Question

A uniform ladder of length 10 m and weighing 20N is placed against a smooth vertical wall with its lower end 8m from wall. In this position the ladder is just to slip. Determine the coefficient of friction between the ladder and the floor and also find the frictional force acting on the ladder at the point of contact between ladder and floor?

Answer 1

Answer:

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Let's start by drawing a diagram of the problem statement.

W is the weight of the ladder, acting downwards on the center of mass of the ladder. N is the normal reaction from the ground on the ladder and f is the friction offered by the ground to the ladder.

Consider the topmost point of the ladder as our reference point.

Now, we assume that the ladder is at rest. So, the torques due to different forces will cancel out each other. Considering our reference point, the torque due to the weight of ladder is W x (1.25)/2

The torque due to normal reaction from the ground is N x 1.25

The torque due to friction is 3(by pythagoras theorem) x f.

Now, there is no force acting in vertical direction on the ladder other than its weight. So, the normal reaction, N=W.

Now, by equating the clockwise and anticlockwise torques, we get f=52.0833333N (nearly equal to 52.1N).