The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.25, determine the largest angle 6 for which the assembly will remain at rest.

Question

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Answer 1

Textbook Question

Chapter 7, Problem 7.4P

The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.25, determine the largest angle 6 for which the assembly will remain at rest.

Chapter 7, Problem 7.4P, The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical

Expert Solution & Answer

To determine

Largest angle θ for which the assembly will remain at rest

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$ .

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 1

Co-efficient of static friction between each bar and wall is $0.25$ .

For impending sliding, the friction force equals its limiting value;

$F=Fmax=μsN$

$μs$ - 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:

FBD of assembly

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 2

Assume $NA,NC$ as the reaction force at point A and C.

Assume $FA,FC$ as the friction force at point A and C.

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

$∑MA=0$

$FC((48 in)cosθ)−(8 lb)((12 in)cosθ)−(10 lb)((36 in)cosθ)=0$

Solve

$48FC−96−360=0 FC=9.5 lb$

Write equilibrium equation in vertical direction.

$↑∑Fy=0$

$FA+FC−(8 lb)−(10 lb)=0 FA=8.5 lb$

Write equilibrium equation in horizontal direction.

$→∑Fx=0$

$NA−NC=0 NA=NC$

The impending sliding will first occur at point C.

$FCNC>FANA$

The reaction force $NC$ at point C

$NC=FC0.25 =9.5 lb0.25 =38 lb$

FBD of bar BC

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 3

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

$∑MB=0$

$FC(( 24 in)cosθ)−NC(( 24 in)sinθ)−(10 lb)(( 12 in)cosθ)=0 9.5(24cosθ)−38(24sinθ)−10(12cosθ)=0 108cosθ−912sinθ=0$

Therefore

$θ=tan−1( 108 912) =6.75°$

Conclusion:

The largest angle θ is equal to $6.75°$ .

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Answer 2

Textbook Question

Chapter 7, Problem 7.4P

The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.25, determine the largest angle 6 for which the assembly will remain at rest.

Chapter 7, Problem 7.4P, The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical

Expert Solution & Answer

To determine

Largest angle θ for which the assembly will remain at rest

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$ .

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 1

Co-efficient of static friction between each bar and wall is $0.25$ .

For impending sliding, the friction force equals its limiting value;

$F=Fmax=μsN$

$μs$ - 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:

FBD of assembly

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 2

Assume $NA,NC$ as the reaction force at point A and C.

Assume $FA,FC$ as the friction force at point A and C.

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

$∑MA=0$

$FC((48 in)cosθ)−(8 lb)((12 in)cosθ)−(10 lb)((36 in)cosθ)=0$

Solve

$48FC−96−360=0 FC=9.5 lb$

Write equilibrium equation in vertical direction.

$↑∑Fy=0$

$FA+FC−(8 lb)−(10 lb)=0 FA=8.5 lb$

Write equilibrium equation in horizontal direction.

$→∑Fx=0$

$NA−NC=0 NA=NC$

The impending sliding will first occur at point C.

$FCNC>FANA$

The reaction force $NC$ at point C

$NC=FC0.25 =9.5 lb0.25 =38 lb$

FBD of bar BC

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 3

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

$∑MB=0$

$FC(( 24 in)cosθ)−NC(( 24 in)sinθ)−(10 lb)(( 12 in)cosθ)=0 9.5(24cosθ)−38(24sinθ)−10(12cosθ)=0 108cosθ−912sinθ=0$

Therefore

$θ=tan−1( 108 912) =6.75°$

Conclusion:

The largest angle θ is equal to $6.75°$ .

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Answer 3

Textbook Question

Chapter 7, Problem 7.4P

The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.25, determine the largest angle 6 for which the assembly will remain at rest.

Chapter 7, Problem 7.4P, The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical

Expert Solution & Answer

To determine

Largest angle θ for which the assembly will remain at rest

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$ .

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 1

Co-efficient of static friction between each bar and wall is $0.25$ .

