solutions to chapter 14 exercise problems - … · - 544 - solutions to chapter 14 exercise...

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Solutions to Chapter 14 Exercise Problems

Problem 14.1

The four-wheeled vehicle shown slides down a steep slope with its rear wheels locked (notmoving relative to the body) and its front wheels rolling freely. If M is the mass of the vehicle, hthe normal distance from its center of mass, G, to the ground, r the wheel radius, and 2c thedistance between the axles, find the acceleration of the vehicle. The angle of the slope is , andthe coefficient of friction between the wheels and the ground is µ The mass and moment ofinertia of each wheel about its axle may be neglected. What is the largest value for the angle atwhich the vehicle will not slide?

c

c

r

r

h

G

θ

a

Front

Solution

A freebody diagram of the vehicle is shown in Fig. 14.1.1a. For the analysis, we will write theequations of dynamics relative to an n-t coordinate system with t parallel to and and n normal tothe direction of motion as shown in Fig. 14.1.1b.

Summing forces in the n and t directions and summing moments about point G gives

Fn =man = 0 = N1 + N2 mgcos (1)

Ft =mat = µN1 +mgsin (2)

MG = I = 0 = N2(c) N1(c) µN1(h) (3)

From Eq. (3),

N2 = N11+µhc[ ] (4)

Combining Eqs. (1) and (4) gives

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c

c

r

r

h

G

θ

mgµ N1N1

N2

c

c

r

r

h

G

θ

mg

µ N1N1

N2

θ

mg sinθ

mg cosθ

n

t

(a)

(b)

Fig. 14.1.1: Freebody diagram for Problem 14.1

N1 + N1 1+µhc[ ] mgcos = 0 N1 =

mgcos

2 +µhc

(5)

From Eqs. (4) and (5),

N2 = N11+µhc[ ]= mgcos

2 +µhc

1+µhc[ ]= mgcos c+ µh

2c+ µh

Combining Eqs. (2), (4), and (5),

mat = µN1 +mgsin = µmgcos

2 +µhc

+mgsin

or

at = g sinµccos2c+ µh

If at = 0 (no sliding),

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tan =µc

2c+ µhand

= tan 1µc

2c +µh

Problem 14.2

The flyball governor shown is started from rest and accelerated slowly about the axis of rotation.At what speed of rotation will it be in the position shown? Friction may be neglected. Ignore themasses of the four links.

AB = GH = 2.50 inEG = FH = 3.76 inCG = DH = 5.54 inAE = BF = 3.76 in

1 lb weight

10 lb weight

1 lb weight

ω

Axis of Rotation Attached to the

Frame

A B

C D

E F

G H

2

34

5 6

7

45˚

Solution:

Because the mechanism is symmetric, we need only consider one half of it. The inertia force oneach weight will be

fI =mar = Wgr 2 = W

2

gHDsin45̊ +AB / 2[ ] =

(1) 2

3865.54sin45̊ +2.5 / 2[ ] = 0.01339 2 (1)

The problem is to determine the angular velocity. Therefore, we will treat the inertia force as anunknown force. Once we determine the magnitude of fI, we can determine the value for theangular velocity from Eq. (1).

A freebody diagram of links 2, 3, and 6 are shown in Fig. 14.2.1. To analyze the system, startwith link 2 and sum forces in the vertical direction . Before doing this, however, note that

F32 = F42

Then, summing forces in the vertical direction for link 2 gives

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Fy = 0 = 10+ 2F32 cos45̊ F32 = 102cos45̊

= 7.071 lb

Because link 3 is a two-force member,

F36 = F32

1 lb weight

10 lb weight

1 lb weight

ω

Axis of Rotation Attached to the

Frame

A B

C D

E F

G H

2

34

5 6

7

45˚

10 lb

A B

2

F32

45˚ 45˚F42

B

F

3

F63

F23

1 lb

D

F

H

645˚

f IF36

F76

F76

x

y

Fig. 14.2.1: Freebody diagrams for Problem 14.2.

as indicated in Fig. 14.2.1. The freebody diagram for link 6 now has only three unknowns.Therefore, we can solve for the unknowns. To find the inertia force, we can sum moments aboutpoint H without solving for F76 . Then,

MH = 0 = F36(FH) + fI (HDcos45̊ ) 1(HDsin45˚)

or

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7.071(3.76)+ fI(5.54cos45̊ ) 1(5.54sin 45̊ ) = 0or

fI =(5.54sin45̊ )+ 7.071(3.76)

(5.54cos45̊ )= 7.787 lb

Therefore, from Eq. (1),

0.01339 2 = 7.787

Then,2 = 7.7870.01339

= 581.5

and= 24.1 rad / s

Problem 14.3

Solve Problem 14.2 assuming a coefficient of friction of 0.3 at each of the six pin joints. Thediameter of each joint is 0.8 in.

Solution:

The inertia force on each weight will be

fI =mar = Wgr 2 = W

2

gHDsin45̊ +AB / 2[ ] =

(1) 2

3865.54sin45̊ +2.5 / 2[ ] = 0.01339 2 (1)

The problem can be approached either graphically or analytically. However, because a nonlinearproblem is involved, it is easiest to solve the problems graphically using friction circles (Chapter11). To determine the general direction of the force vectors, we need to conduct a friction freeanalysis first. Freebody diagrams of the links are shown in Fig. 14.3.1. Note that we are usingthe fact that the mechanism is symmetric so that we need analyze only half of the mechanism.Also, we will use skeletal diagram for the links to simplify the drawings.

To start the analysis, sum forces on link 2. Then,

F = 0 = F32 + F42 10j

The force polygon is shown in Fig. 14.3.2. From the polygon,

F32 = 7.07 lb .

