Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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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 μ s = 0.50 and μ k = 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.

Expert Solution
Check Mark
To determine

i.

If there is any slipping between the belt and the cylinders and the angular acceleration of two cylinders.

Answer to Problem 16.39P

There is slipping between cylinder b and belt.

Angular acceleration of cylinder a = αB = 9.659 rad/s2

Angular acceleration of cylinder A = αA = 61.824 rad/s2

Explanation of Solution

Given:

weight of Cylinder A, wa = 5 lb radius of cylinder A = 4 in.

weight of cylinder B, wb = 20 lb

radius of cylinder B, rB = 8 in.

Magnitude of Force pulling the belt = 3.6 lb

Coefficient of static friction = 0.5

Coefficient of kinetic friction = 0.4

Concept used:

Condition of non-slipping,

Friction force due to static friction must be greater than the pulling force on each cylinder.

Fs> FA, Fs>FB

Fs = μsN> FAFs = μsN> FB In case of slipping, maximum tangential force between a cylinder and belt is Fs , otherwise it FA and FB, respectively for cylinder A and B

Mass moment of inertia for a disk is given by-

I = 12mr2, where r is the radius of disk 

The tangential force acting on a cylinder will provide the angular acceleration to the cylinder. Therefore,

Summation of moments applied on it = Mass Moment of inertia × angular accelerationΣM = Iα 

Calculation:

Mass of cylinder A =  5lb32.2ft/s2  = 0.15528 lb-s2/ft

Mass of cylinder B =  20lb32.2ft/s2  = 0.62112 lb-s2/ft

Mass moment of inertia of cylinder A =  12mArA2 = 12(0.15528 lb-s2/ft)×(412)2 = 8.6267×10-3lb×s2ft

Mass moment of inertia of cylinder B =  12mBrB2 = 12(0.62112 lb-s2/ft)×(8/12)2 = 0.13803 lb×s2ft

tangential acceleration of cylinder A is equal to the tangential acceleration of cylinder BThe relation between acceleration and angular accelerationa = rαTherefore, at = rAαA = rBαBA = 8αBαA = 2αB

For cylinder A,

ΣMA = IAαAFA×rA= IAαAFA×(4/12 ft) = (8.6267×10-3) lb×ft×s2 × (2αB)FA= 0.05176αB lb

For cylinder B,

ΣMB = IBαBFB×rB= IBαB

FB×(8/12 ft) = (0.13803) lb×ft×s2 × (αB)FB = 0.20705αB

the net force on the belt should be summation of P, force by cylinder B and force by cylinder C, all equals to 0.

ΣF = P- FA-FB = 03.6 lb - FA-FB = 0 3.6lb - 0.05176αB - 0.20705αB = 0αB = 13.91 rad/s2αA = 27.8196 rad/s2FA = 0.05176 αB = 0.72 lbFB = 0.20705αB = 2.88 lb  

Maximum friction force to avoid slipping Fs = μsN = 0.5×5lb = 2.5 lb

Condition of slipping is μsN> FB

μsN`= 2.5 lb,FB = 2.88 lbThe condition of non-slipping is not satisfied. Therefore slipping occurs between the belt and the cylinder B

Conclusion:

There is slipping between cylinder b and belt.

Expert Solution
Check Mark
To determine

ii.

As there sliping, find the angular acceleration of two cylinders.

Answer to Problem 16.39P

Angular acceleration of cylinder a = αB = 9.659 rad/s2

Angular acceleration of cylinder A = αA = 61.824 rad/s2

Explanation of Solution

Given:

weight of Cylinder A, wa = 5 lb radius of cylinder A = 4 in.

weight of cylinder B, wb = 20 lb

radius of cylinder B, rB = 8 in.

Magnitude of Force pulling the belt = 3.6 lb

Coefficient of static friction = 0.5

Coefficient of kinetic friction = 0.4

Concept used:

Condition of non-slipping,

Friction force due to static friction must be greater than the pulling force on each cylinder.

Fs> FA, Fs>FB

Fs = μsN> FAFs = μsN> FB In case of slipping, maximum tangential force between a cylinder and belt is Fs , otherwise it FA and FB, respectively for cylinder A and B

Mass moment of inertia for a disk is given by-

I = 12mr2, where r is the radius of disk 

The tangential force acting on a cylinder will provide the angular acceleration to the cylinder. Therefore,

Summation of moments applied on it = Mass Moment of inertia × angular accelerationΣM = Iα 

Calculation:

Mass of cylinder A =  5lb32.2ft/s2  = 0.15528 lb-s2/ft

Mass of cylinder B =  20lb32.2ft/s2  = 0.62112 lb-s2/ft

Mass moment of inertia of cylinder A =  12mArA2 = 12(0.15528 lb-s2/ft)×(412)2 = 8.6267×10-3lb×s2ft

Mass moment of inertia of cylinder B =  12mBrB2 = 12(0.62112 lb-s2/ft)×(8/12)2 = 0.13803 lb×s2ft

Assuming slipping,

coefficient of kinetic friction is applicable μk= 0.4Fs = FB= μkN= 0.4×5 lb  = 2 lb

For cylinder B,

ΣMB = IBαBFB×rB= IBαB2×(812)=  (0.13803) lb×ft×s2 × (αB)αB = 9.659 rad/s2

Next, for the belt,

ΣF = P- FA-FB = 03.6 lb - FA- 2lb = 0  FA = 1.6 lb

Since  FA <  Fs

There is no slipping on cylinder A

ΣMA = IaαAFA×rA= IAαA1.6lb ×(4/12 ft) = (8.6267×10-3) lb×ft×s2 × (αA)αA = 61.824 rad/s2

Conclusion:

Angular acceleration of cylinder a = αB = 9.659 rad/s2

Angular acceleration of cylinder A = αA = 61.824 rad/s2

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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.
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Chapter 16 Solutions

Vector Mechanics For Engineers

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