A 10 kg monkey climbs a uniform ladder with weight 1.20 *10^2 N and length L=3.0
ID: 2068364 • Letter: A
Question
A 10 kg monkey climbs a uniform ladder with weight 1.20 *10^2 N and length L=3.00 m. The ladder rests against the wall and makes an angle of theta= 60.0 degrees 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) Find the normal force exerted on the bottom of the ladder B) and the tension in the rope when the monkey is two-thirds of the way up the ladder. C) What is the maximum distance (d) 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
Net downward force = (10*9.81)+120 = 218.1 N
(a) Equating net upward force and net downward force, we get,
Net normal force exerted on bottom of ladder = 218.1 N
(b) Taking moments about the point where ladder touches the ground,
(10*9.81)*(2/3*3 Cos 60) + 120*(1/2*3 Cos 60) = Rwall*(3 Sin 60) , where Rwall is normal reaction from wall on ladder.
This gives Rwall = 72.4 N
Doing horizontal force balance, tension in rope = Rwall = 72.4 N
(c) (10*9.81)*(d Cos 60) + 120*(1/2*3 Cos 60) = 80*(3 Sin 60) ,
This gives, d = 2.4 m
(d) The friction force would replace the horizontal rope tension.
We would need the coefficient of static friction between ladder and ground in order to calculate the friction force.
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