The freebody diagram for link 6 has three unknowns. To solve the problem graphically, we needto resolve the weight of the counter weight and F36 into a single vector. The resultant we willcall P. Next, resolve P and fI into a single vector. We cannot find the magnitude yet, but weknow that the resultant must go through the point of intersection of the two vectors. After this isdone, there will be only two forces acting on link 6. Therefore, the line of action of F76 mustpass through the intersection of P and fI. Knowing the direction of F76, we can complete theanalysis by summing forces on link 6. That is,

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F = 0 = F76 + F36 1j + fI

From the polygon,

fI = 7.60 lb .

Force Scale

10 lb

10 lb

2

F

B

3

fI

1 lb

H

D

Position Scale2 in

6

10 lb

B2

F42 F32

F

B

3

F23

F63

F

fI

1 lb

H

D

6

F76

Fig. 14.3.1: Freebody diagrams for Problem 14.3.

Next consider the friction case. We now know the general direction of the forces, and from akinematic analysis, we can determine the direction of the friction torques. These tend to opposethe relative motion. The general directions of the relative angular velocities and friction torquesare shown in Fig. 14.3.3.

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To use the friction circle approach, we need to compute the radius of the friction circles. This isgiven by Eq. (11.15) as

Rf = Rsinwhere

= tan 1µ = tan 10.3 =16.7̊

Then,

Rf = Rsin = 0.4sin(16.7) = 0.115 in .

The friction circles are also shown in Fig. 14.3.3. The forces are located relative to the frictioncircles so that the effect of the force is to create the friction torque on the free body of the link.The forces are located on the proper sides of the circle in Fig. 14.3.3.

The freebody diagrams with the forces located properly are shown in Fig. 14.3.4. We can thenfollow the same general procedure as was used in the nonfriction case. That is, start summingforces on link 2. Then,

F = 0 = F32 + F42 10j

The force polygon is shown in Fig. 14.3.4. From the polygon,

10 lb

B2

F42 F32

F42

10 lb

F32

F36

F

fI

1 lb

H

D

6

F76

P

F36

1 lb

fI

PF76

Fig. 14.3.2: Force polygons for friction-free case.

F32 = 7.53 lb .

The freebody diagram for link 6 has three unknowns. To solve the problem graphically, we needto resolve the weight of the counter weight and F36 into a single vector. The resultant we will

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call P. Next, resolve P and fI into a single vector. We cannot find the magnitude yet, but weknow that the resultant must go through the point of intersection of the two vectors. After this isdone, there will be only two forces acting on link 6. Therefore, the line of action of F76 mustpass through the intersection of P and fI. Knowing the direction of F76, we can complete theanalysis by summing forces on link 6. That is,

Force Scale

10 lb

10 lb 2

F

B

3

fI

1 lb

H

D

Position Scale2 in

6

10 lb

B2

F42 F32

F

B

3

F23

F63

Increasing

decreasing

Increasing

F

fI

1 lb

H

D

6

F76

F36

Fig. 14.3.3: Freebody diagrmas for friction case.

F = 0 = F76 + F36 1j + fI

From the polygon,

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fI =8.69 lb .

Therefore, from Eq. (1),

0.01339 2 = 8.69

Then,

2 = 8.690.01339

= 649

and= 25.5 rad / s

F42

10 lb

F32

F76

P

F36

1 lb

fI

P F76

10 lb

B2

F42 F32

F

f I

1 lb

H

D

6

F36

Fig. 14.3.4: Force polygons for friction case.

Problem 14.4

A wheel, of mass m and radius r, rolls without slipping on a horizontal plane. It hits a step ofheight h. If the velocity of the center of the wheel before striking the step is V, directed asshown, find:

1. The magnitude and direction of the velocity of the center of the wheel immediately after theimpact

2. The minimum value of V for which the wheel surmounts the step

3. The impulse exerted on the wheel by the edge of the step at impact

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The impact may be considered to take place over a vanishingly small time interval. The wheel isassumed to remain in contact with the edge of the step after the impact. The wheel may beconsidered to have moment of inertia about its center in the direction of rotation I = mr2.

h

rV

Solution

Part 1

Before impact, the angular momentum about the top edge of the step is

H1 =mV(r h) + IVr

for the wheel,

I =mr2

Therefore,

r

u

h

θ

θ

Fig. 14.4.1: Wheel immediately after impact with the step

H1 =mV(r h) + I = mV(r h) + IVr=mV(r h)+ V

rmr2

= mV r h + r[ ] =mV 2r h[ ](1)

Immediately after impact, the wheel will rotate about the corner of the step. Let the velocity be uas shown in Fig. 14.4.1. Then, the angular momentum about the edge of the step is

H2 =mru + Iur= mur + u

rmr2 = 2mur (2)

By conservation of angular momentum,

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H1 =H2 2mur =mV 2r h[ ]

Then,u =V 2r h[ ] / (2r)

The direction is given by

= cos 1 r hr( )

Part 2

The cylinder will just reach the top of the step, when all of the initial kinetic energy is convertedto potential energy. The kinetic energy just after impact (energy is not conserved during impact)is

KE = 12mu2 + I 2[ ] = 1

2 mu2 + I u

r( )2

[ ] = 12 mu2 +mr2 ur( )2

[ ]= mu2 = mV2 2r h[ ]2 / (4r2)

and the potential energy is zero. After just reaching the top of the step, the potential energy ismgh and the kinetic energy is zero. Therefore,

mgh =mV2 2r h[ ]2 / (4r2)

or the initial velocity required to just reach the top of the step is

V = gh 2r2r h

Part 3:

To determine the impulse exerted on the wheel, we must determine the change in momentumbefore and after the impact. Before the impact

q1 =mVi

After impact (see Fig. 14.4.1),

q2 =mu cos i + sin j( )

From Fig. 14.4.2,

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r

u

θr - h

r2 − r− h( )2

h

Fig. 14.4.2: Geometry at step

cos = r hr

and

sin =r2 r h( )2

r= 2rh h2

r

Therefore,

q2 =mu r hri + 2rh h2

rj( ) =mV 2r h[ ]

2rr hri+ 2rh h2

rj( )

The impulse is

im = q2 q1 =mV 2r h[ ]

2rr hri + 2rh h2

rj( ) mVi

= mV2r2

(r h)(2r h) 2r2{ }i+ (2r h) 2rh h2 j( )

= mV2r2

h(h 3r)i + h(2r h)(3/2) j( )

Problem 14.5

In the mechanism shown below, link 2 rotates at an angular velocity of 20 rad/s (CW) andangular acceleration of 140 rad/s2 (CW). Find the torque that must be applied to link 2 tomaintain equilibrium. Link 2 is balanced so that its center of mass is at the pivot O2. The centerof mass of link 3 is at A, and the mechanism moves in the horizontal plane. Friction may beneglected.

