Question

A monkey purchases a crate of bananas at the warehouse. The total mass of the crate plus bananas is 45.00 kg. The monkey needs to pull the crate across the rough horizontal warehouse floor to the door. The coefficients of friction between the crate and the floor are μ_s = 0.18 and μ_k = 0.15. The monkey attaches a vine around the crate and pulls on the vine at an angle of exactly 40.00^o above the horizontal. He pulls with a force of 70.00 N but the crate will not move.

(i) Calculate the normal force on the crate.

(ii) Calculate the friction force on the crate.

The monkey now increases his applied force, still at an angle of exactly 40.00^o above the horizontal.

(iii) What is the applied force at which the monkey is just able to move the crate?

The monkey now pulls the vine horizontally (0.00^o above the horizontal) with an applied force of 85.0 N.

(iv) What is the acceleration of the crate?

Answer 1

(i)

normal force FN = mg - F*sintheta

Fn = 45*9.81 - 70*sin40

Fn = 396.45 N

(ii)

friction force fs = us*Fn = 0.18*396.45 = 71.361 N

(iii)

if crate just moves

fsmax = F*x

us*(mg - F*sintheta) = F*costheta

0.18*(40*9.81 - F*sin40) = F*cos40

F = 80 N

(iv)

theta = 0

normal force Fn = m*g

friction force fk = uk*Fn = uk*m*g

Fnet = m*a

F - fk = m*a

85 - 0.15*45*9.81 = 45*a

a = 0.42 m/s^2