Question

A 10.0kg uniform ladder that is 2.50m long is placed against a smooth (frictionless) vertical wall and reaches to a height of 2.10m. The angle between the floor and the ladder is 60°. An 80.0kg bucket of concrete is suspended from the top rung of the ladder, right next to the wall.

a) Draw a free-body diagram of the ladder.

b) What is the magnitude of the friction force, that the floor exerts on the ladder?

c) What is the coefficient of static friction?

Answer 1

Given                                                           P                   N1

Mass of ladder (m) = 10 kg

Length of ladder (l) = 2.5 ml                        h    m1g                         N2

Height (h) = 2.1 m                                                      mg

Angle (?) = 60 deg                                                             Ff

Mass of bucket (m1) = 60 kg                                       l2

(b) The net vertical force = 0

m1g + mg – N2 = 0

• N2 = (m+m1)g = (60+10)*9.8 = 686 N

L2 = l cos 60 = 2.5 cos 60 = 1.25 m

Taking moment of force about point P we get

N2*l2 – mg *l2/2 – Ff*h = 0

686*1.25 – 60*9.8*1.25/2 – Ff*2.1 = 0

• Ff = (686*1.25 – 60*9.8*1.25/2)/2.1 = 233.33 N

Force of friction = 233.33 N

(C) Ff = u*N2

=> coefficient of friction (u)= Ff /N2 = 233.33 / 686 = 0.34