 shear press computation, 11–12, 11i–12i Efficiency, 321
Complex link, 3i, 4 toggle clamp computation, 9i, 9–10 Elastic parts, 2
Component method Degree of freedom mechanical press, 17–18, Electric motors, 12
Engines, actuator type, 4, 12
analytical addition, 53–55, 754i, 54t 17i–18i Enveloping worm gear teeth, 284, 285i
analytical subtraction, 59–60, 59i, 59t Diametral pitch Epicyclic train, 288–293, 289i
Components, 4, 12 Equilibrium, 333–338, 334i–337i
Composite bodies, 349–350, 349i gear terminology, 261 Equivalent linkage, 201
Computer methods/programs spur gear selection, 273–279, 273t, 276i, 278i
dynamic analysis programs, 31 Differential screw, description of, 322–323, F
mechanism analysis technique, 24–25
user–written programs, 221–222, 221i 322i–323i, 326 Face width, 260i, 260
value of, 31 Disk cam profile design Fine pitch, 315, 316t
Computer-aided design (CAD) systems Flat belt, 300, 301i
mechanism analysis technique, 24–25 analytical method, 242–249 Flat-faced follower
vector analysis, 44 design limitations, 241–242, 242i
Configuration, 74–75 flat-faced follower, 239–240, 239i analytical profile design, 248
Connecting arm, 19 graphical features, 237i, 237 description of, 224i, 225
Constant acceleration, 228, 229i–230i, 229t in-line knife-edge follower, 238i, 237–238 profile design, 239–240, 239i
Constant angular acceleration, 173–174, 174i in-line roller follower, 238i, 238 Floating link
Constant rectilinear acceleration, 171 offset roller follower, 239, 239i relative acceleration analysis, 191i,
Constant velocity, 228, 228t pivoted roller follower, 240–241, 240i
Contact forces, 331–332, 331i, 339–340 Disk cams, 223–224, 224i 191–192, 193i–194i, 195–196
Contact line, 262 Displacement relative acceleration summary problems,
Contact ratio, 267–268 analytical method, 79–81, 80i
Coriolis acceleration analytical summary problems, 101i–105i, 106 210, 210i
description of, 197–201 description of, 74 relative velocity method, 132–135, 133i–134i
summary problems, 210–212, 210i–211i general summary problems, 101, 101i–105i Follower, 19. See also Cam followers
Cosine, 44, 47 and linear velocity, 123 Follower motion, 224, 224i
Counterclockwise, 126 graphical driver analysis, 74i, 74–75 Follower position, 224–225, 224i
Coupler, 19 graphical problems, 76–78, 76i, 78i–79i, 80 Follower shape, 224i, 225
Coupler curve, 101, 101i graphical slave links analysis, 75–76, 75i, 76i Foot-pound, 328–329
Crank graphical summary problems, 101–106, Force
definition of, 3–4, 3i definition of, 328
eccentric, 14, 14i 101i–105i screw mechanisms, 320i–321i, 320–322
slider-crank mechanism, 22 kinematic analysis, 2 summary problems, 341, 341i
Crank-rockers point of interest, 4 vector, 43
circuits of, 87, 87i types of, 73 Force analysis. See also Dynamic force analysis,
four-bar mechanism, 19–20, 19t, 20i, 22 vector, 43
machine design, 115–117, 117i Working Model problems, 101i–103i, 105i, Static force analysis
summary problems, 120 and acceleration, 170
Crank-shapers 107–108 machine design, 1
analytical methods, 118 Displacement diagrams. See also Cam follower Formula, spreadsheets, 215–218, 217i–218i
graphical procedure, 117–118 Four-bar linkage, 87
machine design, 117–118, 117i displacement diagrams Four-bar mechanisms
summary problems, 120 cycle position analysis, 98–99, 99i–100i algebraic solutions, 142, 190–191
Cross drives, 303, 303i summary problems, 101i–103i, 105i, 107 categories of, 19t, 19–20
Crossed helical gears, 280 velocity curves, 155, 156i, 157–158, 158i circuits of, 87, 87i
Cycle analysis Double crank, 20, 19t, 20i coupler two–point synthesis, 119i, 118–119
analytical position, 96, 97i, 99, 100i Double enveloping worm gear set, 284, 285i description of, 19, 19i
description of, 94 Double rocker, 20, 19t, 20i Grashof ’s theorem, 19
displacement diagram, 101–106, 101i–105i Drafting motion classification, 20–21, 21i, 29, 29i
graphical position, 94–96, 94i–96i as technique, 24 nomenclature, 19
Cycloidal motion, follower motion scheme, vector analysis, 43–44 three-point synthesis, 119, 119i
Driver, 98 transmission angle, 93–94, 94i
230–236, 231t–232t, 231i–236i Driver link user-written programs, 221i, 221–222
Cylinders, actuator type, 4 mechanism analysis, 73 Working Model tutorial, 32–37, 33i–34i, 36i
position analysis, 74i, 74–75
Driver point, 73

