# Homework Help: Painter (Equilibrium)

1. Jun 22, 2009

### turandorf

1. The problem statement, all variables and given/known data
A physics student of mass mm = 100 kg gets a summer job painting houses. On his first project, he builds a platform using large pine board with a mass of mb = 60 kg The board has an overall length of L = 8.5 meters and set on two 1.1 meter tall saw horses so that l = 2.25 meters overhangs each side.
(a) the student stands over the support at point B, calculate the force exerted by the beam on the support at A__.
(b) b)How far from the left end of the beam can the painter stand before the board (and painter) begin to tip over?
(c) He now removes one of the supports and places the other one 1/3 of the way from the left edge. Standing at the end of the board, he has his girl friend place paint cans, each of mass mc = 1.75 kg, on the opposite end. How many cans will the girl have to place on the board to provide the best balance? (You may neglect the small length of the board that both the man and the cans occupy. Assume both are points at the ends of the board.)

2. Relevant equations

Torque=F(r)
F=ma

3. The attempt at a solution
I thought part (a) would be mg*r but that didn't work.

2. Jun 22, 2009

### LowlyPion

In a) if he is standing over the support, then what weights are acting through that point?

For b) the sum of the torques (or moments if you like) will need to be 0, or there will be an incident and some paint to clean off the ground.

Where is the center of mass acting relative to a saw horse?

At tipping then won't his distance from the say horse times his weight need to be greater than the center of mass times its distance from the same point?

3. Jun 22, 2009

### turandorf

For (a) the weight is (100+60)9.81=1569.6N (man+board) Is this just split between the two points? Or do I have to use Torque=F(r)? (in which case r would be 4 meters [distance between B and A] right?)

4. Jun 22, 2009

### LowlyPion

Isn't just half the weight of the board acting through each support? Then the weight of the man is additionally acting over the support at one end?

5. Jun 22, 2009

### turandorf

Oh wow yeah can't see why I couldn't figure that out. (It equaled 294.3N) Thanks for the insight! So in (b), the torque of the man can't exceed 294.3(4+x), right? (Where x is the distance from the support to the painter and the 4 comes from the distance from point A to B.) Then if the torque pf the man equals Mg*x you can just set the two equations equal and sole for x. Let me know if there are any flaws in this. Thanks!

6. Jun 22, 2009

### LowlyPion

In b) you are dealing with a balance. First choose the point that it will pivot about ... like the saw horse at whatever end he stands at. He can move outside that only so long as the weight at the center of mass acting at guess where ... the center of mass ... times its distance from the pivot is not exceeded by his distance away to the outside times his weight.

7. Jun 22, 2009

### turandorf

OK so i pivot around point B. So the weight of the painter times the distance past B has to equal the center of mass times the distance to B which would be 2 m right? Or is the center of mass including the painter as well as the board? I really am confused, sorry.

8. Jun 23, 2009

### LowlyPion

Yes. That's right.

Consider the elements acting separately is fine.

What you are determining is the condition at which the center of mass of the system, painter and board, lays no farther than the support point at the saw horse.