For impending sliding, the friction force equals its limiting value;

$F=Fmax=μsN$

$μs$ - 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:

FBD of assembly

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 2

Assume $NA,NC$ as the reaction force at point A and C.

Assume $FA,FC$ as the friction force at point A and C.

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

$∑MA=0$

$FC((48 in)cosθ)−(8 lb)((12 in)cosθ)−(10 lb)((36 in)cosθ)=0$

Solve

$48FC−96−360=0 FC=9.5 lb$

Write equilibrium equation in vertical direction.

$↑∑Fy=0$

$FA+FC−(8 lb)−(10 lb)=0 FA=8.5 lb$

Write equilibrium equation in horizontal direction.

$→∑Fx=0$

$NA−NC=0 NA=NC$

The impending sliding will first occur at point C.

$FCNC>FANA$

The reaction force $NC$ at point C

$NC=FC0.25 =9.5 lb0.25 =38 lb$

FBD of bar BC

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 3

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

$∑MB=0$

$FC(( 24 in)cosθ)−NC(( 24 in)sinθ)−(10 lb)(( 12 in)cosθ)=0 9.5(24cosθ)−38(24sinθ)−10(12cosθ)=0 108cosθ−912sinθ=0$

Therefore

$θ=tan−1( 108 912) =6.75°$

Conclusion:

The largest angle θ is equal to $6.75°$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 7, Problem 7.4P

The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.25, determine the largest angle 6 for which the assembly will remain at rest.

Chapter 7, Problem 7.4P, The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical

Expert Solution & Answer

To determine

Largest angle θ for which the assembly will remain at rest

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$ .

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 1

Co-efficient of static friction between each bar and wall is $0.25$ .

For impending sliding, the friction force equals its limiting value;

$F=Fmax=μsN$

$μs$ - 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:

FBD of assembly

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 2

Assume $NA,NC$ as the reaction force at point A and C.

Assume $FA,FC$ as the friction force at point A and C.

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

$∑MA=0$

$FC((48 in)cosθ)−(8 lb)((12 in)cosθ)−(10 lb)((36 in)cosθ)=0$

Solve

$48FC−96−360=0 FC=9.5 lb$

Write equilibrium equation in vertical direction.

$↑∑Fy=0$

$FA+FC−(8 lb)−(10 lb)=0 FA=8.5 lb$

Write equilibrium equation in horizontal direction.

$→∑Fx=0$

$NA−NC=0 NA=NC$

The impending sliding will first occur at point C.

$FCNC>FANA$

The reaction force $NC$ at point C

$NC=FC0.25 =9.5 lb0.25 =38 lb$

FBD of bar BC

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 3

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

$∑MB=0$

$FC(( 24 in)cosθ)−NC(( 24 in)sinθ)−(10 lb)(( 12 in)cosθ)=0 9.5(24cosθ)−38(24sinθ)−10(12cosθ)=0 108cosθ−912sinθ=0$

Therefore

$θ=tan−1( 108 912) =6.75°$

Conclusion:

The largest angle θ is equal to $6.75°$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Largest angle θ for which the assembly will remain at rest

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$ .

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 1

Co-efficient of static friction between each bar and wall is $0.25$ .

For impending sliding, the friction force equals its limiting value;

$F=Fmax=μsN$

$μs$ - 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:

FBD of assembly

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 2

Assume $NA,NC$ as the reaction force at point A and C.

Assume $FA,FC$ as the friction force at point A and C.

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

$∑MA=0$

$FC((48 in)cosθ)−(8 lb)((12 in)cosθ)−(10 lb)((36 in)cosθ)=0$

Solve

$48FC−96−360=0 FC=9.5 lb$

Write equilibrium equation in vertical direction.

$↑∑Fy=0$

$FA+FC−(8 lb)−(10 lb)=0 FA=8.5 lb$

Write equilibrium equation in horizontal direction.