Index 373

Frame slave links displacement, 75–76, 75i–76i analytical velocity method, 123,
definition of, 2 vector addition, 48–50, 49i–50i 152–154, 153i
four-bar mechanism, 19 vector magnitudes, 63–65, 63i–65i,
kinematic diagram, 8 analytical velocity method summary
Free-body diagrams, 331–333, 331i–333i 70–71, 70i–71i problems, 162i–164i, 166–167
vector subtraction, 55–57, 55i–57i
Friction Graphical differentiation graphical velocity method, 123, 149–152,
coefficients of, 339, 339t acceleration curves, 202–203, 203i 150i, 152i
force analysis, 1 velocity curves, 157i, 157
screw mechanisms, 320 velocity curves summary problems, graphical velocity method summary
problems, 162i–164i, 166
Friction force, 339 162i–164i, 167
Full joints, 3, 3i Graphical disk cam profile design Instant centers
locating, 142–149, 143i–149i, 146t, 148t
G features of, 237i, 237 summary problems, 162i–164i, 165–166
flat-faced follower, 239–240, 239i
Gear joint, 3, 3i, 5t–6t in-line knife-edge follower, 238i, 237–238 Instantaneous center of rotation, 142
Gear kinematics in-line roller follower, 238i, 238 Interference, 268–269, 268t
offset roller follower, 239, 239i Internal angle
gear function, 271–273, 271i–272i, 293 pivoted roller follower, 240–241, 240i
Gear mesh relationship Grashof ’s theorem, 19 triangle addition problem, 51
Gruebler’s equation triangle subtraction problem, 57–58, 58i
backlash, 270–271 description of, 8 Internal gears, 259, 259i
center distance, 266–267, 266i exceptions to, 18 International Organization for
contact ratio, 267–268 special cases, 16–18, 16i–18i
interference, 268–269, 268t Gyration radius, 348 Standardization (ISO)
operating pressure angle, 271 force unit, 328
undercutting, 269–270, 270i H metric threads, 315
Gear rack, 259, 259i, 262, 279 moment unit, 329
Gear selection, 294 Half joint, 3, 3i Inverted tooth/silent chain, 306i, 307
Gear standardization, 264–266, 264t Harmonic motion, 228–230, 228t–229t, Involute tooth profile, 262–264, 262i, 263i
Gear trains
description of, 286–288, 286i 229i–230i J
design summary problems, 296 Helical gear kinematics, 280–282, 281i,
summary problems, 295i–296i, 295–296 Joint, definition of, 3, 3i
Gear-driven mechanisms, summary 281t–282t Joints
Helical gears, 259, 259i, 294–295
problems, 296, 296i Helix angle, 280, 281i coincident, 16i, 16–17
Gears Herringbone gears, 259, 259i commonly used, 14–15
Higher order joint, 3, 3i
description of, 258–259, 258i–259i Hinge joint, 3, 3i K
terminology, 260i, 260–262, 261t, 262i Home position, 237, 237i
types of, 259–260, 259i, 284 Hydraulic cylinders, 13i, 12–13 Kennedy’s theorem, instant centers,
General triangles, 46–48, 47i, 48i Hydraulic motors, 12 144–146, 147
Geneva mechanism, 250–252, 250i, 252i, 255
Graphical analysis I Kinematic analysis, 1i, 2, 3
cam follower displacement diagrams Kinematic diagram
Idler gears, 288i, 288
summary problems, 252–253 Idler pulley, 301 four-bar mechanism, 19, 19i
cycle position, 94–96, 94i–96i In-line follower, 225, 224i manual water pump, 22
cylindrical cam profile design, 250, 250i In-line knife-edge follower, 238i, 237–238 symbol system, 4, 5t–6t
disk cam profile design, 237–242, 237i–242i In-line roller follower Kinematic diagram problems
disk cam profile design summary problems, beer crusher, 10i, 10–11
analytical profile design, 245–247, 245i, 247i lift table, 15i–16i, 15–16
254i–255i, 254–255 graphical profile design, 238i, 238 mechanical press, 17i–18i, 17–18
displacement, 74–79, 74i–79i In-line slider-crank nose wheel assemble, 20–21, 21i
displacement diagrams summary closed-form analysis, 81–83, 81i–82i outrigger, 13–14, 13i–14i
machine design, 113–114, 114i shear presses, 6–7, 6i–7i, 11–12, 11i–12i
problems,102i–105i, 107 Inch-pound, 329 sketching diagrams, 25–28, 25i–28i
displacement summary problems, 101–106, Included angle toggle clamp, 9i, 9–10
screw efficiency, 321 vice grip, 7, 7i
101i–105i thread feature, 314, 314i Kinematic inversion, 8
driver link displacement, 74i Inertia, 170, 328 Kinematics, 2
instant center location summary problems, Inertia-force method of dynamic Knife-edge follower
analytical profile design, 242–244, 243i–244i
162i–164i, 165 equilibrium, 351 description of, 224i, 225
instant center method, 123, 149–152, Inertial forces
L
150i, 152i description of, 350–355, 351i–352i, 354i
instant center method summary problems, summary problems, 362, 362i Law of cosines, oblique triangles, 46
Inertial torques Law of sines, oblique triangles, 46
162i–164i, 166 description of, 355–361, 355i–357i, 359i Lead, 315–316
limiting positions, 88i, 87–88 summary problems, 362–363, 362i–363i Lead angle
mechanism analysis, 23–24 Input link, 19
relative acceleration analysis, 181–187, Instant center, 142, 142i screw threads, 316
Instant center diagram, 144–145, 145i thread feature, 314, 314i
182i–183i, 185i–186i Instant center method worm gears, 284, 285i
relative acceleration summary problems, Limiting positions
analytical analysis, 91–93, 92i
208–209, 208i–209i definition of, 87, 88i
relative velocity method, 130–137,

