VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter 16.1, Problem 16.39P

A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are μs = 0.50 and μk = 0.40. For P = 3.6 lb, determine (a) whether slipping occurs between the belt and either cylinder, (b) the angular acceleration of each cylinder.

Chapter 16.1, Problem 16.39P, A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force

Fig. P16.39

(a)

Expert Solution
Check Mark
To determine

Find whether slipping occurs between the belt and either cylinder.

Explanation of Solution

The force pulled between cylinders A and B (P) is 3.6lb.

The weight of the cylinder A (WA) is 5lb.

The weight of the cylinder B (WB) is 20lb.

The coefficient of the static friction (μs) is 0.50.

The coefficient of the kinetic friction (μk) is 0.40.

The radius of the cylinder A (rA) is 4in..

The radius of the cylinder B (rB) is 8in..

Calculation:

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

Convert the unit of the radius of the cylinder A (rA):

rA=4in.×1ft12in.=13ft

Convert the unit of the radius of the cylinder B (rB):

rB=8in.×1ft12in.=23ft

Consider that no slipping occurs.

Calculate the acceleration of the belt (abelt):

abelt=rAαA=rBαBabelt=αA3=2αB3

αA3=2αB3αA=2αBαB=αA2

Calculate the mass of the cylinder A (mA):

mA=WAg

Substitute 5lb for WA and 32.2ft/s2 for g.

mA=532.2lbs2/ft

Calculate the mass of the cylinder B (mB):

mB=WBg

Substitute 20lb for WB and 32.2ft/s2 for g.

mB=2032.2lbs2/ft

Calculate the mass moment of inertia of the cylinder A (I¯A):

I¯A=12mArA2

Substitute 532.2lbs2/ft for mA and 13ft for rA.

I¯A=12×532.2×(13)2=8.6266×103lbs2ft

Calculate the mass moment of inertia of the cylinder B (I¯B):

I¯B=12mBrB2

Substitute 2032.2lbs2/ft for mB and 23ft for rB.

I¯B=12×2032.2×(23)2=138.0262×103lbs2ft

Show the free body diagram of the cylinder A as in Figure 1.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 16.1, Problem 16.39P , additional homework tip  1

Here, FA is the horizontal force of the cylinder A and αA is the angular acceleration of the cylinder A.

Refer to Figure 1.

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

MG=IGα(FArA)=I¯AαA

Substitute 13ft for rA, and 8.6266×103lbs2ft for I¯A.

(FA×13)=8.6266×103αA13FA=8.6266×103αAFA=8.6266×103αA×3FA=25.8798×103αA (1)

Show the free body diagram of the cylinder B as in Figure 2.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 16.1, Problem 16.39P , additional homework tip  2

Here, FB is the horizontal force of the cylinder B and αB is the angular acceleration of the cylinder B.

Refer to Figure 2.

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

MG=IGα(FBrB)=I¯BαB

Substitute 23ft for rB, αA2 for αB, and 138.0262×103lbs2ft for I¯B.

(FB×23)=138.0262×103×αA223FB=69.0131×103αAFB=69.0131×103αA×32FB=103.51965×103αA (2)

Show the free body diagram of the belt as in Figure 3.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 16.1, Problem 16.39P , additional homework tip  3

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

Fx=03.6FAFB=0FA+FB=3.6 (3)

Calculate the angular acceleration of the cylinder A (αA):

Substitute 25.8798×103αA for FA and 103.51965×103αA for FB in Equation (3)

25.8798×103αA+103.51965×103αA=3.6129.4125×103αA=3.6αA=3.6129.4125×103αA=27.82rad/s2

Calculate the horizontal force of the cylinder A (FA):

Substitute 27.82rad/s2 for αA in Equation (1).

FA=25.8798×103×27.82=0.72lb

Calculate the horizontal force of the cylinder B (FB):

Substitute 27.82rad/s2 for αA in Equation (2).

FB=103.51965×103×27.82=2.88lb

Calculate the magnitude of the friction force (Fm):

Fm=μsWA

Substitute 5lb for WA and 0.50 for μs.

Fm=0.50×5=2.5lb

The horizontal force of the cylinder B is greater than the magnitude of the friction force (FB>Fm).

Therefore, the slipping occurs between cylinder B and the belt and the slipping not occur between cylinder A and the belt.

(b)

Expert Solution
Check Mark
To determine

Find the angular acceleration of each cylinder (αAandαB).

Answer to Problem 16.39P

The angular acceleration of each cylinder (αAandαB) are 61.8rad/s2_ and 9.66rad/s2_.

Explanation of Solution

The force pulled between cylinders A and B (P) is 3.6lb.

The weight of the cylinder A (WA) is 5lb.

The weight of the cylinder B (WB) is 20lb.

The coefficient of the static friction (μs) is 0.50.

The coefficient of the kinetic friction (μk) is 0.40.

The radius of the cylinder A (rA) is 4in..

The radius of the cylinder B (rB) is 8in..

Calculation:

Refer the part (a).

Consider the slipping occurs at cylinder B.

Therefore, the angular acceleration of the cylinder B is αB12αA

Calculate the horizontal force of the cylinder B (FB):

FB=μkWA

Substitute 5lb for WA and 0.40 for μk.

FB=0.40×5=2lb

Show the free body diagram of the cylinder B as in Figure 4.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 16.1, Problem 16.39P , additional homework tip  4

Here, FB is the horizontal force of the cylinder B and αB is the angular acceleration of the cylinder B.

Refer to Figure 4.

Calculate the angular acceleration of the cylinder B (αB):

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

MG=IGα(FBrB)=I¯BαB

Substitute 2lb for FB, 23ft for rB, and 138.0262×103lbs2ft for I¯B.

(2×23)=138.0262×103×αB43=138.0262×103αBαB=43×138.0262×103αB=9.66rad/s2

Calculate the horizontal force of the cylinder A (FA):

Substitute 2lb for FB in Equation (3).

FA+2=3.6FA=3.62=1.6lb

The horizontal force of the cylinder A is less than the force of the cylinder B due to the static friction (FA<Fm).

There is no slipping between the cylinder A and the belt.

Calculate the angular acceleration of the cylinder A (αA):

Substitute 1.6lb for FA in Equation (1).

1.6=25.8798×103αAαA=1.625.8798×103=61.8lb

Hence, the angular acceleration of each cylinder (αAandαB) are 61.8rad/s2_ and 9.66rad/s2_.

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Chapter 16 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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