posted by Kristi .
A 8.49-m ladder with a mass of 24.2 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 247 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.75 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?
I am reluctant to do this, it is so simple.
= -247*L + weight*L/2
the negative sign direction is in the direction the painter is pulling.
Ladders moment of inertia = thin rod rotating about an end http://en.wikipedia.org/wiki/List_of_moments_of_inertia mL/3