posted by jeff .
A bench is made of a 3 meter long plank with a mass of 12kg and 2 sawhorses(each 0.2m from an end).There are two boxes on the bench. A 15kg box is 0.8 meter from the left hand side. A second box is 0.5 m from the right hand side. What would the mass of the second box have to be for the normal force exerted by the right sawhorse to be 3 times larger than that exerter by the left sawhorse?
You need three equations to solve for the unknown mass M and the forces at each sawhorse, F1 and F2
One of the equations is F2 = 3 F1
Another equation can be
F1 + F2 = 4 F1 = (12 + 15 + M)g
The third equation should be a moment equation. I suggest you take it about the point of contact of the sawhorse on the right.
F1*2.6 - 15g*2.0 - 12g*1.3 - Mg*0.3 = 0
You should draw yourself a figure of the loaded plank to see where the lever arm length coefficients came from.
Now you have two equations in two unknowns, M and F1. Solve for M. The g factor can be canceled out when solving for M.