The free-body diagram of the forces acting on the ladder.

Question

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Answer 1

Question

Chapter 14, Problem 82PQ

(a)

To determine

The free-body diagram of the forces acting on the ladder.

(a)

Expert Solution

Answer to Problem 82PQ

The free-body diagram of the forces acting on the ladder is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 1

Explanation of Solution

A free-body diagram is a graphical tool used to illustrate the different forces acting on a particular object. It helps to solve complex physical problems. The free-body diagram of the ladder in the given situation is drawn in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 2

The forces acting on the ladder are the weight, the normal force, the tension and the force of static friction. In the figure weight is represented as $Fg$ , normal force is denoted as $FN$ , tension is denoted as $FT$ and the force of static friction is denoted by $Fs$ .

Conclusion:

Thus, the free-body diagram of the forces acting on the ladder is drawn in figure 1.

(b)

To determine

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium.

(b)

Expert Solution

Answer to Problem 82PQ

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Explanation of Solution

Take the lower end of the ladder as the pivot point. This will eliminate the torque due to normal force and the torque due to force of static friction.

Since the ladder is in rotational equilibrium, the net torque about the lower end of the ladder must be zero.

Write the condition for the rotational equilibrium.

$∑τlower end=0$ (I)

Here, $∑τlower end$ is the magnitude of the net torque about the lower end of the ladder.

Write the equation for $∑τlower end$ .

$∑τlower end=τN+τs+τg+τT$

Here, $τN$ is the magnitude torque due to the normal force about the lower end of the ladder, $τs$ is the magnitude torque due to the static friction about the lower end of the ladder, $τg$ is the magnitude torque due to the weight about the lower end of the ladder and $τT$ is the magnitude torque due to the tension about the lower end of the ladder.

Put the above equation in equation (I).

$τN+τs+τg+τT=0$ (II)

Write the expression for $τN$ .

$τN=0$

Write the expression for $τs$ .

$τs=0$

Write the expression for $τg$ .

$τg=−Mg(L2cosθ)$

Here, $M$ is the mass of the ladder, $g$ is the acceleration due to gravity, $L$ is the length of the ladder and $θ$ is the angle the ladder makes with the floor.

Write the expression for $τT$ .

$τT=FT(Lsinθ)$

Here, $FT$ is the magnitude of the tension in the rope.

Put the above four equations in equation (II) and rewrite it for $FT$ .

$0+0+(−Mg(L2cosθ))+FT(Lsinθ)=0FTLsinθ=MgL2cosθFT=Mg2cosθsinθ=Mg2cotθ$ (III)

Conclusion:

Therefore, the tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

(c)

To determine

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium.

(c)

Expert Solution

Answer to Problem 82PQ

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Explanation of Solution

Since the ladder is in translational equilibrium, the net force in $x$ and $y$ directions must be zero.

Write the conditions for the translational equilibrium.

$∑Fx=0$ (IV)

Here, $∑Fx$ is the net force in $x$ direction.

$∑Fy=0$ (V)

Here, $∑Fy$ is the net force in $y$ direction.

Write the equation for $∑Fx$ .

$∑Fx=−FT+Fs$ (VI)

Here, $Fs$ is the force of static friction.

Write the equation for $Fs$ .

$Fs=μsFN$

Here, $μs$ is the coefficient of static friction and $FN$ is the normal force.

Put the above equation in equation (VI).

$∑Fx=−FT+μsFN$

Put the above equation in equation (IV) and rewrite it for $FT$ .

$−FT+μsFN=0FT=μsFN$ (VII)

Write the equation for $∑Fy$ .

$∑Fy=FN−Fg$ (VIII)

Here, $Fg$ is the weight of the ladder.

Write the equation for $Fg$ .

$Fg=Mg$

Put the above equation in equation (VIII).

$∑Fy=FN−Mg$

Put the above equation in equation (V) and rewrite it for $FN$ .

$FN−Mg=0FN=Mg$

Put the above equation in equation (VII).

$FT=μsMg$ (IX)

Conclusion:

Therefore, the expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

(d)

To determine

The coefficient of static friction in terms of the angle $θ$ .

(d)

Expert Solution

Answer to Problem 82PQ

The coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Explanation of Solution

Equate equations (III) and (IX).

$Mg2cotθ=μsMgμs=cotθ2$

Conclusion:

Therefore, the coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

(e)

To determine

The after effect of moving the ladder slightly so as to reduce the angle $θ$ .

(e)

Expert Solution

Answer to Problem 82PQ

The ladder will slip if it is moved slightly to reduce the angle $θ$ .

