Question

Two builders carry a sheet of drywall up a ramp.

Assume that W = 2.00 m, L = 3.10 m, theta = 23.0

Answer 1

This is basically a geometry problem.

Consider a bisector parallel to the W side that goes through the center of mass CM. as the sheet is rotated, the point on the bottom of the board that is directly below (vertically) the CM slides toward the back. The angle between vertical and our bisector is ?; we have a little triangle consisting of a vertical line through the CM, the lower half of the bisector, and the length along the bottom of the sheet representing the shift in the CM along the bottom. Let's call that "x"; then
tan? = x / (W/2) = 2x / W, which rearranges to x = Wtan? / 2
Hopefully you could follow that, and hopefully I'm right!
Now it's basic statics. Sum the moments about the vertical through the CM.
On the left, we've got the unknown force F and distance (L/2 - x)cos?.
On the right we've got the known force P and distance (L/2 + x)cos?. Then
F(L/2 - x)cos? = P(L/2 + x)cos? ? cos? cancels; sub for x from above
F(L/2 - Wtan?/2) = P(L/2 + Wtan?/2) ? multiply through by 2
F(L - Wtan?) = P(L + Wtan?) ? and, finally,
F = P(L + Wtan?) / (L - Wtan?)
This looks like useful general result. For our values, then,
F = 181N * (3.1m + 2.0m*tan23) / (3.1m - 2.0m*tan23) = 318 N

B) Now F = P * (2.0m + 3.1m*tan23) / (2.0m - 3.1m*tan23) = 4.85 * P
But what is P? It can't be the same as before. The front guy carries less, the rear guy carries more. We know from A that the weight of the sheetrock is
F + P = 181N + 318N = 499 N, so
P = 499N - F. Plug that in:
F = 4.85(499N - F) = 2420N - 4.85F
5.85F = 2420N
F = 414 N
(so the front guy gets only 85 N !)