Three boxes are placed on a seesaw. Box #1 (8 kg) is placed at the very end of the left side of the 3.4 m long seesaw. Box #2 (13 kg) is placed at the other end of the seesaw. The pivot point of the seesaw is 0.2 m to the left of the middle of the seesaw.

How far away from the pivot point would you need to place Box #3 (31 kg) for the seesaw to be balanced?

Would you need to place Box #3 to the left or right of the pivot for the seesaw to be balanced?

What is the net torque if Box #3 were placed at the pivot point?

1. 1.4GA - 1.8GB + xGC = 0
   ==> 1.4A - 1.8B + xC = 0
   ==> x = (-1.4A + 1.8B)/C
   ==> x = 0.3935483871
   0.3935483871 m

2. left

3. 1.4GA - 1.8GB + .2GC
   = G(1.4A - 1.8B + .2C)
   = 9.8*(1.4*8 - 1.8*13 + .2*31) (assuming the seesaw is on earth)
   = -58.8 N*m

Thanks!