Explanation of Solution

The expression for the angle $θ$ obtained in part (d), $cotθ2$ is valid only at the critical angle $θ$ . The critical angle is the angle when the ladder is on the verge of slipping and the frictional force has its maximum possible value. The angle $θ$ will be reduced when the ladder is moved to the left.

The expression for the tension force obtained in part (b), $FT=Mg2cotθ$ shows that as the value of $θ$ decreases, the tension force will also increases. For the ladder to be in translational equilibrium, the net force in the $x$ direction must be zero. Increasing the tension force will require a larger friction force to balance this condition. But the static friction force is already at its maximum value. This implies, when the ladder is slightly moved to reduce the angle $θ$ , the ladder will slip.

Conclusion:

Thus, the ladder will slip if it is moved slightly to reduce the angle $θ$ .

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Answer 2

Question

Chapter 14, Problem 82PQ

(a)

To determine

The free-body diagram of the forces acting on the ladder.

(a)

Expert Solution

Answer to Problem 82PQ

The free-body diagram of the forces acting on the ladder is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 1

Explanation of Solution

A free-body diagram is a graphical tool used to illustrate the different forces acting on a particular object. It helps to solve complex physical problems. The free-body diagram of the ladder in the given situation is drawn in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 2

The forces acting on the ladder are the weight, the normal force, the tension and the force of static friction. In the figure weight is represented as $Fg$ , normal force is denoted as $FN$ , tension is denoted as $FT$ and the force of static friction is denoted by $Fs$ .

Conclusion:

Thus, the free-body diagram of the forces acting on the ladder is drawn in figure 1.

(b)

To determine

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium.

(b)

Expert Solution

Answer to Problem 82PQ

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Explanation of Solution

Take the lower end of the ladder as the pivot point. This will eliminate the torque due to normal force and the torque due to force of static friction.

Since the ladder is in rotational equilibrium, the net torque about the lower end of the ladder must be zero.

Write the condition for the rotational equilibrium.

$∑τlower end=0$ (I)

Here, $∑τlower end$ is the magnitude of the net torque about the lower end of the ladder.

Write the equation for $∑τlower end$ .

$∑τlower end=τN+τs+τg+τT$

Here, $τN$ is the magnitude torque due to the normal force about the lower end of the ladder, $τs$ is the magnitude torque due to the static friction about the lower end of the ladder, $τg$ is the magnitude torque due to the weight about the lower end of the ladder and $τT$ is the magnitude torque due to the tension about the lower end of the ladder.

Put the above equation in equation (I).

$τN+τs+τg+τT=0$ (II)

Write the expression for $τN$ .

$τN=0$

Write the expression for $τs$ .

$τs=0$

Write the expression for $τg$ .

$τg=−Mg(L2cosθ)$

Here, $M$ is the mass of the ladder, $g$ is the acceleration due to gravity, $L$ is the length of the ladder and $θ$ is the angle the ladder makes with the floor.

Write the expression for $τT$ .

$τT=FT(Lsinθ)$

Here, $FT$ is the magnitude of the tension in the rope.

Put the above four equations in equation (II) and rewrite it for $FT$ .

$0+0+(−Mg(L2cosθ))+FT(Lsinθ)=0FTLsinθ=MgL2cosθFT=Mg2cosθsinθ=Mg2cotθ$ (III)

Conclusion:

Therefore, the tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

(c)

To determine

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium.

(c)

Expert Solution

Answer to Problem 82PQ

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Explanation of Solution

Since the ladder is in translational equilibrium, the net force in $x$ and $y$ directions must be zero.

Write the conditions for the translational equilibrium.

$∑Fx=0$ (IV)

Here, $∑Fx$ is the net force in $x$ direction.

$∑Fy=0$ (V)

Here, $∑Fy$ is the net force in $y$ direction.

Write the equation for $∑Fx$ .

$∑Fx=−FT+Fs$ (VI)

Here, $Fs$ is the force of static friction.

Write the equation for $Fs$ .

$Fs=μsFN$

Here, $μs$ is the coefficient of static friction and $FN$ is the normal force.

Put the above equation in equation (VI).

$∑Fx=−FT+μsFN$

Put the above equation in equation (IV) and rewrite it for $FT$ .

$−FT+μsFN=0FT=μsFN$ (VII)

Write the equation for $∑Fy$ .

$∑Fy=FN−Fg$ (VIII)

Here, $Fg$ is the weight of the ladder.

Write the equation for $Fg$ .

$Fg=Mg$

Put the above equation in equation (VIII).

$∑Fy=FN−Mg$

Put the above equation in equation (V) and rewrite it for $FN$ .

$FN−Mg=0FN=Mg$

Put the above equation in equation (VII).

$FT=μsMg$ (IX)

Conclusion:

Therefore, the expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

(d)

To determine

The coefficient of static friction in terms of the angle $θ$ .