O2A = CA = 100 mm, m3= 0.74 kg, m4= 0.32 kg

IG2= .00205 N-s2-m, IG3=.0062 N-s2-m

- 556 -

A

B

4

32 G3 C G4,O 2 G 2,

15˚T12

Solution:

Velocity Analysis:

1vA2 = 1vA2 /O2 = 1vA3 = 1 2 rA2/O2

1vC3 =1vC4 =1vA3 + 1vC3 /A3 (1)

Now,

1vA3 = 12 rA2/O2 = 20 100 = 2000mm / s ( to rA2 /O2 )

1vC3 in horizontal direction

1vC3 /A3 =1 3 rC3/A3 1vC3 /A3 = 1 3 rC3 /A3 ( to rC3 /A3)

Solve Eq. (1) graphically with a velocity polygon. From the polygon,

1vC3 =1035.3mm / s

Also,

1vC3 /A3 = 2000mm / sor

13 =

1vC3/A3rC3/A3

= 2000100

= 20 rad / s CCW

Acceleration Analysis:

1aC3 =1aC4 =1aA3 +1aC3/A3

1aC3 =1 aA2 /O2r +1 aA2 /O2

t +1 aC3/ A3r +1 aC3 /A3

t (2)

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A

C2

4

3

D

o

1000 mm/s

Velocity Polygon

a2 a3,

c3, c4

1vC 3

1vA 31vC3 /A3

o'

20,000 mm/sAcceleration Polygon

2

a 2'a 3',

1aA2 / O 2r

1aA2/O2t

1aC3

c 3'

1aC3 / A 3r

1aC3 / A 3

t

G4

G3

g3',

g4',

Fig. 14.5.1: Position, velocity, and acceleration polygon for Problem 14.5

Now,

1aC3 in horizontal direction

1aA2 /O2r =1 2

12 rA2/O2( ) 1a A2/O2

r = 122 rA /O2 = 202 100 = 40,000 mm / s2

in the direction opposite to rA/O2

1aA2 /O2t =1 2 rA2/O2 1aA2 /O2

t = 1 2 rA/O2 = 140 100 =14,000mm / s2 ( to rA/O2)1aC3/A3

r =1 313 rC3/ A3( ) 1a C3/ A3

r = 132 rC3/A3 = 202 100 = 40,000mm / s2

in the direction opposite to rC3/A3

1aC3/ A3t =1 3 rC3 /A3 1aC3/ A3

t = 1 3 rC3/A3 ( to rC3/A3)

Solve Eq. (2) graphically with a acceleration polygon. From the polygon,

1aC3 =1aG4 = 70,028mm / s2and

1aA3 = 1aG3 = 42,400 mm / s2

Also,

- 558 -

1 3 =1aC3/A3t

rC3/A3= 14000100

=140 rad / s2 CCW

Inertia Force Analysis:

We can now conduct the inertia force analysis. There will be an inertia force associated witheach center of gravity. We will represent these forces with a lower case f to distinguish betweeninternal inertia forces and externally applied forces. The forces are

f12 = m21aG2 = 0f13 = m3 1aG3 = 0.74(42,400) = 31,380 kg mm / s2 =31.380 kg m / s2 = 31.380 N(opposite1aG3 )f14 = m41aG4 = 0.32(70,028) = 22,409 kg mm / s2 = 22.409 kg m / s2 = 22.409 N(opposite 1aG4 )

Links 2 and 3 have angular accelerations so only these links have inertia moments. The inertiamoments are given by

M12 = I2 1 2 = 0.00205(140) = 0.287N m (opposite 1 2 or CCW)M13 = I31 3 = 0.0062(140) = 0.868N m (opposite 1 3 or CW)

The inertia forces are shown in Fig. 14.5.2. The orientation angles for the forces and links areshown in Fig. 14.5.2.

A freebody diagram for each of the links is shown in Fig. 14.5.2. From the free body diagrams,it is clear that no single free body can be analyzed separately because in each case, fourunknowns result. Therefore, we must write the equilibrium equations for each freebody andsolve the equations as a set.

For the freebody diagram for link 4, assume initially that all of the unknown forces are in thepositive x and y directions. Then a negative result will indicate that the forces are in the negativedirection. Note that F14x = 0 because there is no friction between the slider and the frame.Summing forces in the X and Y directions gives

Fx = 0 F34x + f14 = 0 F34x = 22.409N

Fy = 0 F14y + F34

y = 0(3)

Between links 3 and 4,

F43x = F34

x

F43y = F34

y (4)

Now move to the free body diagram for link 3. Summing forces in the X and Y directions gives:

Fx = 0 F23x + F43x + f13cos34.30 = 0 = F23x + F43x +31.38cos34.30 F23x + F43x = 25.92 N

