Slightly off on the answer to a problem, but why?

[Brianjw]

Two friends are carrying a crate of mass 200 kg up a flight of stairs. The crate has length 1.25 m and height 0.50 m, and its center of gravity is at its center. The stairs make a 45.0 angle with respect to the floor. The crate also is carried at an 45.0 angle , so that its bottom side is parallel to the slope of the stairs . The force each person applies is vertical.

What is the magnitude of the upper person's force?

I keep getting 1176, but the answer is supposed to be 588 which is just half of my answer?

Also, for the next question:

What is the magnitude of the force applied by the bottom person?

I'm not able to get the 1370N that's its supposed to be, I'm trying to use the sum of the torques about the point of F2 which is the upper person's force, and that doesn't seem to work

[arildno]

Hm.. you must have made a mistake in your torque equation about the top person.
Using "t" and "b" as subscripts for the persons, \theta as the angle to the horizontal, "l" as the length, "h" as the height, "m" for the mass, "g" as the acceleration due to gravity, \vec{t} as the unit tangent along the bottom side, \vec{n} as the unit normal, we have the following relations:
\vec{t}=\cos\theta\vec{i}+\sin\theta\vec{k}
\vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{k}
\vec{k}=\sin\theta\vec{t}+\cos\theta\vec{n}
\vec{n}\times\vec{t}=\vec{j}
Balance of forces:
F_{b}+F_{t}=mg
Balance of torques about the top person:
-l\vec{t}\times{F}_{b}\vec{k}+(-\frac{l}{2}\vec{t}+\frac{h}{2}\vec{n})\times(-mg\vec{k})=\vec{0}

This should give you the correct answers.