(d)

Expert Solution

Answer to Problem 82PQ

The coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Explanation of Solution

Equate equations (III) and (IX).

$Mg2cotθ=μsMgμs=cotθ2$

Conclusion:

Therefore, the coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

(e)

To determine

The after effect of moving the ladder slightly so as to reduce the angle $θ$ .

(e)

Expert Solution

Answer to Problem 82PQ

The ladder will slip if it is moved slightly to reduce the angle $θ$ .

Explanation of Solution

The expression for the angle $θ$ obtained in part (d), $cotθ2$ is valid only at the critical angle $θ$ . The critical angle is the angle when the ladder is on the verge of slipping and the frictional force has its maximum possible value. The angle $θ$ will be reduced when the ladder is moved to the left.

The expression for the tension force obtained in part (b), $FT=Mg2cotθ$ shows that as the value of $θ$ decreases, the tension force will also increases. For the ladder to be in translational equilibrium, the net force in the $x$ direction must be zero. Increasing the tension force will require a larger friction force to balance this condition. But the static friction force is already at its maximum value. This implies, when the ladder is slightly moved to reduce the angle $θ$ , the ladder will slip.

Conclusion:

Thus, the ladder will slip if it is moved slightly to reduce the angle $θ$ .

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Answer 3

Question

Chapter 14, Problem 82PQ

(a)

To determine

The free-body diagram of the forces acting on the ladder.

(a)

Expert Solution

Answer to Problem 82PQ

The free-body diagram of the forces acting on the ladder is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 1

Explanation of Solution

A free-body diagram is a graphical tool used to illustrate the different forces acting on a particular object. It helps to solve complex physical problems. The free-body diagram of the ladder in the given situation is drawn in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 2

The forces acting on the ladder are the weight, the normal force, the tension and the force of static friction. In the figure weight is represented as $Fg$ , normal force is denoted as $FN$ , tension is denoted as $FT$ and the force of static friction is denoted by $Fs$ .

Conclusion:

Thus, the free-body diagram of the forces acting on the ladder is drawn in figure 1.

(b)

To determine

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium.

(b)

Expert Solution

Answer to Problem 82PQ

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Explanation of Solution

Take the lower end of the ladder as the pivot point. This will eliminate the torque due to normal force and the torque due to force of static friction.

Since the ladder is in rotational equilibrium, the net torque about the lower end of the ladder must be zero.

Write the condition for the rotational equilibrium.

$∑τlower end=0$ (I)

Here, $∑τlower end$ is the magnitude of the net torque about the lower end of the ladder.

Write the equation for $∑τlower end$ .

$∑τlower end=τN+τs+τg+τT$

Here, $τN$ is the magnitude torque due to the normal force about the lower end of the ladder, $τs$ is the magnitude torque due to the static friction about the lower end of the ladder, $τg$ is the magnitude torque due to the weight about the lower end of the ladder and $τT$ is the magnitude torque due to the tension about the lower end of the ladder.

Put the above equation in equation (I).

$τN+τs+τg+τT=0$ (II)

Write the expression for $τN$ .

$τN=0$

Write the expression for $τs$ .

$τs=0$

Write the expression for $τg$ .

$τg=−Mg(L2cosθ)$

Here, $M$ is the mass of the ladder, $g$ is the acceleration due to gravity, $L$ is the length of the ladder and $θ$ is the angle the ladder makes with the floor.

Write the expression for $τT$ .

$τT=FT(Lsinθ)$

Here, $FT$ is the magnitude of the tension in the rope.

Put the above four equations in equation (II) and rewrite it for $FT$ .

$0+0+(−Mg(L2cosθ))+FT(Lsinθ)=0FTLsinθ=MgL2cosθFT=Mg2cosθsinθ=Mg2cotθ$ (III)

Conclusion:

Therefore, the tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

(c)

To determine

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium.

(c)

Expert Solution

Answer to Problem 82PQ

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Explanation of Solution

Since the ladder is in translational equilibrium, the net force in $x$ and $y$ directions must be zero.

Write the conditions for the translational equilibrium.

$∑Fx=0$ (IV)

Here, $∑Fx$ is the net force in $x$ direction.

$∑Fy=0$ (V)

Here, $∑Fy$ is the net force in $y$ direction.

Write the equation for $∑Fx$ .

$∑Fx=−FT+Fs$ (VI)

Here, $Fs$ is the force of static friction.

Write the equation for $Fs$ .

$Fs=μsFN$

Here, $μs$ is the coefficient of static friction and $FN$ is the normal force.

Put the above equation in equation (VI).

$∑Fx=−FT+μsFN$

Put the above equation in equation (IV) and rewrite it for $FT$ .