Fy = 0 F23y + F43

y + f13sin34.30 = 0 = F23y + F43

y + 31.38sin34.30 F23y + F43

y = 17.64 N(5)

and summing moments about point A gives

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3

Force Scale

13

C

4

G4f1415˚

2M12

T12

15˚

A

20 N

G3

M13

f

34.3˚

C 4 f14

F34x

F34y

14F

2

M12

T12 F12x

F12y

A

G3 F32x

F32y

A

3G3

M13

C

15˚34.3˚

F23

f13

x

F23y

x

F43y

F43

O2

O2

Fig. 14.5.2: Force diagrams for Problem 14.5

MA = 0 M13 + F43x (ACsin15) + F43y (AC cos15) = 0

= 0.868(1000)+ F43x(100sin15) + F43y (100cos15)

F43x (25.88) + F43y (96.59) = 868

(6)

Now using Eqs. (4),

F23x F34x = 25.85

F23y F34

y = 17.65F34x (25.88) F34

y (96.59) = 868(7)


F32x = F23

x

F32y = F23

y (8)

For link 2, the equilibrium equations are

Fx = 0 F32x + F12x = 0

Fy = 0 F32y + F12

y = 0(9)

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and summing moments about point O2 gives

MO2 = 0 F32x (ADsin15) + F32y (ADcos15) +T12 + M12 = 0

= F32x (100sin15) + F32y (100cos15)+ T12 + 0.287(1000) = 0

F32x (25.88) + F32y (96.59) + T12 = 287

(10)

Now using Eqs. (8),

Fx = 0 F23x + F12x = 0

Fy = 0 F23y + F12

y = 0

MO2 = 0 F23x (25.88) F23y(96.59) + T12 = 287

(11)

Equations (3), (7), and (11) can be written in matrix form for solution. This gives,

0 1 0 0 0 0 0 01 0 1 0 0 0 0 00 1 0 1 0 0 0 00 0 1 0 1 0 0 00 25.88 96.59 0 0 0 0 00 0 0 1 0 1 0 00 0 0 0 1 0 1 00 0 0 25.88 96.59 0 0 1

F14y

F34x

F34y

F23x

F23y

F12x

F12y

T12

=

22.409025.8517.6586800287

(12)

Equation (12) can be easily solved using MATLAB. The results are

F14x = 0 NF14y = 2.98 N

F34x = F43x = 22.41N

F34y = F43

y = 2.98 N

F23x = F32x = 48.26N

F23y = F32

y = 20.63 N

F12x = 48.26 NF12y = 20.63N

T12 = 1030.9 N mm

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Problem 14.6

Find the external torque (T12) that must be applied to link 2 of the mechanism illustrated to driveit at 2=1,800 rad/s CCW and 2=0 rad/s2. Link 2 is in a horizontal position. and it is balancedso that its center of mass is at the pivot O2. The mechanism moves in the horizontal plane, andfriction may be neglected.

W3=0.708 lb IG3 = 0.0154 lb-s2-in

W4=0.780 lb IG4 = 0.0112 lb-s2-in

G3G4

ω2

2

3

4

AO2 = 3.0 inO O2 = 7.0 in

AG = 4.0 inBO4 = 6.0 in

AB = 8.0 in

4

BG4 = 3.0 in

3

O4

G2

O2A

B

T12

Solution:

Velocity Analysis:

1vA2 = 1vA2 /O2 = 1vA3 = 1 2 rA2/O2

1vB3 = 1vB4 = 1vA3 +1vB3/ A3 =1vO4 +1vB4/O4 = 1vB4 /O4 (1)

Now,

1vA3 = 12 rA2/O2 =1800 3 = 5400 in / s

1vB4 /O4 = 1 4 rB4 /O4 1vB4/O4 = 1 4 rB4 /O4 ( to rB4/O4)

1vB3 /A3 = 1 3 rB3/A3 1vB3/A3 = 1 3 rB3/ A3 ( to rB3/ A3)


1vB3 /A3 = 4283.4 in / sand

1vB4 /O4 = 3195.6 in / s

Then,

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13 =

1vB3/ A3rB3/A3

= 4283.48

= 535.4 rad / s CCW

and

14 =

1vB4/O4rB4/O4

= 3195.66

= 532.6 rad / s CCW


1aB3 =1aB4 = 1aA3 + 1aB3/ A3 =1aO4 + 1aB4/O4 = 1aB4 /O4

1aB3 =1aA2/O2r + 1aA2 /O2

t + 1aB3/ A3r + 1aB3 /A3

t = 1aB4 /O4r + 1aB4 /O4

t (2)Now,

1aA2 /O2r =1 2

12 rA2/O2( ) 1a A2/O2

r = 122 rA/O2 =18002 3 = 9,720,000 in / s 2


1aA2 /O2t =1 2 rA2/O2 = 0

1aB3 /A3r = 1 3 1 3 rB3 /A3( ) 1a B3/ A3

r = 1 32 rB3 /A3 = 535.42 8 = 2,293,000 in / s 2

in the direction opposite to rB3/ A3

1aB3/A3t =1 3 rB3 /A3 1aB3/A3

t = 1 3 rB3 /A3 ( to rB3/ A3)

1aB4 /O4r =1 4 1 4 rB4 /O4( ) 1a B4 /O4

r = 1 42 rB4/O4 = 532.62 6 =1,702,000 in / s 2

in the direction opposite to rB4 /O4

1aB4 /O4t =1 4 rB4/O4 1aB4 /O4

t = 1 4 rB4 /O4 ( to rB4 /O4)


1aG3 = 7,700,000 in / s 2

1aG4 = 2,867,000 in / s 2

1aB3 /A3t = 4,151,000 in / s 2

1aB4 /O4t = 5,533,000 in / s 2

Also,

13 =

1aB3/A3t

rB3/ A3=4,151,000

8= 518,900 rad / s2 CCW

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G3G4

ω2

2

3

4

AO2 = 3.0 inO O2 = 7.0 in

AG = 4.0 inBO4 = 6.0 in

AB = 8.0 in

4

BG4 = 3.0 in

3

O4

G2

O2A

B

T12

3000 in/s

Velocity Polygon

a2 a3,

b3, b4

1vB3

1v A3

1vB3/A 3

o

5,000,000 in/sAcceleration Polygon

2

a'2 a'3,1aA2/O2

r

1aB 3/A 3

t

o'

b'

