- #1
physicsgrouch
- 9
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Two builders carry a sheet of drywall up a ramp. They exert vertical forces at the lower corners of the sheet. Assume that W = 2.00m, L = 3.10m, theta = 15.0 degrees, and that the lead builder carries a weight of 123.0N (27.7lb). What is the weight carried by the builder at the rear ?
[tex]\tau[/tex] = F[tex]\bot[/tex]*r
I know that there is no rotation of the drywall, and it is a thin, rectangular plate. The lever arm of the leading builder is (L/2) * cos(15). The torque of this builder is thus 123*cos(15)*(1/2)*(L/2) = 184 N*m.
This means that the torque of the other builder must be the same. However, I can't figure out what the lever arm of this other builder is; is it the same as the leader? It looks like the corners produce different lever arms at an angle, but I'm not sure how to prove that. I thought that the lever arm would be shorter by a distance of W*sin(15), but this is wrong. What am I missing?
[tex]\tau[/tex] = F[tex]\bot[/tex]*r
I know that there is no rotation of the drywall, and it is a thin, rectangular plate. The lever arm of the leading builder is (L/2) * cos(15). The torque of this builder is thus 123*cos(15)*(1/2)*(L/2) = 184 N*m.
This means that the torque of the other builder must be the same. However, I can't figure out what the lever arm of this other builder is; is it the same as the leader? It looks like the corners produce different lever arms at an angle, but I'm not sure how to prove that. I thought that the lever arm would be shorter by a distance of W*sin(15), but this is wrong. What am I missing?