$−FT+μsFN=0FT=μsFN$ (VII)

Write the equation for $∑Fy$ .

$∑Fy=FN−Fg$ (VIII)

Here, $Fg$ is the weight of the ladder.

Write the equation for $Fg$ .

$Fg=Mg$

Put the above equation in equation (VIII).

$∑Fy=FN−Mg$

Put the above equation in equation (V) and rewrite it for $FN$ .

$FN−Mg=0FN=Mg$

Put the above equation in equation (VII).

$FT=μsMg$ (IX)

Conclusion:

Therefore, the expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

(d)

To determine

The coefficient of static friction in terms of the angle $θ$ .

(d)

Expert Solution

Answer to Problem 82PQ

The coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Explanation of Solution

Equate equations (III) and (IX).

$Mg2cotθ=μsMgμs=cotθ2$

Conclusion:

Therefore, the coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

(e)

To determine

The after effect of moving the ladder slightly so as to reduce the angle $θ$ .

(e)

Expert Solution

Answer to Problem 82PQ

The ladder will slip if it is moved slightly to reduce the angle $θ$ .

Explanation of Solution

The expression for the angle $θ$ obtained in part (d), $cotθ2$ is valid only at the critical angle $θ$ . The critical angle is the angle when the ladder is on the verge of slipping and the frictional force has its maximum possible value. The angle $θ$ will be reduced when the ladder is moved to the left.

The expression for the tension force obtained in part (b), $FT=Mg2cotθ$ shows that as the value of $θ$ decreases, the tension force will also increases. For the ladder to be in translational equilibrium, the net force in the $x$ direction must be zero. Increasing the tension force will require a larger friction force to balance this condition. But the static friction force is already at its maximum value. This implies, when the ladder is slightly moved to reduce the angle $θ$ , the ladder will slip.

Conclusion:

Thus, the ladder will slip if it is moved slightly to reduce the angle $θ$ .

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Answer 4

Question

Chapter 14, Problem 82PQ

(a)

To determine

The free-body diagram of the forces acting on the ladder.

(a)

Expert Solution

Answer to Problem 82PQ

The free-body diagram of the forces acting on the ladder is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 1

Explanation of Solution

A free-body diagram is a graphical tool used to illustrate the different forces acting on a particular object. It helps to solve complex physical problems. The free-body diagram of the ladder in the given situation is drawn in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 2

The forces acting on the ladder are the weight, the normal force, the tension and the force of static friction. In the figure weight is represented as $Fg$ , normal force is denoted as $FN$ , tension is denoted as $FT$ and the force of static friction is denoted by $Fs$ .

Conclusion:

Thus, the free-body diagram of the forces acting on the ladder is drawn in figure 1.

(b)

To determine

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium.

(b)

Expert Solution

Answer to Problem 82PQ

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Explanation of Solution

Take the lower end of the ladder as the pivot point. This will eliminate the torque due to normal force and the torque due to force of static friction.

Since the ladder is in rotational equilibrium, the net torque about the lower end of the ladder must be zero.

Write the condition for the rotational equilibrium.

$∑τlower end=0$ (I)

Here, $∑τlower end$ is the magnitude of the net torque about the lower end of the ladder.

Write the equation for $∑τlower end$ .

$∑τlower end=τN+τs+τg+τT$

Here, $τN$ is the magnitude torque due to the normal force about the lower end of the ladder, $τs$ is the magnitude torque due to the static friction about the lower end of the ladder, $τg$ is the magnitude torque due to the weight about the lower end of the ladder and $τT$ is the magnitude torque due to the tension about the lower end of the ladder.

Put the above equation in equation (I).

$τN+τs+τg+τT=0$ (II)

Write the expression for $τN$ .

$τN=0$

Write the expression for $τs$ .

$τs=0$

Write the expression for $τg$ .

$τg=−Mg(L2cosθ)$

Here, $M$ is the mass of the ladder, $g$ is the acceleration due to gravity, $L$ is the length of the ladder and $θ$ is the angle the ladder makes with the floor.

Write the expression for $τT$ .

$τT=FT(Lsinθ)$

Here, $FT$ is the magnitude of the tension in the rope.

Put the above four equations in equation (II) and rewrite it for $FT$ .

$0+0+(−Mg(L2cosθ))+FT(Lsinθ)=0FTLsinθ=MgL2cosθFT=Mg2cosθsinθ=Mg2cotθ$ (III)

Conclusion:

Therefore, the tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

(c)

To determine

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium.

(c)

Expert Solution

Answer to Problem 82PQ

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Explanation of Solution

Since the ladder is in translational equilibrium, the net force in $x$ and $y$ directions must be zero.

Write the conditions for the translational equilibrium.

