Static equlibrium on a diving board

[ph123]

A diving board of length 3.00 m is supported at a point a distance 1.00 m from the end, and a diver weighing 550 N stands at the free end. The diving board is of uniform cross section and weighs 255 N. Find the force at the end that is held down.

In part a, I found the normal force at the support point, 1m from the end held down.

N = [550N(3m) + 255N(1.5m)]/1.0m = 2030N

Part b wants the force at the end held down. I don't really know how to approach this part, since the end held down is at my x=0m point. I found the center of mass to be 2.52m. I know the torque calculated should be about the support point.

Sum torques = N(L/3) + N(0) - W_b(L) - W_d(L/2)
where W_b = weight of the board
W_d = weight of the diver

But since the end held down is at zero, this term drops out and I'm left with what I got for the first part.

[hage567]

I don't really know how to approach this part, since the end held down is at my x=0m point.

You want the distance from your pivot point to where the force is acting. Just because the end of the board is at x=0, doesn't mean that the distance to that point is zero.

[andrevdh]

Your choice of calculating the torques about the fixed point eliminates the force you want to calculate, as you discovered. The trick is shift this point to another place. Since the board is in rotational equilibrium the sum of the torques about any point needs to be zero.