130i–131i, 133i–136i
relative velocity method summary

problems, 162–164, 162i–164i

374 Index Metric thread, 314–315, 315i, 317t Offset slider-crank, 221i, 221
Minor diameter, 314, 314i closed-form analysis, 84–87, 84i–85i
Limiting positions (Continued) Miter gears, 260, 259i, 283 machine design, 114–115, 115i
graphical analysis, 88i, 87–88 Mobility (M)
graphical analysis problems, 88–91, 88i–90i Off-set slider crank Mechanism, 221i, 221
summary problems, 101i–105i, 106–107 equation, 8 Open, 32
equation exceptions, 18 Open-loop linkages, 8, 8i
Line of center, 149, 150i special cases, 16–18, 16i–18i Operating pressure angle, 271
Line of contact, 262, 262i Mobility (M) calculation Output link, 19
Line of proportion, 149, 150i can crusher, 11
Linear acceleration, 170–173 four-bar mechanism, 19 P
Linear cam, 224, 224i graphical displacement analysis, 76
Linear displacement, 73 lift table, 16 Parallel axis theorem, 248i, 348–349
Linear motion, 125i, 124–125 manual water pump, 22 Parallelogram mechanisms, 22, 23i
Linear velocity mechanical press, 17–18 Phase, 94
outrigger, 14 Pin, 3, 3i
and angular velocity, 126–127, 127i shear press, 12 Pin joint
definition of, 123 sketching diagrams, 25i–29i, 29
general point, 124, 124i toggle clamp, 9–12, 14 kinematic diagram symbol, 5t
rectilinear points, 124i, 123–124 Module, 261, 261t four-bar mechanism tutorial, 33–35, 34i
Link, angular position, 72 Moment slider-crank mechanism tutorial, 38–41, 40i
Linkage acceleration. See also Timing charts description, 328–331, 329i–330i Pin-in-a-slot joint, commonly used, 14, 15i
Linkage, definition of, 2 summary problems, 341, 341i Pinion, 266, 266i. See also Gear rack
Linkage velocity analysis. See also Timing charts Moment of inertia Piston, 3, 3i
Links basic shapes, 346–348, 347i, 347t Pitch
acceleration analysis, 170, 173–174, 210, 210i composite bodies, 349i, 349–350 thread features, 315, 315i
angular velocity, 125–126, 126i description of, 346, 346i worm gears, 284, 285i
commonly used, 14 experiential determination, Pitch circle, 260, 260i, 262, 262i
definition of, 2 Pitch curve, 237, 237i
four-bar mechanism tutorial, 32–33, 33i 350, 350i Pitch diameter
relative velocity method, 130i, 130–135, parallel axis theorem, 348i, 348–349 belt drive, 301i–302i, 302–303
radius of gyration, 348 chain drive, 308, 308i
131i, 133i–134i summary problems, 361–362, 361i gear terminology, 260, 260i
resizing, 33 Motion thread feature, 314, 314i
slider-crank mechanism tutorial, 37–38, 38i laws of, 331, 350 Pitch line
types of, 3 mechanism analysis, 1, 1i, 73 gear terminology, 262, 262i
Locked mechanism, 8, 8i Motor spur gear kinematics, 271, 271i