$∑Fx=0$ (IV)

Here, $∑Fx$ is the net force in $x$ direction.

$∑Fy=0$ (V)

Here, $∑Fy$ is the net force in $y$ direction.

Write the equation for $∑Fx$ .

$∑Fx=−FT+Fs$ (VI)

Here, $Fs$ is the force of static friction.

Write the equation for $Fs$ .

$Fs=μsFN$

Here, $μs$ is the coefficient of static friction and $FN$ is the normal force.

Put the above equation in equation (VI).

$∑Fx=−FT+μsFN$

Put the above equation in equation (IV) and rewrite it for $FT$ .

$−FT+μsFN=0FT=μsFN$ (VII)

Write the equation for $∑Fy$ .

$∑Fy=FN−Fg$ (VIII)

Here, $Fg$ is the weight of the ladder.

Write the equation for $Fg$ .

$Fg=Mg$

Put the above equation in equation (VIII).

$∑Fy=FN−Mg$

Put the above equation in equation (V) and rewrite it for $FN$ .

$FN−Mg=0FN=Mg$

Put the above equation in equation (VII).

$FT=μsMg$ (IX)

Conclusion:

Therefore, the expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

(d)

To determine

The coefficient of static friction in terms of the angle $θ$ .

(d)

Expert Solution

Answer to Problem 82PQ

The coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Explanation of Solution

Equate equations (III) and (IX).

$Mg2cotθ=μsMgμs=cotθ2$

Conclusion:

Therefore, the coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

(e)

To determine

The after effect of moving the ladder slightly so as to reduce the angle $θ$ .

(e)

Expert Solution

Answer to Problem 82PQ

The ladder will slip if it is moved slightly to reduce the angle $θ$ .

Explanation of Solution

The expression for the angle $θ$ obtained in part (d), $cotθ2$ is valid only at the critical angle $θ$ . The critical angle is the angle when the ladder is on the verge of slipping and the frictional force has its maximum possible value. The angle $θ$ will be reduced when the ladder is moved to the left.

The expression for the tension force obtained in part (b), $FT=Mg2cotθ$ shows that as the value of $θ$ decreases, the tension force will also increases. For the ladder to be in translational equilibrium, the net force in the $x$ direction must be zero. Increasing the tension force will require a larger friction force to balance this condition. But the static friction force is already at its maximum value. This implies, when the ladder is slightly moved to reduce the angle $θ$ , the ladder will slip.

Conclusion:

Thus, the ladder will slip if it is moved slightly to reduce the angle $θ$ .

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

Answer 5

Chapter 14, Problem 82PQ

(a)

To determine

The free-body diagram of the forces acting on the ladder.

(a)

Expert Solution

Answer to Problem 82PQ

The free-body diagram of the forces acting on the ladder is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 1

Explanation of Solution

A free-body diagram is a graphical tool used to illustrate the different forces acting on a particular object. It helps to solve complex physical problems. The free-body diagram of the ladder in the given situation is drawn in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 2

The forces acting on the ladder are the weight, the normal force, the tension and the force of static friction. In the figure weight is represented as $Fg$ , normal force is denoted as $FN$ , tension is denoted as $FT$ and the force of static friction is denoted by $Fs$ .

Conclusion:

Thus, the free-body diagram of the forces acting on the ladder is drawn in figure 1.

(b)

To determine

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium.

(b)

Expert Solution

Answer to Problem 82PQ

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Explanation of Solution

Take the lower end of the ladder as the pivot point. This will eliminate the torque due to normal force and the torque due to force of static friction.

Since the ladder is in rotational equilibrium, the net torque about the lower end of the ladder must be zero.

Write the condition for the rotational equilibrium.

$∑τlower end=0$ (I)

Here, $∑τlower end$ is the magnitude of the net torque about the lower end of the ladder.

Write the equation for $∑τlower end$ .

$∑τlower end=τN+τs+τg+τT$

Here, $τN$ is the magnitude torque due to the normal force about the lower end of the ladder, $τs$ is the magnitude torque due to the static friction about the lower end of the ladder, $τg$ is the magnitude torque due to the weight about the lower end of the ladder and $τT$ is the magnitude torque due to the tension about the lower end of the ladder.

Put the above equation in equation (I).

$τN+τs+τg+τT=0$ (II)

Write the expression for $τN$ .

$τN=0$

Write the expression for $τs$ .

$τs=0$

Write the expression for $τg$ .

$τg=−Mg(L2cosθ)$

Here, $M$ is the mass of the ladder, $g$ is the acceleration due to gravity, $L$ is the length of the ladder and $θ$ is the angle the ladder makes with the floor.

Write the expression for $τT$ .

