Equilibrium problem: sliding door

1. Sep 13, 2010

ZacBones

1. The problem statement, all variables and given/known data

There is a sliding door lifted at a constant rate by two cables, A and B. The sides of the door fit loosely into grooves in the walls.

Dimensions of door: 9 ft tall by 10 ft wide
Weight of the door: 200 lbs
Cable A is attached at the top of the door, 2 ft from the left side
Cable B is attached at the top of the door, 2 ft from the right side
The coefficient of kinetic friction between the door and grooves is 0.3

Suppose cable B breaks. What force must cable A exert to continue raising the door at a constant rate?

2. Relevant equations

Equilibrium equations:

$$\sum$$Fx = 0, $$\sum$$Fy = 0, $$\sum$$M = 0

Friction equation (N is force normal to surface):

f=$$\mu$$ * N

3. The attempt at a solution

It seems as if I don't have enough information to solve the problem.

To highlight the difficulty I have chosen to sum the moments about the center of mass of the door.

$$\sum$$Fx = Nleft - Nright = 0

$$\sum$$Fy = TA - 200 - (.3)Nleft - (.3)Nright = 0

$$\sum$$M = 5*0.3*Nleft - 5*0.3*Nright - 3*TA = 0

What I immediately discover is that the two normal forces are equal, and therefore from the last equation TA = 0. Obviously this is not the correct answer!

I thought that perhaps I could add a couple to the final equation to reflect the fact that the unbalanced door is trying to rotate and can't due to the door frame, but then I end up with four unknowns :-/

Thanks,
-Zac