Long and short addendum system, 269 actuator types, 4 Pitch point, 260
Lowering a load (screw drive), 321 four-bar mechanism tutorial, 35–36 Pivot link, 118i, 118
Lubrication, chain drive kinematics, 309 slider-crank mechanism tutorial, 41 Pivoted followers, 224, 224i
Multiple threads, 315–316, 317i Pivoted roller follower
M Multiple-strand roller chain, 306, 306i analytical profile design, 249i, 248–249
Multi-strand chains, 307, 307t profile design, 240–241, 240i
Machine design Multi-V-belt, 300, 301i Planar mechanism, 2
crank-shaper, 117–118, 117i Planet gear, 288, 289i
crank-rocker, 115, 116i, 117, 120 N Planetary gear analysis
slider-crank ratio, 113–115, 114i–115i, 120 equation, 291–292
three point synthesis, 119, 119i, 121 Newton, force magnitude unit, 328
time ratio problems, 109–110, 120 Newton, Sir Isaac, 170, 331 summary problems, 292–293
two position links, 118–121, 118i–119i Normal acceleration superposition, 289

Machines, 1 acceleration analysis, 174–177, summary problems, 289–291
Magnitude, 73, 73i 175i–176i Planetary gear trains
Magnitude direction
description of, 173, 173i description of, 288–293, 289i, 290t–291t, 291i
triangle addition, 52 summary problems, 206–207, 207i summary problems, 297, 297i
triangle subtraction, 57, 58i Normal circular pitch, 281, 281i Plate cams, 224, 224i
Major diameter, 314, 314i Normal diametral pitch, 281 Pneumatic cylinders, 13i, 12–13
Manual force, 13 Normal force, 339 Point (P), 73i, 72–73
Mass, 344, 361–362, 361i–362i Normal module, 281 Point of interest, 4
Mass moment of inertia Normal pressure angle, 281 Point of interest path, 37, 37i
basic shapes, 346, 347i, 347t, 348 Normal section, 281, 281i Point position measurement, 36–37
composite bodies, 349i, 349–350 Numerical differentiation Points
description of, 346, 346i acceleration curves, 204–205, 205i linear acceleration, 173
experiential determination, 350, 350i velocity analysis, 159–160, 160i linear velocity, 124, 124i
parallel axis theorem, 348i, 348–349 relative acceleration analysis, 274, 191–196,
radius of gyration, 348 O
summary problems, 361–362, 361i–362i 191i–194i
Mechanism Oblique triangle, 46–48, 46i–48i relative acceleration summary problems,
basic components of, 2–4, 3i Offset follower, 225, 224i
definition of, 1–2 Offset roller follower 210, 210i
degrees of freedom computation, 74 Points/floating link, 132–135, 133i–134i
motion analysis, 73 analytical profile design, 248 Points/multiple, relative velocity method,
phases of, 80 graphical profile design, 239, 239i
vectors, 43 Offset sidebar roller chain, 306i, 307 135–137, 135i–136i
Mechanism analysis, 2, 23 Points/one link, relative velocity method,

