Solving Drywall Weight Problems

[fruitl00p]

Homework Statement

Two builders carry a sheet of drywall up a ramp. Assume that W=1.80 m, L= 3.30m, and that the lead builder carries a vertical weight of 183.0 N (41.1 lbs). What is the vertical weight carried by the builder at the rear?

The builder at the rear gets tired and suggest that the drywall should be held by its narrow side. What is the weight he must now carry?

Homework Equations

I think:

Fy = 0
Fx= 0

torque=0

The Attempt at a Solution

Well I don't have a godd idea as to how to do this and what I am doing.

First I said Fy= F1 + F2 - W= 0
torque = F2*l2 - Wlw = 0

l2= 3.30 cos 25
lw= .75 cos 25

W= F2(l2/lw)= 183 N (3.30 cos 25/ .75 cos 25)
W= 805.2 N

F1= W-F2 = 805.2 - 183 = 622.2 N

However my answer is wrong. Since I can't get the first part correct I haven't moved on to the second.

What am I doing wrong?

 

[Mentz114]

You have not described the problem sufficiently. Where did you get the 25deg angle ? A diagram would help.

 

[fruitl00p]

Oh, I'm sorry, the 25 degrees came from the angle of the ramp. However, I finally just figured out how to do the problem.

Thanks for pointing that out though