Painter on a balance beam (rotational motion and Torque)

[wsc6c]

Homework Statement

A man of mass mm = 90 kg decides to paint his house. To do this, he builds a platform using a uniform beam with a mass of mb = 91 kg and a length of L = 7 meters. The beam is supported by two sawhorses,

----------------------
A B
(distance from A to left is 2 m and distance from far left to B is 5m)

a) If the man stands over the support at point B, calculate the force exerted by the beam on the support at A.

b)How far from the end of the beam (the end closest to support A) does the person have to stand to unbalance the beam

c)Later that day, after thinking about how cool rotational dynamics really is, the man decides to conduct an experiment. He removes one of the supports and places the other as shown in the diagram. Standing at the end of the board, he has his daughter place paint cans, each of mass mc = 2.5 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 actual length of the board that both the man and the cans occupy. Assume both are points at the ends of the board.) Also, take the axis of rotation about the sawhorse.

Homework Equations

Torque=rF sin(theta)

The Attempt at a Solution

I attempted to sum all of the forces. Since I picked my axis at B that eliminates the torque of the man because it =0. If i assume that the force of the beam is in the center then r=1.5. I can't seem to get the right answer.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Welcome to PF!

Hi wsc6c! Welcome to PF!

… If i assume that the force of the beam is in the center then r=1.5. I can't seem to get the right answer.

Sorry, I can't work out which question you're answering, or how you got 1.5.

Show us your full calculations, and then we can see what went wrong, and we'll know how to help!

 

[wsc6c]

I got the first answer so that is fine. Now, for the second question would you set both the torque of the man and the torque of the beam to be larger than the force of A and then solve for the distance from the axis in the torque equation of the man?
I got 1.5 because it is t, the distance from the center of the beam to B. If the total distance from the left to B is 5 then the distance from A to B is 3. 3/2=1.5

 

[tiny-tim]

Hi wsc6c!

I got the first answer so that is fine. Now, for the second question would you set both the torque of the man and the torque of the beam to be larger than the force of A and then solve for the distance from the axis in the torque equation of the man?
I got 1.5 because it is t, the distance from the center of the beam to B. If the total distance from the left to B is 5 then the distance from A to B is 3. 3/2=1.5

ok, so you're answering …

b)How far from the end of the beam (the end closest to support A) does the person have to stand to unbalance the beam

The position of B doesn't matter … if the beam is about to unbalance, then the reaction at B will be zero.

You have to equate the moments of the other three forces (and if you take moments about A, you can ignore the reaction there anyway, so that only leaves the weights of the man and of the beam) …

so what do you get? ​