130–132, 130i–131i

Index 375

Points of interest Resultant Spherical-faced follower, description of,
four-bar mechanism tutorial, 33, 34i definition of, 49 224i, 225
slider-crank mechanism tutorial, 55i graphical addition problems, 49, 49i, 50
graphical subtraction problem, 56–57, 57i Spreadsheets
Position, 2, 72 acceleration curves, 205i
Position analysis Resultant components, 53–55, 54i cycle position analysis, 97, 98i
Resultant force, summary problems, 341, 341i cycloidal motion, 235i
analytical displacement, 80i, 80–81 Resultant magnitude displacement curve, 155, 156i
graphical displacement, 74–76, 74i–76i displacement diagram, 100i
graphical displacement problems, triangle addition problem, 51 general description, 215–220, 215i–220i
triangle subtraction problem, 58, 58i in-line roller follower design, 247i
76–79, 76i–79i Reversible gearset, 285 knife-edge follower design, 244i
in-line slider-crank, 81, 81i Revolute joint, 3, 3i mechanism analysis technique, 24
limiting positions, 91–93, 92i Richardson method (numerical differentiation) summary problems, 222
offset slider crank, 84, 84i acceleration curves, 204 velocity analysis, 160i
purpose of, 72i, 72 velocity curves, 159
Position vector, 73i, 72 Right triangle, 44–46, 44i–46i Sprocket, 307, 307i, 308t
Pound, 328 Ring gear, 288, 289i Spur gear geometry, summary problems, 293
Pressure angle Rocker, 3i, 4 Spur gear kinematics
description of, 241, 241i Rocker arm link, 4
gear terminology, 262, 262i Roller chain, 306, 306i gear function, 271–273, 271i–272i
spur gear selection, 274–279, 276i, 278i Roller follower, 224i, 225 summary problems, 293
Primary centers Rotation Spur gear selection
locating problems, 145–149, 145i–149i, 146t, instantaneous center of, 142, 142i diametral pitch, 273–279, 273t, 276i, 278i
kinematic analysis, 2 pressure angle, 274–279, 276i, 278i
148t relative velocity method, 130–132, 130i–131i set center distance summary problems, 294
rules of, 203i, 143–144, 143i–144i teeth number, 274–279, 276i, 278i
Primary joints, 3, 3i S Spur gears
Prime circle, 237, 237i gear type, 259, 259i
Prismatic joint, 3, 3i Scalar quantities, 43 terminology, 260i, 260–262, 261t
Pulleys, 301i, 301, 301t Scotch Yoke mechanism, 23, 23i Square thread, 314–315, 315i
Pythagorean theorem, 45 Screw actuators, 13 Static equilibrium
Screw forces/torques, 320–322, 320i–321i, 326 conditions of, 333
Q Screw joints, 15, 15i definition of, 328
Screw kinematics, 316–320, 318i–320i Static machine forces, summary problems,
Quick-return mechanisms, 23, 23i Screw mechanisms, 314
Screw thread, 315, 314i–315i, 323–324 341–343, 341i–343i