$→∑Fx=0$

$NA−NC=0 NA=NC$

The impending sliding will first occur at point C.

$FCNC>FANA$

The reaction force $NC$ at point C

$NC=FC0.25 =9.5 lb0.25 =38 lb$

FBD of bar BC

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 3

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

$∑MB=0$

$FC(( 24 in)cosθ)−NC(( 24 in)sinθ)−(10 lb)(( 12 in)cosθ)=0 9.5(24cosθ)−38(24sinθ)−10(12cosθ)=0 108cosθ−912sinθ=0$

Therefore

$θ=tan−1( 108 912) =6.75°$

Conclusion:

The largest angle θ is equal to $6.75°$ .

Answer 8

Expert Solution & Answer

To determine

Largest angle θ for which the assembly will remain at rest

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$ .

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 1

Co-efficient of static friction between each bar and wall is $0.25$ .

For impending sliding, the friction force equals its limiting value;

$F=Fmax=μsN$

$μs$ - 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:

FBD of assembly

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 2

Assume $NA,NC$ as the reaction force at point A and C.

Assume $FA,FC$ as the friction force at point A and C.

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

$∑MA=0$

$FC((48 in)cosθ)−(8 lb)((12 in)cosθ)−(10 lb)((36 in)cosθ)=0$

Solve

$48FC−96−360=0 FC=9.5 lb$

Write equilibrium equation in vertical direction.

$↑∑Fy=0$

$FA+FC−(8 lb)−(10 lb)=0 FA=8.5 lb$

Write equilibrium equation in horizontal direction.

$→∑Fx=0$

$NA−NC=0 NA=NC$

The impending sliding will first occur at point C.

$FCNC>FANA$

The reaction force $NC$ at point C

$NC=FC0.25 =9.5 lb0.25 =38 lb$

FBD of bar BC

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 3

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

$∑MB=0$

$FC(( 24 in)cosθ)−NC(( 24 in)sinθ)−(10 lb)(( 12 in)cosθ)=0 9.5(24cosθ)−38(24sinθ)−10(12cosθ)=0 108cosθ−912sinθ=0$

Therefore

$θ=tan−1( 108 912) =6.75°$

Conclusion:

The largest angle θ is equal to $6.75°$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Largest angle θ for which the assembly will remain at rest

Answer 12

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$ .

Answer 13

Explanation of Solution

Given information:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 1

Co-efficient of static friction between each bar and wall is $0.25$ .

For impending sliding, the friction force equals its limiting value;

$F=Fmax=μsN$

$μs$ - 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:

FBD of assembly

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 2

Assume $NA,NC$ as the reaction force at point A and C.

Assume $FA,FC$ as the friction force at point A and C.

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

$∑MA=0$

$FC((48 in)cosθ)−(8 lb)((12 in)cosθ)−(10 lb)((36 in)cosθ)=0$

Solve

$48FC−96−360=0 FC=9.5 lb$

Write equilibrium equation in vertical direction.

$↑∑Fy=0$

$FA+FC−(8 lb)−(10 lb)=0 FA=8.5 lb$

Write equilibrium equation in horizontal direction.

$→∑Fx=0$

$NA−NC=0 NA=NC$

The impending sliding will first occur at point C.

$FCNC>FANA$

The reaction force $NC$ at point C

$NC=FC0.25 =9.5 lb0.25 =38 lb$

FBD of bar BC

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 7, Problem 7.4P , additional homework tip 3

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

$∑MB=0$

$FC(( 24 in)cosθ)−NC(( 24 in)sinθ)−(10 lb)(( 12 in)cosθ)=0 9.5(24cosθ)−38(24sinθ)−10(12cosθ)=0 108cosθ−912sinθ=0$

Therefore

$θ=tan−1( 108 912) =6.75°$

Conclusion:

The largest angle θ is equal to $6.75°$ .

Answer 14

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The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.25, determine the largest angle 6 for which the assembly will remain at rest.

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Answer to Problem 7.4P

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