$τT=FT(Lsinθ)$

Here, $FT$ is the magnitude of the tension in the rope.

Put the above four equations in equation (II) and rewrite it for $FT$ .

$0+0+(−Mg(L2cosθ))+FT(Lsinθ)=0FTLsinθ=MgL2cosθFT=Mg2cosθsinθ=Mg2cotθ$ (III)

Conclusion:

Therefore, the tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

(c)

To determine

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium.

(c)

Expert Solution

Answer to Problem 82PQ

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Explanation of Solution

Since the ladder is in translational equilibrium, the net force in $x$ and $y$ directions must be zero.

Write the conditions for the translational equilibrium.

$∑Fx=0$ (IV)

Here, $∑Fx$ is the net force in $x$ direction.

$∑Fy=0$ (V)

Here, $∑Fy$ is the net force in $y$ direction.

Write the equation for $∑Fx$ .

$∑Fx=−FT+Fs$ (VI)

Here, $Fs$ is the force of static friction.

Write the equation for $Fs$ .

$Fs=μsFN$

Here, $μs$ is the coefficient of static friction and $FN$ is the normal force.

Put the above equation in equation (VI).

$∑Fx=−FT+μsFN$

Put the above equation in equation (IV) and rewrite it for $FT$ .

$−FT+μsFN=0FT=μsFN$ (VII)

Write the equation for $∑Fy$ .

$∑Fy=FN−Fg$ (VIII)

Here, $Fg$ is the weight of the ladder.

Write the equation for $Fg$ .

$Fg=Mg$

Put the above equation in equation (VIII).

$∑Fy=FN−Mg$

Put the above equation in equation (V) and rewrite it for $FN$ .

$FN−Mg=0FN=Mg$

Put the above equation in equation (VII).

$FT=μsMg$ (IX)

Conclusion:

Therefore, the expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

(d)

To determine

The coefficient of static friction in terms of the angle $θ$ .

(d)

Expert Solution

Answer to Problem 82PQ

The coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Explanation of Solution

Equate equations (III) and (IX).

$Mg2cotθ=μsMgμs=cotθ2$

Conclusion:

Therefore, the coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

(e)

To determine

The after effect of moving the ladder slightly so as to reduce the angle $θ$ .

(e)

Expert Solution

Answer to Problem 82PQ

The ladder will slip if it is moved slightly to reduce the angle $θ$ .

Explanation of Solution

The expression for the angle $θ$ obtained in part (d), $cotθ2$ is valid only at the critical angle $θ$ . The critical angle is the angle when the ladder is on the verge of slipping and the frictional force has its maximum possible value. The angle $θ$ will be reduced when the ladder is moved to the left.

The expression for the tension force obtained in part (b), $FT=Mg2cotθ$ shows that as the value of $θ$ decreases, the tension force will also increases. For the ladder to be in translational equilibrium, the net force in the $x$ direction must be zero. Increasing the tension force will require a larger friction force to balance this condition. But the static friction force is already at its maximum value. This implies, when the ladder is slightly moved to reduce the angle $θ$ , the ladder will slip.

Conclusion:

Thus, the ladder will slip if it is moved slightly to reduce the angle $θ$ .

Answer 6

Chapter 14, Problem 82PQ

(a)

To determine

The free-body diagram of the forces acting on the ladder.

(a)

Expert Solution

Answer to Problem 82PQ

The free-body diagram of the forces acting on the ladder is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 1

Explanation of Solution

A free-body diagram is a graphical tool used to illustrate the different forces acting on a particular object. It helps to solve complex physical problems. The free-body diagram of the ladder in the given situation is drawn in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 2

The forces acting on the ladder are the weight, the normal force, the tension and the force of static friction. In the figure weight is represented as $Fg$ , normal force is denoted as $FN$ , tension is denoted as $FT$ and the force of static friction is denoted by $Fs$ .

Conclusion:

Thus, the free-body diagram of the forces acting on the ladder is drawn in figure 1.

Answer 7

Chapter 14, Problem 82PQ

(a)

To determine

The free-body diagram of the forces acting on the ladder.

Answer 8

(a)

Expert Solution

Answer to Problem 82PQ

The free-body diagram of the forces acting on the ladder is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 1

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

A free-body diagram is a graphical tool used to illustrate the different forces acting on a particular object. It helps to solve complex physical problems. The free-body diagram of the ladder in the given situation is drawn in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 14, Problem 82PQ , additional homework tip 2

The forces acting on the ladder are the weight, the normal force, the tension and the force of static friction. In the figure weight is represented as $Fg$ , normal force is denoted as $FN$ , tension is denoted as $FT$ and the force of static friction is denoted by $Fs$ .

Conclusion:

Thus, the free-body diagram of the forces acting on the ladder is drawn in figure 1.