2 b'

3,

B3 /A3ra1

g'4

1aB4 /O4

t1aB4 /O4

r

g'3

Fig. 14.6.1: Position, velocity, and acceleration polygon for Problem 14.6.and

14 =

1aB4/O4t

rB4 /O4=5,533,000

6= 922,200 rad / s2 CW



f12 = m21aG2 = 0

f13 = m3 1aG3 =0.708386

(7,700,000) =14,100 lb (opposite1aG3 )

f14 = m41aG4 =0.780386

(2,867,000) = 5,800 lb (opposite 1aG4 )

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Links 3 and 4 have angular accelerations so these links have inertia moments. The inertiamoments are given by

M13 = I31 3 = 0.0154(518,900) = 7,990 in lb (opposite 1 3 or CW)M14 = I4 1 4 = 0.0112(922,200) = 10,300 in lb (opposite 1 4 or CCW)


A freebody diagram for each of the links is shown in Fig. 14.6.2. From the free body diagrams,it is clear that no single free body can be analyzed separately. Therefore, we must write theequilibrium equations for each freebody and solve the equations as a set.

For the freebody diagram for link 4, assume initially that all of the unknown forces are in thepositive x and y directions. Then a negative result will indicate that the forces are in the negativedirection. Summing forces in the X and Y directions gives

Fx = 0 F14x + F34x f14cos20.19̊ = 0 F14x + F34x = 5800cos20.19̊ = 5,440 lb

Fy = 0 F14y + F34

y f14 sin20.19̊ = 0 F14y + F34

y = 5800sin20.19̊ = 2,000 lb(3)

MO4 = 0 M14 F34x (BO4sin52.55̊ ) F34y (BO4 cos52.55̊ )

+ f14 cos20.19̊ (G4O4sin52.55̊ ) + f14sin20.19̊ (G4O4cos52.55̊ ) = 0

10,300 F34x (6sin52.55̊ ) F34y (6cos52.55̊ )

+ 5800cos20.19̊ (3sin52.55̊ ) + 5800sin20.19̊ (3cos52.55̊ ) = 0

F34x (4.763) F43y(3.648) = 26,920


F43x = F34

x

F43y = F34

y (4)


Fx = 0 F23x + F43x f13cos7.25̊ = 0 = F23x + F43x 14,100cos7.25̊ F23x + F43x = 13,990 lb

Fy = 0 F23y + F43

y f13sin7.25̊ = 0 = F23y + F43

y 14,100sin7.25̊ F23y + F43

y =1,780 lb(5)


MA = 0 M13 F43x (ABsin36.39˚) + F43y (ABcos36.39˚)

+ f13cos7.25̊ (AG3sin36.39̊ ) f13sin7.25̊ (AG3cos36.39̊ ) = 0

= 7,990 F43x (8sin36.39̊ ) + F43y (8cos36.39̊ )

+14,100cos7.25̊ (4sin36.39̊ ) 14,100sin7.25̊ (4cos36.39̊ ) = 0

F43x (4.74) + F43y (6.44) = 19,470

(6)

Now using Eqs. (4),

- 565 -

G3

G4

2

34

O4

G2

O2A

B

T 12

5000 lbForce Scale

M13

f1 3 M14f1 4

G4

4

O4

B

M14f1 4

F34x

F34y

F14x

F14y

T

2 G2 O2A

12

F12x

F12y

F32x

F32y

G3

3

A

B

M13

f1 3

F23x

F23y

x

F43y

F43

36.39˚ 52.55˚

20.19˚7.25˚

Fig. 14.6.2: Force diagrams for Problem 14.6.

F23x F34x =13,990

F23y F34

y =1,780F34x (4.74) F34

y (6.44) = 19,470(7)


F32x = F23

x

F32y = F23

y (8)


Fx = 0 F32x + F12x = 0

Fy = 0 F32y + F12

y = 0(9)

- 566 -


MO2 = 0 F32x (AO2sin0˚)+ F32y (AO2cos0̊ ) +T12 = 0

= F32y (3cos0̊ ) +T12 = 0F32y (3) + T12 = 0

(10)

Now using Eqs. (8),

Fx = 0 F23x + F12x = 0

Fy = 0 F23y + F12

y = 0

MO2 = 0 F23y (3) + T12 = 0

(11)


1 0 1 0 0 0 0 0 00 1 0 1 0 0 0 0 00 0 4.763 3.648 0 0 0 0 00 0 1 0 1 0 0 0 00 0 0 1 0 1 0 0 00 0 4.74 6.44 0 0 0 0 00 0 0 0 1 0 1 0 00 0 0 0 0 1 0 1 00 0 0 0 0 3 0 0 1

F14x

F14y

F34x

F34y

F23x

F23y

F12x

F12y

T12

=

5,4442,00026,92013,9901,78019,470000

(12)


F14x = 3310 lbF14y = 2594 lb

F34x = F43x = 2134 lb

F34y = F43

y = 4594 lb

F23x = F32x =16,124lb

F23y = F32

y = 6374 lb

F12x =16,124 lbF12y = 6374 lb

T12 =19,121 in lb

- 567 -

Problem 14.7

Find the external torque (T12) that must be applied to link 2 of the mechanism illustrated in orderto drive it at 2=210 rad/s CCW and 2=0 rad/s2. Link 2 is balanced so that its center of mass isat the pivot O2. The mechanism moves in the horizontal plane and friction may be neglected.

