International Edition---engineering Mechanics: Statics, 4th Edition
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN: 9781305501607
Author: Andrew Pytel And Jaan Kiusalaas
Publisher: CENGAGE L
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Chapter 7, Problem 7.3P

Two identical chairs, each weighing 14 lb, are stacked as shown. The center of gravity of each chair is denoted by G. The coefficient of static friction is 0.2 at B (the contact point between the chairs) and 0.35 at A, C, and D. Determine the smallest force P that would cause sliding.

Chapter 7, Problem 7.3P, Two identical chairs, each weighing 14 lb, are stacked as shown. The center of gravity of each chair

Expert Solution & Answer
Check Mark
To determine

Smallest force 'P' that would cause sliding.

Answer to Problem 7.3P

Smallest force 'P' that would cause sliding is 9.52lb.

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.3P , additional homework tip  1

Weight of each chair is 14lb.

Co-efficient of static friction at point B is 0.2.

Co-efficient of static friction at point A,C and D is 0.35.

If there is no relative motion between two surfaces that are in contact, The relationship between normal force N and friction force F would be:

  F=μsN

  μs - Co-efficient of static friction

Steps to follow in the equilibrium analysis of a body are:

1. Draw the free body diagram.

2. Write the equilibrium equations.

3. Solve the equations for the unknowns.

Calculation:

Assume impending sliding at point B, C and D.

Assume FB,FC,FD as the friction force at point B, C and D.

Assume NB,NC,ND as the reaction force at point B, C and D.

  FB=0.2NBFC=0.35NCFD=0.35ND

FBD of left chair

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.3P , additional homework tip  2

For the equilibrium of above section, the bending moment about point A is equal to zero.

  MA=0

  0.2NB(20in)NB(24in)+(14lb)(12in)=0NB=8.4lb

FBD of right chair

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.3P , additional homework tip  3

Write equilibrium equation in vertical direction.

  Fy=0

  NC+NDNB14lb=0NC+ND8.4lb14lb=0NC+ND=22.4lb

Write equilibrium equation in horizontal direction.

  Fx=0

  P0.35(NC+ND)0.2NB=0P0.35(22.4lb)0.2(8.4lb)=0

Solve

  P=9.52lb

Check for sliding at point A.

FBD of left chair

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.3P , additional homework tip  4

Write equilibrium equation in vertical direction.

  Fy=0

  NA+NB14lb=0NA+8.4lb14lb=0NA=5.6lb

Write equilibrium equation in horizontal direction.

  Fx=0

  FA0.2NB=0FA=0.2(8.4lb)FA=1.68lb

Coefficient friction μA at point A

  μA=FANA=1.68lb5.6lb=0.3<(μA)S

Conclusion:

Smallest force 'P' that would cause sliding is 9.52lb.

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Chapter 7 Solutions

International Edition---engineering Mechanics: Statics, 4th Edition

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