# Torque and equilibrium

I am a little stuck on the concept involved in part b) of the following problem:

A $5 m$ long road gate has a mass of $m = 15 kg$. The mass is evenly distributed in the gate. The gate is supported at point A by a rod attached to the ground below the gate. Point A is located $20 cm$ from the left end. The gate rotates around a point B located $80 cm$ from the right end, also supported by a rod attached to the ground. At the right end of the gate, a block of mass $25 kg$ is placed. The block's mass center is $15 cm$ from the right end.

a) Find the forces at A and B

b) How much force must be applied downward at the right end of the gate in order to swing the gate open?

OK, so for part a) I simply used the fact that since we are in equilibrium we must have zero net forces and zero net torque. This gave me $F_A = 23 N$ and $F_B = 369 N$, which is correct. In part b), I know the answer is supposed to be $113 N$, and I have been told that in order to find this I have to find out what force I have to apply to the gate in order to get $F_A = 0$. But this is a bit conceptually confusing to me. If the system is in equilibrium initially, wouldn't then just the slightest bit of force added cause the system to lose its equilibrium and start moving?

If anyone can explain this to me, I would be very grateful.

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The assumption is that there is a frictional force at point A that resists swinging of the gate. The amount of frictional force available is proportional to the downward force at that point. It's just like if you have a block on the ground. You push on the block horizontally, the block resists moving due to the frictional force. The heavier the block, the more force you need to overcome friction. If you could make the block weightless, then it would move freely.

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Thanks a lot! It make sense then. I just didn't interpret the force at A as a static frictional force based on the way the problem was stated. Appreciate it a lot! – user12277 Oct 5 '12 at 23:35