A diving board 3.00m long is held down at the left end and is supported underneath at a point 1.00m to the right of the hold. A diver weighing 500N stands at the other end. The diving board is of uniform cross section and weighs 280N. Find (a) the force at the support point, and (b) the force at the left-hand side (the hold).
2. Relevant equations
From the hold, ∑τ = 0
3. The attempt at a solution
I know that the answers are 1920N and 1140N respectively. I think that we are suppose to use the formula, (refering from the hold)
∑τ = F(normal from the support)*1.00m + ( – F(diver)*3.00m) + ( – F(diving board)*1.50m(center of mass)) = 0
∑τ = Fn*1.00m – 500N*3.00m – 280N*1.50m = 0
Fn = (500N*3.00m + 280N*1.50m)/1.00m = 1920N
So my issue is that we were taught the chapter before to use ∑τ(z) = I(cm)*α(z) for rigid bodies with uniform mass density. So why don’t they use I(cm)*α(z) for the diving board, where α(z)= g/r ? This of course gives a much different answer.
I checked similar problems like this from other sources, and they don’t use I(cm)*α(z) either.