Question

A ladder shown in figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder at the middle. The angle between two legs is 60°. A person of mass 40 kg is sitting on the top. The tension in cross bar is nearly (g = 10 m/s2)​

Answer 1

Given:

A ladder is given and a crossbar connects the two legs of the ladder at the middle.

The angle between two legs is 60°

A person of mass 40 kg is sitting on the top.

To Find:

The tension in cross bar

Solution:

If we consider, the vertical equilibrium of the ladder ,

then, we have weight of the man in the downward direction and normal reaction from the ground on both the legs.

Let W be weight of the man ,

• W = 40Kg g = 400N

Let N be the normal reaction.

Therefore N + N = W = 400N

• N = 200N.

Now let consider the rotational equilibrium of the leg.

We can consider it with respect to the top most point.

We know Torque = Force x perpendicular distance

Then for equilibrium,

• R x 0 + T x l/2 cos (30) = Wx0 + Nxlxsin(30)
• T = 2N tan(30) = 400/√3 N

The tension in cross bar is nearly  is 400/√3 N