Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 16.1, Problem 16.34P

Each of the double pulleys shown has a mass moment of inertia of 15 lb·ft·s2 and is initially at rest. The outside radius is 18 in., and the inner radius is 9 in. Determine (a) the angular acceleration of each pulley, (b) the angular velocity of each pulley after point A on the cord has moved 10 ft.

Chapter 16.1, Problem 16.34P, Each of the double pulleys shown has a mass moment of inertia of 15 lbfts2 and is initially at rest.

Fig. P16.34

(a)

Expert Solution
Check Mark
To determine

Find the angular acceleration of the pulley 1 (α1).

Find the angular acceleration of the pulley 2 (α2).

Find the angular acceleration of the pulley 3 (α3).

Find the angular acceleration of the pulley 4 (α4).

Answer to Problem 16.34P

The angular acceleration of the pulley 1 (α1) is 8.00rad/s2_.

The angular acceleration of the pulley 2 (α2) is 6.74rad/s2_.

The angular acceleration of the pulley 3 (α3) is 4.24rad/s2_.

The angular acceleration of the pulley 4 (α4) is 5.83rad/s2_.

Explanation of Solution

The mass moment of inertia of the double pulleys (IO) is 15lbfts2.

The outside radius of the pulley (R) is 18in..

The inner radius of the pulley (r) is 9in..

The finial angular velocity of the pulley (ω) is 0.

The load of the pulley 1 (W1) is 160lb.

The load of the pulley 2 (W2) is 160lb.

The left side load of the pulley 3 (W3) is 460lb.

The right side load of the pulley 3 (W4) is 300lb.

The load of the pulley 4 (W5) is 80lb.

Calculation:

Consider the acceleration due to gravity (g) is 32.2ft/s2.

Case 1:

Convert the unit of the outside radius (R):

R=18in.×1ft12in.R=1.5ft

Convert the unit of the inner radius (r):

r=9in.×1ft12in.r=0.75ft

Show the free body diagram of the double pulley 1 as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.34P , additional homework tip  1

Here, α is the angular acceleration of the pulley.

Refer to Figure 1.

Calculate the angular acceleration of the pulley 1 (α1):

Calculate the moment about point O by applying the equation of equilibrium:

MO=IOα(160×0.75)=15α1α1=160×0.7515α1=8rad/s2

Hence, the angular acceleration of the pulley 1 (α1) is 8.00rad/s2_.

Case 2:

Calculate the mass of the pulley 2 (m2):

m2=W2g

Substitute 160lb for W2 and 32.2ft/s2 for g.

m2=16032.2=4.9689lbs2/ft

Show the free body diagram of the double pulley 2 as in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.34P , additional homework tip  2

Refer to Figure 2.

Calculate the moment about point O by applying the equation of equilibrium:

MO=IOα(160×0.75)=15α2+(m2×0.75α2×0.75)120=15α2+0.5625m2α2120=(15+0.5625m2)α2 (1)

Calculate the angular acceleration of the pulley 2 (α2):

Substitute 4.9689lbs2/ft for m2 in Equation (1).

120=[15+(0.5625×4.9689)]α217.795α2=120α2=12017.795α2=6.74rad/s2

Hence, the angular acceleration of the pulley 2 (α2) is 6.74rad/s2_.

Case 3:

Calculate the left side mass of the pulley 3 (m3):

m3=W3g

Substitute 460lb for W3 and 32.2ft/s2 for g.

m3=46032.2=14.2857lbs2/ft

Calculate the right side mass of the pulley 3 (m4):

m4=W4g

Substitute 300lb for W4 and 32.2ft/s2 for g.

m4=30032.2=9.3168lbs2/ft

Show the free body diagram of the double pulley 3 as in Figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.34P , additional homework tip  3

Refer to Figure 3.

Calculate the moment about point O by applying the equation of equilibrium:

MO=IOα(460×0.75)(300×0.75)=[15α3+(m3×0.75α3×0.75)+(m4×0.75α3×0.75)]120=15α3+0.5625m3α3+0.5625m4α3120=(15+0.5625m3+0.5625m4)α3 (2)

Calculate the angular acceleration of the pulley 3 (α3):

Substitute 14.2857lbs2/ft for m3 and 9.3168lbs2/ft for m4 in Equation (2).

