Torque On a Clock Hanging from a Nail

[tul725]

Homework Statement

A square clock of inertia m is hung on a nail driven into a wall (Figure 1) . The length of each side of the square is ℓ, the thickness is w, and the top back edge of the clock is a distance d from the wall. Assume that the wall is smooth and that the center of mass of the clock is at the geometric center.
Obtain an expression for the magnitude of the normal force exerted by the wall on the clock. Use the notation l for the length ℓ.
Express your answer in terms of the variables m, l, w, d, and acceleration due to gravity g.

https://imgur.com/a/EMf0x

Homework Equations

Torque = r_⊥*F_Gravity - F_wall⊥r = 0

The Attempt at a Solution

I divided the rectangle into 2, lengthwise, and into 2 width-wise. The point at where these lines meet is the center of mass of the clock. I set the point of the nail and the corner of the rectangle as the axis of rotation. I tried to find the lever arm distance of the gravitational force acting on the center of mass of the clock by using trigonometry to find angles. For the normal force of the wall I decided to find the tangential component of the normal force since I already had the lever arm distance in terms of l. I figured since the clock is at rest then the sum of Torque would be equal to zero. The final answer I get is:

https://imgur.com/FGhGKmB
This is not the right answer, and I can't really find where I went wrong.

 

[haruspex]

The method is fine, but hard to say where you are going wrong without seeing your working.
I have not tried to unravel all your trig functions. I get $\frac{mg}2(1-\frac d{\sqrt{l^2-d^2}}(\frac wl +2))$. Try to see if that matches.

 

[haruspex]

just noticed a couple of problems with your answer. The overall expression has the wrong dimension. Looks like you forgot a divisor l.
And the denominator appears to be an angle.

 

[tul725]

just noticed a couple of problems with your answer. The overall expression has the wrong dimension. Looks like you forgot a divisor l.
And the denominator appears to be an angle.

Thanks for the response. You are right, I changed the bottom to L * cos(sin^-1 (d/l)). I still don't get a right answer though. I tried your equation and it didn't seem to be correct either. If I have time later tonight, I'll try clearly writing out my work and taking a picture.

 

[haruspex]

Thanks for the response. You are right, I changed the bottom to L * cos(sin^-1 (d/l)). I still don't get a right answer though. I tried your equation and it didn't seem to be correct either. If I have time later tonight, I'll try clearly writing out my work and taking a picture.

Did you try to determIne whether my equation is the same as your corrected one? I see that they agree on the condition for the force being zero.
A typed in version of your algebra would be preferable. The upload image feature is really for diagrams amd textbook extracts.