Mechanics question involving moments

[xllx]

Homework Statement

A plank of wood is lifted off the ground so it has one side on the ground (A) and the rest of the wood at a slope at 15degrees to the horizontal. The length of the wood (AC) is 2.5m. A force is applied at C at 65degrees to the horizontal to the left keeping it in equilibrium. The mass of the wood is 90g and the centre of mass is at B (this was 1.875m from A when the wood was flat, so presuming that it is at the same point?). Show that the lifting force is 649N

Homework Equations

Moment= Force x Perpendicular distance
Total clockwise moments=total anticlockwise

The Attempt at a Solution

Well I've tried taking moments about A so:
Fsin65 x 2.5 = 90gsin15x1.875

But this doesn't give the right answer.
Do I need to consider the reaction force at A? Or friction?
Any help at all would be greatly appreciated. Many thanks.

 

[N-Gin]

We have two equilibrium conditions:

\sum_{i}\vec{F_{i}}=0

\sum_{i}\vec{\tau_{i}}=0

Forces on the body are actually not relevant, but it's good to consider them. Besides gravitational force \vec{F_{g}} and force \vec{F} acting at point C there is also a reaction force \vec{N} on A and friction \vec{F_{fr}} acting to the right. Vector sum of all these forces equals to zero.

Another thing to consider is the net torque (which is also zero since body is not rotating). Since we don't know anything about reaction force and friction, we will calculate the torque relative to point A in order to eliminate them. (Distances of those forces from point A are zero, so the torques are also zero).

Torque is defined as

\vec{\tau}=\vec{r}\times\vec{F}

\tau=rF\sin{\varphi}

So we have

\vec{\tau_{B}}+\vec{\tau_{C}}=0

Algebraically

\tau_{B}=\tau_{C}

mgl\cos{\alpha}=FL\sin{(\alpha+\beta)}

Where m=90 kg, L=2.5 m, l=1.875 m, \alpha=15^{\circ}, \beta=65^{\circ}. Angles are easily understood by looking at the geometry of the problem.

Finally, lifting force is

F=\frac{mgl\cos{\alpha}}{L\sin{(\alpha+\beta)}}