120=[15+(0.5625×14.2857)+(0.5625×9.3168)]α328.2764α3=120α3=12028.2764α3=4.24rad/s2

Hence, the angular acceleration of the pulley 3 (α3) is 4.24rad/s2_.

Case 4:

Calculate the left side mass of the pulley 4 (m5):

m5=W5g

Substitute 80lb for W5 and 32.2ft/s2 for g.

m5=8032.2=2.4845lbs2/ft

Show the free body diagram of the double pulley 4 as in Figure 4.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.34P , additional homework tip  4

Refer to Figure 4.

Calculate the moment about point O by applying the equation of equilibrium:

MO=IOα(80×1.5)=15α4+(m5×1.5α4×1.5)120=15α3+2.25m5α4120=(15+2.25m5)α3 (3)

Calculate the angular acceleration of the pulley 4 (α4):

Substitute 2.4845lbs2/ft for m5 in Equation (3).

120=[15+(2.25×2.4845)]α420.59α4=120α4=12020.59α4=5.83rad/s2

Hence, the angular acceleration of the pulley 4 (α4) is 5.83rad/s2_.

(b)

Expert Solution
Check Mark
To determine

Find the angular velocity of the pulley 1 (ω1).

Find the angular velocity of the pulley 2 (ω2).

Find the angular velocity of the pulley 3 (ω3).

Find the angular velocity of the pulley 4 (ω4).

Answer to Problem 16.34P

The angular velocity of the pulley 1 (ω1) is 14.61rad/s_.

The angular velocity of the pulley 2 (α2) is 13.41rad/s_.

The angular velocity of the pulley 3 (α3) is 10.64rad/s_.

The angular velocity of the pulley 4 (α4) is 8.82rad/s_.

Explanation of Solution

The mass moment of inertia of the double pulleys (IO) is 15lbfts2.

The outside radius of the pulley (R) is 18in..

The inner radius of the pulley (r) is 9in..

The finial angular velocity of the pulley (ω) is 0.

The load of the pulley 1 (W1) is 160lb.

The load of the pulley 2 (W2) is 160lb.

The left side load of the pulley 3 (W3) is 460lb.

The right side load of the pulley 3 (W4) is 300lb.

The load of the pulley 4 (W5) is 80lb.

The moved distance of the point A (l) is 10ft.

Calculation:

Refer part (a).

Case 1:

Calculate the angle of the pulley 1 (θ1):

θ1=lr

Substitute 10ft for l and 0.75ft for r.

θ1=100.75=13.333rad

Calculate the angular velocity of the pulley 1 (ω1):

ω12=2α1θ1

Substitute 8rad/s2 for α1 and 13.333rad for θ1

ω12=2×8×13.333ω1=213.328ω1=14.61rad/s

Hence, the angular velocity of the pulley 1 (ω1) is 14.61rad/s_.

Case 2:

Calculate the angle of the pulley 2 (θ2):

θ2=lr

Substitute 10ft for l and 0.75ft for r.

θ2=100.75=13.333rad

Calculate the angular velocity of the pulley 2 (ω2):

ω22=2α2θ2

Substitute 6.74rad/s2 for α2 and 13.333rad for θ2

ω22=2×6.74×13.333ω2=179.72884ω2=13.41rad/s

Hence, the angular velocity of the pulley 2 (ω2) is 13.41rad/s_.

Case 3:

Calculate the angle of the pulley 3 (θ3):

θ3=lr

Substitute 10ft for l and 0.75ft for r.

θ3=100.75=13.333rad

Calculate the angular velocity of the pulley 3 (ω3):

ω32=2α3θ3

Substitute 4.24rad/s2 for α3 and 13.333rad for θ3

ω32=2×4.24×13.333ω3=113.06384ω3=10.64rad/s

Hence, the angular velocity of the pulley 3 (ω3) is 10.64rad/s_.

Case 4:

Calculate the angle of the pulley 4 (θ4):

θ4=lR

Substitute 10ft for l and 1.5ft for R.

θ4=101.5=6.6667rad

Calculate the angular velocity of the pulley 4 (ω4):

ω32=2α3θ3

Substitute 5.83rad/s2 for α4 and 6.6667rad for θ4

ω42=2×5.83×6.6667ω4=77.733722ω4=8.82rad/s

Hence, the angular velocity of the pulley 4 (ω4) is 8.82rad/s_.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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