# Homework Help: EeekRotational Dynamics

1. Nov 20, 2005

### TastyTyr

A uniform plank of length 5.8 m and weight 216 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 441 N walk on the overhanging part of the plank before it just begins to tip?

This is what I did:

I took the weight of the plank and multiplied by the hanging support. I next set that equal to the weight of the person times x distance

So it looks like:

216(1.1)=441x
I solved for x and got
x=0.539

but it's wrong...what do I do?

2. Nov 20, 2005

### Physics Monkey

This problem begins with the realization that just before the plank tips over, the normal force from the left support vanishes. To find the position x of equilibrium, you should balance the force and the torque on the plank. Which of these two equations will be useful for finding x?

3. Nov 20, 2005

### TastyTyr

..and so the sum of all forces and the sum of torque should both equal zero
but do I use the torque equation at all?
T=R*F or T=F*L

was what I did at first completely wrong??

4. Nov 20, 2005

### Physics Monkey

Well, the force equation doesn't involve x, so it can't be used to find x. I think you had the right idea above, but you excuted it incorrectly. You need to figure out the torque exerted by the left side of the board, the right side of the board, and the person. All these torques should balance in equilibrium. To find the torque you need to know the force that is creating the torque. Once you have this force, you can figure out the moment arm either by doing a simple integral or simply by guessing.