Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
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Chapter 14, Problem 82PQ

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 82PQ

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

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, 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, Extended Version With Modern Physics, 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
Check Mark

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
Check Mark

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=FNFg                                                                                                   (VIII)

Here, Fg is the weight of the ladder.

Write the equation for Fg .

  Fg=Mg

Put the above equation in equation (VIII).

  Fy=FNMg

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

  FNMg=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
Check Mark

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
Check Mark

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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Chapter 14 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

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