Asked by <3

A 8.78 -m ladder with a mass of 23.8 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 236 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.90 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?
Answered by drwls
About what rotation axis do you want the moment of inertia? You need to specify it. Is it at one of the ends? The center?
Answered by <3
With the ladder laying flat on the ground, the axis is on the left and it is being pulled upward on the right. But the center of gravity confuses me. I don't know how that comes into play. These topics are still new to me.
Answered by Elena
See my answer to a similar question below under related questions.
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