EBK COLLEGE PHYSICS
EBK COLLEGE PHYSICS
11th Edition
ISBN: 8220103600385
Author: Vuille
Publisher: Cengage Learning US
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Question
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Chapter 8, Problem 94AP

(a)

To determine

Free body diagram for the ladder.

(a)

Expert Solution
Check Mark

Answer to Problem 94AP

EBK COLLEGE PHYSICS, Chapter 8, Problem 94AP , additional homework tip  1

Explanation of Solution

The following figure shows the free body diagram.

EBK COLLEGE PHYSICS, Chapter 8, Problem 94AP , additional homework tip  2

(b)

To determine

The normal force exerted by the bottom of the ladder.

(b)

Expert Solution
Check Mark

Answer to Problem 94AP

The normal force exerted by the bottom of the ladder is 218N .

Explanation of Solution

The sum of forces on the ladder is Fy=0=nFwmmg and this expression can be rearranged for the normal force exerted by the bottom of the ladder.

Given info: The weight of the ladder is 1.20×102N , mass of the monkey is 10.0kg , and acceleration due to gravity is 9.80m/s2 .

The formula for the normal force exerted by the bottom of the ladder is,

nF=w+mmg

  • w is weight of the ladder.
  • mm is mass of the monkey.
  • g is acceleration due to gravity.

Substitute 1.20×102N for w , 10.0kg for mm , and 9.80m/s2 for g to find nF .

nF=1.20×102N+(10.0kg)(9.80m/s2)=218N

Thus, the normal force exerted by the bottom of the ladder is 218N .

Conclusion:

Therefore, the normal force exerted by the bottom of the ladder is 218N .

(c)

To determine

The tension in the rope when the monkey is two-thirds of the way up the ladder.

(c)

Expert Solution
Check Mark

Answer to Problem 94AP

The tension in the rope when the monkey is two-thirds of the way up the ladder is 72.4N .

Explanation of Solution

The sum of torques on the ladder is τ=0=w(L/2)cosθmmg(2L/3)cosθ+nwLsinθ and now this expression is rearranged for the normal force on the ladder due to the wall. The forces which act on the ladder along horizontal direction is Fx=Tnw=0 .

Given info: The weight of the ladder is 1.20×102N, angle of the ladder with horizontal is 60.0° , mass of the monkey is 10.0kg , and acceleration due to gravity is 9.80m/s2 .

The formula for the normal force on the ladder by the wall is,

nw=(1/2)wcotθ+(2/3)mmgcotθ

  • w is weight of the ladder.
  • θ is the angle of the ladder with horizontal.
  • mm is mass of the ladder.
  • g is acceleration due to gravity.

Substitute 1.20×102N for w , 60.0 for θ , 10.0kg for mm, and 9.80m/s2 for g to find nw .

nw=(1/2)(1.20×102N)(cot60.0°)+(2/3)(10.0kg)(9.80m/s2)(cot60.0°)=72.4N

Thus, the normal force on the ladder due to the surface of the wall is equal to the tension in the rope is 72.4N .

Conclusion:

Therefore, the tension in the rope when the monkey is two-thirds of the way up the ladder is 72.4N.

(d)

To determine

The maximum distance d that the monkey can climb up the ladder.

(d)

Expert Solution
Check Mark

Answer to Problem 94AP

The maximum distance d that the monkey can climb up the ladder is 2.41m .

Explanation of Solution

The tension in the rope when it ready to break is Tmax=80.0N and the resultant torque on the ladder is 0=w(L/2)cosθmmgxcosθ+nwLsinθ , from this expression, the distance of the monkey is obtained by solving the following expression 0=w(L/2)cosθmmgxcosθ+nwLsinθc.

Given info: The weight of the ladder is 1.20×102N, angle of the ladder with horizontal is 60.0° , mass of the monkey is 10.0kg , length of the ladder is 3.00m , the maximum tension in the rope for  which it can withstand is 80.0N , and acceleration due to gravity is 9.80m/s2 .

The formula for the maximum distance d that the monkey can climb up the ladder is,

d=w(L/2)+TmaxLtanθmmg

  • w is weight of the ladder.
  • L is length of the ladder.
  • Tmax is maximum tension in the rope.
  • θ is the angle of the ladder with horizontal surface.
  • mm is mass of the monkey.
  • g is acceleration due to gravity.

Substitute 1.20×102N for w , 60.0 for θ , 10.0kg for mm, 80.0N for Tmax , 3.00m for L , and 9.80m/s2 for g to find d .

d=(1.20×102N)(3.00m/2)+(80.0N)(3.00m)(tan60.0°)(10.0kg)(9.80m/s2)=2.41m

Thus, the maximum distance d that the monkey can climb up the ladder is 2.41m .

Conclusion:

Therefore, the maximum distance d that the monkey can climb up the ladder is 2.41m .

(e)

To determine

The required alternate analysis if the rope would be removed and the surface be enough rough.

(e)

Expert Solution
Check Mark

Explanation of Solution

There exists frictional force if the surface would be rough enough and this force acts opposite to the force that being applied on an objects. It is defined as f=μsn where μs and n are coefficient of static friction and the normal force.

When the rope tide with lower end of the ladder, the tension in that rope tend to be towards the wall, similarly, when it removed, the frictional force would be directed towards the wall and it do the job of the tension in the rope, the remaining analysis would same as in the previous parts.

For the maximum distance climbed by the monkey is determined by the condition of maximum frictional force it means, the monkey can climb maximum distance if the static friction is large.

Conclusion:

Therefore, to solve this problem, the information of coefficient of static friction is needed.

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Chapter 8 Solutions

EBK COLLEGE PHYSICS

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