A 9.0-kg monkey climbs a uniform ladder with weight 120 N and length L = 3.5 m a

Question

A 9.0-kg monkey climbs a uniform ladder with weight 120 N and length L = 3.5 m as shown in the figure below. The ladder rests against the wall and makes an angle of ? = 60.0°with the ground. The upper and lower ends of the ladder rest on frictionless surfaces. The lower end is connected to the wall by a horizontal rope that is frayed and can support a maximum tension of only 80.0 N.

(a) Draw an extended force diagram for the ladder.

(b) Find the tension in the rope when the monkey is two-thirds of the way up the ladder.

(c) Find the maximum distance d (along the ladder) that the monkey can climb up the ladder before the rope breaks.

(d) If the horizontal surface were rough and the rope were removed, how would your analysis of the problem change? What other information would you need to answer parts (b) and (c)?

Explanation / Answer

b)

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

applying moments about the upper end:

for equilibrium anticlockwise = clockwise

9g(3.5/3)cos60° + 120(3.5/2)cos60° + T(3.5)sin60° =(9(9.81) + 120 )(3.5)cos60°

=> 156.5+3.03 T = 364.35

T = 68.59 N



(c) Find the maximum distance d that the monkey can climb up the ladder before the rope breaks.

max distance d corresponds to T = 80 N = normal reaction R(w) for horizontal equilibrium

applying moments about ladder foot:

R(w)(3.5)sin60° = 120(3.5/2)(1/2) + 9(9.81)dcos60°

=> 80(3.03) = 120(0.875) + 44.1 d

=> d ~= 3.115 m

d)

The coefficient of friction would be required to solve further

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