A Question about Torque Used for Static Equilibrium

[student34]

Homework Statement

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).

Homework Equations

From the hold, ∑τ = 0

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.

 

[SteamKing]

Homework Statement

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).

Homework Equations

From the hold, ∑τ = 0

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.

Because the diving board is 'held down' in static equilibrium. ∑τ(z) = I(cm)*α(z) is used for bodies undergoing rotational acceleration, which is clearly not the case here.

 

[student34]

Because the diving board is 'held down' in static equilibrium. ∑τ(z) = I(cm)*α(z) is used for bodies undergoing rotational acceleration, which is clearly not the case here.

Yes, but I thought that Einstein's equivalence principle implies that acceleration is equivalent to force. For example, I thought that a body accelerating at 9.81 m/s^2 in space is equivalent to the body on Earth at rest.

If that is true, then how is the force of g on all particles of the diving board different than angular acceleration of a moving diving board driven by the support force?

 

[SteamKing]

The diving board may deflect locally under the action of the forces applied to it (i.e., the board is not a perfectly rigid body), but it is not able to rotate bodily since it is restrained from doing so. This is a simple problem in statics: there is no need to drag Einstein into it.

N.B.: a body accelerating at 9.81 m/s^2 in space is an example of rectilinear, not rotational, motion. You analyze rectilinear motion using different equations than those for analyzing rotational motion. The equations for the two types of motion may be analogous, but they are not the same.