W3= 3.4 lb IG3 = 0.1085 lb-s2-inW4= 2.86 lb

O2

G2

A

BG3

G42

3

4

AO2 = 3.0 inBG = 6.0 inAB = 12.0 in

3

T12

45˚

ω2

Solution:

Velocity Analysis:

1vA2 = 1vA2 /O2 = 1vA3 = 1 2 rA2 /O2

1vB3 = 1vB4 = 1vA3 +1vB3/ A3 (1)Now,

1vA3 = 12 rA2/O2 = 210 3 = 630 in / s ( to rA2 /O2 )

1vB3 in horizontal direction

1vB3 /A3 =1 3 rB3 /A3 1vB3/A3 = 1 3rB3 /A3 ( to rB3 /A3)


1vB3 = 525 in / s

Also,

1vB3 /A3 = 452.7 in / sor

13 =

1vB3/ A3rB3/A3

= 452.712

= 37.72 rad / s CW


1aB3 =1aB4 = 1aA3 + 1aB3/ A3

1aB3 =1 aA2 /O2r +1 aA2 /O2

t +1 aB3 /A3r +1 aB3/A3

t (2)

- 568 -

Now,1aB3 in horizontal direction

O2

G2

A

BG3

G42

3

4

T12

45˚

ω2

300 in/s

Velocity Polygon

a2 a3,

b3, b4

1vB3

1vA3

1vB3/A 3

o'

50,000 in/sAcceleration Polygon

2

a2'a3',

1aA2/O2r

1aB3

1aB 3 /A 3r

1aB3/A3

t

b3' g4',

o

Fig. 14.7.1: Position, velocity, and acceleration polygon for Problem 14.7.

1aA2 /O2r =1 2 1 2 rA2 /O2( ) 1a A2 /O2

r = 1 22 rA/O2 = 2102 3 =132,300 in / s 2


1aA2 /O2t =1 2 rA2/O2 1aA2/O2

t = 1 2 rA/O2 = 01aB3 /A3

r = 1 3 1 3 rB3 /A3( ) 1a B3/ A3r = 1 3

2 rB3 /A3 = 37.722 12 =17,070 in / s 2

in the direction opposite to rC3/A3

1aB3/A3t =1 3 rB3/ A3 1aB3 /A3

t = 1 3 rB3/A3 ( to rB3 /A3)


1aB3 = 1aG4 = 94,100 in / s 2

and1aB3 /A3t = 91,610 in / s 2

- 569 -

Determine the acceleration of G3 by image. Then,

1aG3 =104,740 in / s 2Also,

1 3 =1aB3/A3t

rB3/A3=91,61012

= 7634 rad / s2 CCW



f12 = m21aG2 = 0

f13 = m3 1aG3 =3.4386

(104,740) = 922.6 lb (opposite 1aG3 )

f14 = m41aG4 =2.86386

(70,028) = 518.9 lb (opposite 1aG4 )

Only links 3 has an angular acceleration so only this link has an inertia moment. The inertiamoment is given by

M13 = I31 3 = 0.1085(7634) = 828.3 in lb (opposite 1 3 or CW)


A freebody diagram for each of the links is shown in Fig. 14.7.2. From the free body diagrams,it is clear that no single free body can be analyzed separately because in each case, fourunknowns result. Therefore, we must write the equilibrium equations for each freebody andsolve the equations as a set.

For the freebody diagram for link 4, assume initially that all of the unknown forces are in thepositive x and y directions. Then a negative result will indicate that the forces are in the negativedirection. Note that F14x = 0 because there is no friction between the slider and the frame.Summing forces in the X and Y directions gives

Fx = 0 F34x + f14 = 0 F34x = 518.9 lb

Fy = 0 F14y + F34

y = 0(3)


F43x = F34

x

F43y = F34

y (4)


- 570 -

M13

B 4 f14

F34x

F34y

14F

F12x

F12y F32

x

F32y

F23 f13x

F23y

x

F43y

F43

BG3

G4

4

O2

G2

A

2

3

T12

45˚

ω2 Force Scale500 lbs

M13

A

B

G33

O2

A

2T12

10.19˚

26.48˚

f13

f14

Fig. 14.7.2: Force diagrams for Problem 14.7.

Fx = 0 F23x + F43x + f13cos26.48 = 0 = F23x + F43x + 922.6cos26.48 F23x + F43x = 825.8 lb

Fy = 0 F23y + F43

y + f13sin26.48 = 0 = F23y + F43

y + 922.6sin26.48 F23y + F43

y = 411.4 lb(5)


MA = 0 M13 + F43x (ABsin10.19) + F43y (ABcos10.19)

+ f13cos26.48(AG3sin10.19) + f13sin26.48(AG3cos10.19) = 0

= 828.3 + F43x (12sin10.19)+ F43y(12cos10.19)

+ 922.6cos26.48(6sin10.19)+ 922.6sin26.48(6cos10.19) = 0F43x (2.123) + F43

y (11.81) = 2477.6

(6)

Now using Eqs. (4),

F23x F34x = 825.8 lb

F23y F34

y = 411.4 lbF34x (2.123) F34

y (11.81) = 2477.6(7)


F32x = F23

x

F32y = F23

y (8)

- 571 -


Fx = 0 F32x + F12x = 0

Fy = 0 F32y + F12

y = 0(9)


MO2 = 0 F32x (AO2sin45) + F32y (AO2 cos45) +T12 = 0

= F32x (3sin45)+ F32y (3cos45) + T12 = 0

F32x (2.121)+ F32y (2.121) +T12 = 0

(10)

Now using Eqs. (8),

Fx = 0 F23x + F12x = 0

Fy = 0 F23y + F12

y = 0

MO2 = 0 F23x (2.121) F23y (2.121)+ T12 = 0

(11)


0 1 0 0 0 0 0 01 0 1 0 0 0 0 00 1 0 1 0 0 0 00 0 1 0 1 0 0 00 2.123 11.81 0 0 0 0 00 0 0 1 0 1 0 00 0 0 0 1 0 1 00 0 0 2.121 2.121 0 0 1

