- #1

Cornbreadddd

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## Homework Statement

## Homework Equations

The relevant equations are obviously going to be the rotational equilibrium expressions regarding balancing clockwise and counterclockwise movement as it relates to T = Fx lever-arm.

## The Attempt at a Solution

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First...here is the answer that the back of the book gives:

First...here is the answer that the back of the book gives:

- "
**B is correct**. The axis of rotation is the point where the rope attached to the board. The hanging weight creates a counter-clockwise torque equal to 3 kg x 0.2 m. The weight of the board creates a clockwise torque at the distance from the rope attachment to the board's center of mass, which is 0.3 m. The net torque is zero, so the clockwise torque equals the counterclockwise torque, so 3 kg x 0.2 m = 0.3 m times the weight of the board. Therefore, the weight of the board is 2 kg."

(0.8 x mass of board) = (0.2 x mass of board) + (0.2 x 3kg) ----> mass = 1kg

If the answer in the book is correct...and only that extra 0.3 meters between the rope (center of rotation) and center of mass is relevant...then how come the other 0.5 meters to the right of the board doesn't throw everything out of equilibrium...I mean it is extra weight after all, right?

It's clear I have some sort of fundamental conceptual hole in my understanding of torque as it pertains to equilibrium, but I'm having trouble filling that hole in...any help would be very appreciated, thanks!