# Man on the end of a beam (rotational equilibrium problem)

## Homework Statement

1. If a man weighing 800 Newtons is standing on a uniform beam an unknown distance from the supports. The beam is of length 5m and weighs 225 Newtons, and one support is positioned at the left end of the beam, the other is placed 1m from the left end and is exerting 600 Newtons up upon the beam. (http://heatherwebb.com/rotationalequilibriumdiagram.jpg [Broken] - I did the parts in red)

a) Draw the free body diagram on the given picture, show all forces, known and unknown, along with all distances.

b) If the beam is in equilibrium, what distance is the man standing from the left support?

## Homework Equations

Torque (left support) = Torque (man)

## The Attempt at a Solution

Here is the diagram and what I've done to it, http://heatherwebb.com/rotationalequilibriumdiagram.jpg [Broken].
I know that the pivot exerts a force of 600N. Isn't that the normal force? Shouldn't I use that and the two known forces to determine the force at the left-most support? Then use that in the Torque equation?
It's not making sense to me because the force of the left-most support comes out to be negative.

## The Attempt at a Solution

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## Answers and Replies

BruceW
Homework Helper
The drawing is correct. (The person could be at some place on the beam, not necessarily at the end).

I've worked out the force from the left-most support, and it is upwards. Maybe you mis-calculated? The best way to do it would be to define up as positive, so then all upward forces are positive and all downward forces are negative. Then all the forces added together should equal zero.

And yes, you're right that you need to use torque equation to see where the person must stand for there to be zero total torque. But actually, you don't need to know the force exerted by the left-most support to find this. Remember you can define the torque around any point, so what point will make this calculation easiest?