F14y

F34x

F34y

F23x

F23y

F12x

F12y

T12

=

518.90825.8411.42477.6000

(12)


F14x = 0 lbF14y = 303.1 lb

F34x = F43x = 518.9 lb

F34y = F43

y = 303.1lb

F23x = F32x = 1344.7 lb

F23y = F32

y = 108.3 lb

F12x = 1344.7 lbF12y = 108.3 lb

T12 = 2622.3 in lb

- 572 -

Problem 14.8

In the mechanism shown, the center of mass of link 3 is at G3, which is located at the center oflink 3. The mass of link 3 is 0.5 kg. Its moment of inertia about G3 is 0.0012 N-s2-m. Theweights and moments of inertia of members 2 and 4 may be neglected. Link 2 is driven at aconstant angular velocity of 50 rad/s CW by the torque applied to link 2. The mechanism movesin the horizontal plane, and friction may be neglected.

1. Find the magnitudes and directions of the inertia force and inertia torque acting on link 3.

2. Find the magnitudes and directions of the forces exerted on link 3 by link 2 at A and by link 4at B . You may use either a graphical solution or numerical solution of the dynamicequilibrium equations.

ω = 50 rad/s2

oa

b

A

B

2

1

34

G 3

O A = 20 mm AB = 70 mm O B = 60 mmO O = 80 mmA B

A

B

OBOA

90˚

1000 mm/s

513 mm/s756 mm/s

o'

b'

a'

50 m/s 2

8.16 m/s 2

4.37 m/s 2

37.1 m/s 2

36.9 m/s 2

(Velocity Polygon)

(Acceleration Polygon)

T12

- 573 -

Solution:

The basic position, velocity, and acceleration analysis have already been conducted. We needonly determine the acceleration of G3 by image and the angular acceleration of link 3. To dothis, we will redraw the acceleration diagram to scale as shown in Fig. 14.8.1. From thepolygon,

1aG3 = 39.8 m / s2= 39,800 mm / s2

1aB3 /A3t = 37.1 m / s2= 37,100 mm / s2

Also,

13 =

1aB3/A3t

rB3/ A3=37,10070

= 530 rad / s2 CCW

o'

b'

a'

50 m/s2

8.16 m/s 2

4.37 m/s2

37.1 m/s 2

36.9 m/s 2

ω = 50 rad/s2

A

B

2

1

3 4

G3

OBOA

90˚

12.5 m/sAcceleration Polygon

2

g'3

1aB3/A3t

1aG3

39.8 m/s2

Fig. 14.8.1: Position, velocity, and acceleration polygon for Problem 14.8.

- 574 -


We can now conduct the inertia force analysis. There will be an inertia force associated with thecenter of gravity of link 3. We will represent this force with a lower case f to distinguishbetween internal inertia forces and externally applied forces. The force is

f13 = m3 1aG3 = 0.5(37.1) =18.55 kg m / s2 =18.55N (opposite1aG3 )

Links 3 has an angular acceleration so this links has an inertia moment. The inertia moment isgiven by

M13 = I31 3 = 0.0012(530) = 0.636 N m (opposite 1 3 or CW)

The inertia forces are shown in Fig. 14.8.2 along with a freebody diagram of each link. From thefree body diagrams it is clear that this problem can be solved graphically. To do this, we mustreplace the inertia force and inertia moment by offsetting the inertia force by a distance h. Thedistance h is give by

h = M13f13

= 0.63618.55

= 0.0343m = 34.3mm

The solution is shown in Fig. 14.8.2. We know the direction of F14 and F34 because link 4 is atwo-force member. We then know the direction of F43 because

F43 = F34

Using the procedure given in Section 11.5 of the text, we can find the direction of F23 becausethe force F23 must pass through point A and the intersection of f13 (after it is offset by h) andF43 . Knowing the direction of F23 , we can sum forces vectorially on link 3 and determine themagnitudes of F43 and F23 . Then,

F = f13 + F23 + F43 = 0

From the force polygon,

F23=22.2N

and

F43=7.54 N

The directions are given on the force polygon for link 3. Link 2 can now be analyzed for forcesif necessary; however, this is not required for this problem.

- 575 -

A

B

21

34

G3

OBOA

90˚

M13

f 13

Force Scale

10 N

A

2

OA

14

B

4

OB F

F34A

B

3

G 3

90˚

M 13

f 13

F43

F23

F32

F12T 13

h

f 13

F43

F23

Force polygon for link 3

Fig. 14.8.2: Force diagrmas for Problem 14.8.

Problem 14.9

Link AB of the geared five-bar linkage shown drives CCW against a load torque T15 = 25 in-lb. If1 2 = 0.001 rpm CW, find the driving torque T12. The mechanism moves in the horizontal plane,and friction may be neglected. The gears 2 and 3 are represented by their pitch circles. Bothgears turn on bearings supported by the tie link, 4. The weight of link 2 is small and can beneglected. Gear 3 is 0.2 in thick and may be treated as a solid disk.

- 576 -

2

34

5

TT 12

15

r

r

A

B

C D3

E

2

θ 2

AB = 1.0 inBC = 2.0 inCD = 1.0 in

AE = 2.0 inDE = 2.0 in

r = 0.5 inr = 1.5 in23

= 135˚2= 90˚

θABC∠

W = 0.235 lb3

Solution

This problem can be solved as a dynamics problem by performing a velocity, acceleration, andinertia force analysis; however, because of the slow speed of link 2, the accelerations will beextremely small. Therefore, the inertia forces throughout the system will be negligible comparedto the applied forces. As a result, this problem can be treated as a statics problem. Because themechanism operates in the horizontal plane, only the applied torque, T15 needs to be considered.

