Painter (another Equilibirum problem)

[Bryon]

Homework Statement

A physics student of mass mm = 95 kg gets a summer job painting houses. On his first project, he builds a platform using large pine board with a mass of mb = 50 kg The board has an overall length of L = 8.5 meters and set on two 1 meter tall saw horses so that l = 2.25 meters overhangs each side as shown above.


Later that day, after thinking about how cool rotational dynamics really is, the student decides to conduct an experiment. He 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 = 1.59 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.)

https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-12-Equilibrium/painter/painter_3.jpg

Homework Equations

torque = rFsin(theta)

The Attempt at a Solution

torque(A) = (1/3)mass(man)g - (2/3)*g*mass(cans) - (1/2)*g*mass(board) = 0
torque(A) = (1/3)*95*9.81 - (2/3)*9.81*m(cans) - (1/2)*9.81*50 = 0

I think for the mass of the cans I got 9.94 (finding the number of cans is easy), and I don't think its correct. I think I'm missing something, any ideas?

 

[rl.bhat]

Check your calculation.
You can rewrite the second equation by canceling 9.81. It becomes
1/3*95 - 1/2*50 = 2/3*m
Now solve for m.

 

[kerimek]

Your calculations for the torque are correct. However, the mass of the cans should be 1.59 kg each, not 9.94 kg. This means that the number of cans needed for the best balance would be 6, since (2/3)*6*1.59 kg = 6.36 kg, which is equal to (1/3)*95 kg. This would provide a perfectly balanced system, with no net torque.