Help with a physics problem

Question

Answer 1

Let

m = 10 kg be the mass of the monkey,

W = 120 N be the weight of the ladder,

L = 3 m be the length of the ladder,

θ = 60º be the angle of the ladder,

x be the distance the monkey has climbed up the ladder,

g = 9.81 m/s² be gravitational acceleration.

a.

The ladder is subject to the following forces

- W (acting down) applied distance L/2 up the ladder

- M = mg (acting down) the weight of the monkey, applied distance x up the ladder,

- N (acting up) the normal force of the ground against the ladder,

- R (acting left to right) the tension in the rope,

- P (acting right to left) the force of the wall pushing on the ladder.

b.

For the ladder not to move, all the forces acting up must equal all the force acting down:

N = M + W

Substituting actual numbers

N = 10×9.81 + 120 = 218.1

c.

Consider the torques around the point where the ladder meets the wall.

All the torques trying to rotate it clock-wise must equal the torques

trying to rotate it counter-clockwise.

NLcosθ = RLsinθ + WLcosθ/2 + M(x-L)cosθ

R = (NLcosθ - WLcosθ/2 - M(L-x)cosθ)/Lsinθ = (N - W/2 - M(1 - x/L))cotθ

Substituting actual numbers, where x = 2

R = (218.1 - 120/2 - 10×9.81/3)cot(60º) = 72.4 N

d.

We solve the above equation for x with R = 80

80 = (218.1 - 120/2 - 10×9.81(1 - x/3))cot(60º)

x = 2.4 m

e.

The coefficient of friction is the ration between the horizontal force, R,

and the vertical force, N

R/N = 80/218.1 = 0.37