The problem then can be solved most easily using conservation of power and instant centers.From conservation of power,

T12 1 2 +T15 1 5 = 0 (1)

To use instant centers, we must draw the mechanism to scale. From Eq. (1), power is involved inlinks 2 and 5. Therefore, we I12, I15, and I25. Using the procedure discussed in Chapter 2, theinstant centers are located as shown in Fig. 14.9.1. Because I25 lies outside of both I12 and I15,both 1 2 and 1 5 will be in the same direction. Therefore, for Eq. (1) to be satisfied, the torquesT12 and T15 must be in opposite directions. As a result, T12 will be CW, and we need onlydetermine the magnitude of T12 . From Eq. (1),

T12 = T151 51 2

.

From the relationships for instant centers,

vI25 = 1 5(rI25 /I15) = 1 2(rI25 /I12)and

1 51 2

=rI25/I12rI25/ I15

Therefore,

T12 = T151 51 2

= T15rI25/I12rI25/ I15

.

From Fig. 14.9.1,

rI25/ I15 = 4.819 inand

- 577 -

rI25/ I12 = 2.833 in

Therefore,

T12 = T15rI25/I12rI25/I15

= 25 2.8334.819

=14.70 in lb

andT12 =14.70 in lb CW

A E

B

CD

2

3

4

5

I12 I15

I35

I34

I24

I23

12

3

4

5

I25

Fig. 14.9.1: Instant center locations for Problem 14.9.

Problem 14.10

A punch press similar to that of Example 14.5 is to punch holes of diameter up to 0.75 in throughsteel plate up to 0.375 in thick. The shear strength of the steel will range up to 60,000 psi. Therated speed of the motor is 1500 rpm, and a 10 % drop in motor speed is allowable. If holes areto be punched at a maximum rate of 1 per second, find the requisite motor power and flywheelinertia.

Solution

The maximum punch force is

F = d t

- 578 -

where = 60,000 psi is the shear strength of the material, d is the diameter of the hole, and t isthe plate thickness. That is, the maximum punch force is simply the shear area multiplied by theshear strength. Then,

F = d t = (0.75)(0.375)(60,000) = 53,014 lb

The area under the curve of punch force versus depth of penetration is the energy used in thepunching operation. It can be measured experimentally, but as discussed in the text,

E =Ft

2(14.16)

is a frequently used approximation. Therefore,

E = Ft2=53,014(0.375)

2= 9,940 in lb= 828.3 ft lb

At a punching rate of once every second the average power required is

P =828.3 ft lb / (1)s =828.3 ft lb / s =1.506 hp

This is the power for which the motor should be sized.

Now, if the rated motor speed is 1500 rpm, we can assume that the maximum motor speed willbe this value so

2 =1,500 2

60= 157 rad/s

Also, we can assume that the motor speed quickly drops to its minimum value during the punchstroke and that it is then built back up approximately uniformly to the maximum value in theremainder of the cycle time. Therefore

= 1 + 2

2

is an adequate approximation to the average motor speed, .

Since the allowable motor speed variation is 10%, c = 0.10 and so, applying Eq. (14.13),

0.10 = 2 1 =2 2 1( )

2 + 1or

2 + 1 = 20 2 1( )

and1 = (19 / 21) 2

Substitution of 2 = 157 gives

1 = (19 / 21) 2 = (19 / 21)157 = 142 rad/s

- 579 -

Also, applying Eq. (14.14),

E =Iw2 2

212( )

or

Iw = 2 E22

12( )=

2(828.3)1572 1422( )

= 0.369 ft-lb s2

Problem 14.11

A uniform rectangular plate is suspended from a rail by means of two bogies as shown. Theplate is connected to the bogies by means of frictionless hinge joints at A and at B. At time t=0the pin of joint B breaks, allowing the plate to swing downward. Write the equations of motionof the plate as it starts to move. Hence find its initial angular acceleration and the initialacceleration of point A.

You may assume that the rollers which support point A are frictionless and that they remain incontact with the rail. You may also assume that the angular displacement from the initialposition is small. The moment of inertia of a uniform rectangular plate with sides 2a and 2babout an axis normal to its plane passing through its centroid is m[(2a)2+(2b) 2]/12, where m isthe mass of the plate.

A BG

a a

b

b

Solution

First draw a free-body diagram (FBD) of the bogie. If it is assumed to be weightless, the freebody diagram is as follows:

From this FBD, it is apparent that the forces must all be vertical. Now draw a FBD for the place.

A B

W

FA

G

- 580 -

Summing forces in the y direction,

Fy = FA W =maGy

Summing forces in the x direction,

Fx = 0 =maGx aGx = 0

Summing moments about the center of mass of the plate,

MG = FAa = I =112

m (2a)2 + (2b)2[ ] =13m a2 + b2[ ]

Because the angular velocity is zero, the acceleration relationship is as shown in the following:

AB

G

aA

aG/A

ab r aG/A

aA

aG

In equation form,

aG = aA +aG /A

and

aGx = 0 = aA + aG/ Abr= aA + r

br

aA = b

aGy = aG /Aar= r a

r= a

Combining terms,

FA =W +maGy =W +ma =mg +ma =m(g a )

where g is the acceleration of gravity. From the moment equation

FA =13am a2 +b2[ ]

Then,

13am a2 + b2[ ] = m(g a )

or

- 581 -

13a

a2 + b2[ ] + a =13a

a2 + b2[ ]+ a{ } = gThen

=g

13a

a2 + b2[ ] +a{ }=

3aga2 + b2[ ]+ 3a2{ }

=3ag

4a2 + b2{ }

in the CW direction. Also,

aA = b =3abg

4a2 + b2{ }

From the figure,

aA =3abg4a2 + b2{ }

to the right. Also,

FA =13am a2 +b2[ ] =

13am a2 + b2[ ] 3ag

4a2 +b2{ }=gm a2 + b2[ ]4a2 + b2{ }

solutions to chapter 14 exercise problems - … · - 544 - solutions to chapter 14 exercise...

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