Answer 13

(b)

To determine

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium.

(b)

Expert Solution

Answer to Problem 82PQ

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Explanation of Solution

Take the lower end of the ladder as the pivot point. This will eliminate the torque due to normal force and the torque due to force of static friction.

Since the ladder is in rotational equilibrium, the net torque about the lower end of the ladder must be zero.

Write the condition for the rotational equilibrium.

$∑τlower end=0$ (I)

Here, $∑τlower end$ is the magnitude of the net torque about the lower end of the ladder.

Write the equation for $∑τlower end$ .

$∑τlower end=τN+τs+τg+τT$

Here, $τN$ is the magnitude torque due to the normal force about the lower end of the ladder, $τs$ is the magnitude torque due to the static friction about the lower end of the ladder, $τg$ is the magnitude torque due to the weight about the lower end of the ladder and $τT$ is the magnitude torque due to the tension about the lower end of the ladder.

Put the above equation in equation (I).

$τN+τs+τg+τT=0$ (II)

Write the expression for $τN$ .

$τN=0$

Write the expression for $τs$ .

$τs=0$

Write the expression for $τg$ .

$τg=−Mg(L2cosθ)$

Here, $M$ is the mass of the ladder, $g$ is the acceleration due to gravity, $L$ is the length of the ladder and $θ$ is the angle the ladder makes with the floor.

Write the expression for $τT$ .

$τT=FT(Lsinθ)$

Here, $FT$ is the magnitude of the tension in the rope.

Put the above four equations in equation (II) and rewrite it for $FT$ .

$0+0+(−Mg(L2cosθ))+FT(Lsinθ)=0FTLsinθ=MgL2cosθFT=Mg2cosθsinθ=Mg2cotθ$ (III)

Conclusion:

Therefore, the tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Answer 14

(b)

To determine

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium.

Answer 15

(b)

Expert Solution

Answer to Problem 82PQ

The tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Take the lower end of the ladder as the pivot point. This will eliminate the torque due to normal force and the torque due to force of static friction.

Since the ladder is in rotational equilibrium, the net torque about the lower end of the ladder must be zero.

Write the condition for the rotational equilibrium.

$∑τlower end=0$ (I)

Here, $∑τlower end$ is the magnitude of the net torque about the lower end of the ladder.

Write the equation for $∑τlower end$ .

$∑τlower end=τN+τs+τg+τT$

Here, $τN$ is the magnitude torque due to the normal force about the lower end of the ladder, $τs$ is the magnitude torque due to the static friction about the lower end of the ladder, $τg$ is the magnitude torque due to the weight about the lower end of the ladder and $τT$ is the magnitude torque due to the tension about the lower end of the ladder.

Put the above equation in equation (I).

$τN+τs+τg+τT=0$ (II)

Write the expression for $τN$ .

$τN=0$

Write the expression for $τs$ .

$τs=0$

Write the expression for $τg$ .

$τg=−Mg(L2cosθ)$

Here, $M$ is the mass of the ladder, $g$ is the acceleration due to gravity, $L$ is the length of the ladder and $θ$ is the angle the ladder makes with the floor.

Write the expression for $τT$ .

$τT=FT(Lsinθ)$

Here, $FT$ is the magnitude of the tension in the rope.

Put the above four equations in equation (II) and rewrite it for $FT$ .

$0+0+(−Mg(L2cosθ))+FT(Lsinθ)=0FTLsinθ=MgL2cosθFT=Mg2cosθsinθ=Mg2cotθ$ (III)

Conclusion:

Therefore, the tension in the rope in terms of $M,g$ and $θ$ if the ladder is in rotational equilibrium is $Mg2cotθ$ .

Answer 20

(c)

To determine

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium.

(c)

Expert Solution

Answer to Problem 82PQ

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Explanation of Solution

Since the ladder is in translational equilibrium, the net force in $x$ and $y$ directions must be zero.

Write the conditions for the translational equilibrium.

$∑Fx=0$ (IV)

Here, $∑Fx$ is the net force in $x$ direction.

$∑Fy=0$ (V)

Here, $∑Fy$ is the net force in $y$ direction.

Write the equation for $∑Fx$ .

$∑Fx=−FT+Fs$ (VI)

Here, $Fs$ is the force of static friction.

Write the equation for $Fs$ .

$Fs=μsFN$

Here, $μs$ is the coefficient of static friction and $FN$ is the normal force.

Put the above equation in equation (VI).

$∑Fx=−FT+μsFN$

Put the above equation in equation (IV) and rewrite it for $FT$ .

$−FT+μsFN=0FT=μsFN$ (VII)

Write the equation for $∑Fy$ .

