Finding the Maximum Length of a .110 kg Remote

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A .110 kg remote that is 21.0 cm long rests on a table. It has a length L1 overhanging the edge of a table. A force of .365 N pushes the power button (located at the end of the remote overhanging the table). How far can the remote extend beyond the table and not tip over? (the mass is evenly distributed)

So far, I've come up with
M2L2g=.365 N+M1L1g
M1+M2=.110 kg
L1+L2=.21 m

and if you change eq. 2 to incoroporate g
(M1+M2)g=1.078 N

It would seem I need one more equation to solve for 4 variables (although I only need one of them solved). I think I'm overlooking something obvious. Any ideas? Thanks.
 
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If L1 is hanging over the edge of the table, that means
L2 = 21.0cm - L1.
Since the mass is evenly distributed, you can say that
L1/21.0 = M1/0.110kg.
Also, M2 = 0.110kg - M1.

Now you just have 1 variable.
 


It seems like you have the right equations set up, but you are missing the equation for torque. Torque is equal to the force applied (in this case, 0.365 N) multiplied by the distance from the point of rotation (the edge of the table) to the point where the force is applied (L1). This can be expressed as T = F * L1.

In order to prevent the remote from tipping over, the torque from the force pushing the power button must be balanced by the torque from the weight of the remote. This can be expressed as T = M2 * L2 * g, where M2 is the mass of the overhanging part of the remote and L2 is the distance from the edge of the table to the end of the remote.

So, your third equation would be T = F * L1 = M2 * L2 * g.

Now, you have three equations with three unknowns (L1, L2, and M2), and you can solve for the maximum value of L1 that will prevent the remote from tipping over. I hope this helps!