R Screw-driven acceleration, 326, 324i–325i Straight-line mechanisms, 22
Screw-driven displacement, 324–325, 324i–325i
Rack, 259, 259i, 279 Screw-driven velocity, 325–326, 324i–325i Peaucellier-Lipkin linkage, 22i
Rack and pinion kinematics, 279–280, 280i, 294 Self-locking, 316 Watt linkage, 22i
Radian, description of, 125–126 Serpentine drives, 303, 303i Stroke, 88i, 88
Rectilinear points Servomotors, 12 Stub teeth, 269
Shaft angle, 283, 283i Sun gear, 288, 289i
linear acceleration, 170–171 Sheaves, 301i, 301, 301t Superposition method, 289
linear velocity, 123–124, 124i, Simple link, 3, 3i, 5t Swing arm followers, 224, 224i
Rectilinear translation, 130–132, 130i–131i Simulation Synthesis, 109
Relative acceleration
components, 179–181, 180i, 197–201, four-bar mechanism, 32–37, 33i–37i T
slider-crank mechanism tutorial, 37–41,
197i–200i Tangent, right triangle, 44
description of, 177–179, 178i, 179t 38i–40i Tangential acceleration
summary problems, 207i, 207–208 Sine, 44, 46
Relative acceleration analysis Sketching, practice problems, 25–28, 25i–28i acceleration analysis, 174–175
analytical method, 188i, 188–190, 190t Slave links, 75–76, 75i–76i description of, 173, 173i, 175–177, 176i
analytical summary problems, 208i–209i, Slider-crank summary problems, 206–207, 207i
Tangential velocity, 126
209–210 transmission angle, 93–94, 94i Teeth. See also Involute tooth profile
graphical method, 181–187, 182i–183i, Slider joint, symbol, 6t chain drive, 308, 308i
Slider-crank mechanisms gear terminology, 260–262
185i–186i spur gear selection, 273–279, 274t–275t,
graphical summary problems, 208–209, algebraic solutions, 142, 190–191
description of, 22 276i, 278i
208i–209i limiting positions, 88i worm gears, 284, 285i
Relative displacement, 317 machine design, 113–115, 114i–115i Thread features, 314, 314i
Relative motion summary problems, 120 Thread forms, 314–315, 315i, 316t–317t
Working Model tutorial, 37–41, 38i–40i Thread number, 315–316, 317i
definition of, 177 Sliding friction force, 339t, 339i–340i, 339–341 Three-point synthesis, machine design, 119,
description of, 128, 177 Sliding joint, 3, 3i
Relative velocity, 128–129, 128i, 129i Sliding link, 38, 38i–39i 119i, 121
Relative velocity method Slot joints, 38, 39i Throw angle, 115
analytical method, 137–141, 138i, 140i Slug, derivation of, 344 Time ratio, 109–110, 120
analytical method summary problems, Software. See Computers methods/programs Timing belt, 301, 301i
Solenoids, actuator type, 4 Timing charts
162i–164i, 164–165 Speed, 2
graphical analysis, 130–137, 130i–131i, problems, 112–113
uses of, 110–111, 111i
133i–136i Tip-to-tail method, 48
graphical analysis summary problems,