$∑Fy=FN−Fg$ (VIII)

Here, $Fg$ is the weight of the ladder.

Write the equation for $Fg$ .

$Fg=Mg$

Put the above equation in equation (VIII).

$∑Fy=FN−Mg$

Put the above equation in equation (V) and rewrite it for $FN$ .

$FN−Mg=0FN=Mg$

Put the above equation in equation (VII).

$FT=μsMg$ (IX)

Conclusion:

Therefore, the expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Answer 21

(c)

To determine

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium.

Answer 22

(c)

Expert Solution

Answer to Problem 82PQ

The expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Since the ladder is in translational equilibrium, the net force in $x$ and $y$ directions must be zero.

Write the conditions for the translational equilibrium.

$∑Fx=0$ (IV)

Here, $∑Fx$ is the net force in $x$ direction.

$∑Fy=0$ (V)

Here, $∑Fy$ is the net force in $y$ direction.

Write the equation for $∑Fx$ .

$∑Fx=−FT+Fs$ (VI)

Here, $Fs$ is the force of static friction.

Write the equation for $Fs$ .

$Fs=μsFN$

Here, $μs$ is the coefficient of static friction and $FN$ is the normal force.

Put the above equation in equation (VI).

$∑Fx=−FT+μsFN$

Put the above equation in equation (IV) and rewrite it for $FT$ .

$−FT+μsFN=0FT=μsFN$ (VII)

Write the equation for $∑Fy$ .

$∑Fy=FN−Fg$ (VIII)

Here, $Fg$ is the weight of the ladder.

Write the equation for $Fg$ .

$Fg=Mg$

Put the above equation in equation (VIII).

$∑Fy=FN−Mg$

Put the above equation in equation (V) and rewrite it for $FN$ .

$FN−Mg=0FN=Mg$

Put the above equation in equation (VII).

$FT=μsMg$ (IX)

Conclusion:

Therefore, the expression for the tension in the rope in terms of $μs,M$ and $g$ by considering the ladder in translational equilibrium is $μsMg$ .

Answer 27

(d)

To determine

The coefficient of static friction in terms of the angle $θ$ .

(d)

Expert Solution

Answer to Problem 82PQ

The coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Explanation of Solution

Equate equations (III) and (IX).

$Mg2cotθ=μsMgμs=cotθ2$

Conclusion:

Therefore, the coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Answer 28

(d)

To determine

The coefficient of static friction in terms of the angle $θ$ .

Answer 29

(d)

Expert Solution

Answer to Problem 82PQ

The coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Equate equations (III) and (IX).

$Mg2cotθ=μsMgμs=cotθ2$

Conclusion:

Therefore, the coefficient of static friction in terms of the angle $θ$ is $cotθ2$ .

Answer 34

(e)

To determine

The after effect of moving the ladder slightly so as to reduce the angle $θ$ .

(e)

Expert Solution

Answer to Problem 82PQ

The ladder will slip if it is moved slightly to reduce the angle $θ$ .

Explanation of Solution

The expression for the angle $θ$ obtained in part (d), $cotθ2$ is valid only at the critical angle $θ$ . The critical angle is the angle when the ladder is on the verge of slipping and the frictional force has its maximum possible value. The angle $θ$ will be reduced when the ladder is moved to the left.

The expression for the tension force obtained in part (b), $FT=Mg2cotθ$ shows that as the value of $θ$ decreases, the tension force will also increases. For the ladder to be in translational equilibrium, the net force in the $x$ direction must be zero. Increasing the tension force will require a larger friction force to balance this condition. But the static friction force is already at its maximum value. This implies, when the ladder is slightly moved to reduce the angle $θ$ , the ladder will slip.

Conclusion:

Thus, the ladder will slip if it is moved slightly to reduce the angle $θ$ .

Answer 35

(e)

To determine

The after effect of moving the ladder slightly so as to reduce the angle $θ$ .

Answer 36

(e)

Expert Solution

Answer to Problem 82PQ

The ladder will slip if it is moved slightly to reduce the angle $θ$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

The expression for the angle $θ$ obtained in part (d), $cotθ2$ is valid only at the critical angle $θ$ . The critical angle is the angle when the ladder is on the verge of slipping and the frictional force has its maximum possible value. The angle $θ$ will be reduced when the ladder is moved to the left.

The expression for the tension force obtained in part (b), $FT=Mg2cotθ$ shows that as the value of $θ$ decreases, the tension force will also increases. For the ladder to be in translational equilibrium, the net force in the $x$ direction must be zero. Increasing the tension force will require a larger friction force to balance this condition. But the static friction force is already at its maximum value. This implies, when the ladder is slightly moved to reduce the angle $θ$ , the ladder will slip.