Question

A 9.19 -m ladder with a mass of 24.9 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 268 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.72 rad/s2 about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?

Answer 1

Given :-

L = 9.19 m

m = 24.6 kg

F = 268 N

= 1.72 rad/s^2

a)

weight of the ladder is

W = mg

W = 24.6 kg x 9.81 m/s^2

W = 241.326 N

The weigth of the ladder acts at the center of the ladder that is L/2 from the each end

applied force acts at the top of the ladder

considering counterclockwise torques are positive and apply torque about the bottom end

= (-W x L/2) + F x L

= (-241.326 x 9.19/2) + (268 x 9.19)

= -1100.446 + 2462.92

= 1362.474 N-m

b)

= I*

I = /

I =  1362.474 N-m / 1.72 rad/s^2

I = 792.14 kg.m^2