162–164, 162i–164i
velocity analysis, 123, 137–141, 138i, 140i
velocity image, 137, 137i

376 Index Vector components, 52i, 52 graphical methods summary problems,
Vector diagram 162–164, 162i–164i
Torque
definition of, 328, 329i graphical addition problems, 49–50, instant center analytical method,
screw mechanisms, 320–322, 320i–321i 49i–50i 123, 152, 153i

Total acceleration, 175–177, 176i graphical subtraction problems, 55–56, 56i instant center graphical analysis, 123,
Trace, 101 magnitude determination, 65, 63i–65i 149–152, 150i, 152i
Trace point, 237, 237i triangle addition problem, 51
Translating followers, 224, 224i triangle subtraction problem, 58–59, 58i Velocity curves
Transmission angle Vector equations, 60, 60i analytical summary problems, 162i–164i,
analytical magnitude determination, 167–168
definition, 93 description of, 155–157, 156i
mechanical advantage, 93 66–67, 71 graphical differentiation, 157–158, 157i–158i
Transverse section, 281, 281i application of, 62 graphical summary problems, 162i–164i, 167
Triangles formation of, 60–62, 61i–62i numerical differentiation, 159–161, 160i
summary problems, 67–69, 67i–68i graphical magnitude determination,
trigonometric relationships, 44, 44i Velocity image, 137, 137i
types of, 44, 46 63–65, 70–71 Velocity profile
vector addition, 50–51, 51i subtraction, 59
vector subtraction, 57–60, 58i–60i, 59t Vector magnitude acceleration, 171–172, 172i
Trigonometry, 44 analytical determination, 66–67, 66i, 66t, linear motion, 125i, 124–125
Triple rocker, 19t, 20, 20i Velocity ratio
Truss, 8, 8i 71, 71i belt drive kinematics, 303
Two-armed synchro loader, 4, 5t graphical determination, 63–65, 63i–65i, chain drive kinematics, 309
Two-force member, 333–338, 334i–337i spur gear kinematics, 271–273, 271i–272i
Two-point synthesis 70–71, 70i–71i Volume, 323
coupler of four-bar mechanism, 119i, 118–119 Vector problems Volumetric acceleration, 323
design class, 118 Volumetric flow, 323
pivot link, 118–119, 118i–119i analytical addition, 51–55, 51i–52i,
single pivot summary problems, 120–121 54i, 54t, 69i, 69 W
two pivots summary problems, 121
analytical magnitude determination, 66–67, Weight, 344
U 66i, 66t, 70–71, 70i–71i Whole depth, 260
Windshield wiper system, design concept, 1, 1i
Undercutting, 269–270, 270i analytical subtraction, 57–60, 58i–60i, 59t, Wipe pattern, components, 1
Unified thread, 314–315, 315i, 316t 69i, 70 Working Model software
United States Customary System
equations, 61–62, 70, 70i acceleration, 208i–209i, 213
force unit, 328 graphical addition, 49–50, 49i–50i, 69i, 69 computer simulation software, 31
mass/weight, 344 graphical magnitude determination, 63–65, displacement problems, 101i–105i,
moment unit, 328–329
threads per inch, 314 63i–64i, 70i–71i, 70–71 107–108
graphical subtraction, 55–57, 55i–57i, 69i, 69 four-bar mechanism tutorial, 32–37,
V triangle addition, 51i, 51, 53
triangle subtraction, 59–60, 59i–60i, 59t 33i–37i
V-belt, belt type, 300–302, 301i–302i Vectors practice problems, 41–42, 41i–42i
Vector analysis addition methods, 43, 48–55, 48i–52i, 54i, 54t purchase information, 32
mechanism characteristics, 43 slider-crank mechanism, 37–41, 38i–40i
analytical magnitude determination, 66–67, oblique/general triangles, 46–48 velocity problems, 162i–164i, 168
66i, right triangle, 44–46, 44i Worm, 284
66t, 71, 71i subtraction methods, 55–60, 55i–60i, 59t Worm gear kinematics, 284–286, 285i
triangle types, 44–48, 44i–48i, 47i–48i Worm gears, 259i, 260, 295
component addition, 53–55, 54i, 54t Velocity Worm pitch diameter, 284, 285i
component subtraction, 59–60, 59i–60i, 59t kinematic analysis, 2 Worm wheel, 284
graphical addition, 48–50, 49i–50i point of interest, 4
graphical magnitude determination, 63–65, summary general problems, 161, 161i X
summary relative problems, 162, 162i
63i–65i vector property, 43 X–axis
graphical subtraction, 55–57, 55i–57i Working Model, 162i–164i, 168 angle addition, 53
triangle addition, 50–51, 51i, 53–55, Velocity analysis angle subtraction, 59t
algebraic solutions, 142 component determination, 66, 66t
54i, 54t, 69i, 69 analytical methods, 137–141, 138i, 140i rotational equation, 155
triangle subtraction, 58i, 57–58 description of, 123
graphical methods, 130–137,

130i–131i, 133i–136i