How to Calculate Torque and Tension in a Diving Board System?

[fp252]

Hello, I've been unable to figure this question about tension and torque out.

1. The question is: calculate the torque about the front support on the right of a diving board exerted by a 70 kg person 3.0 m from that support.

This is a crude diagram of a person on a diving board, where
D = a diver, and || means a diving board support:

____________________________________________D
||<----1.0m---->||<-----------3.0m-------------->
||- - - - - - - - - ||
||- - - - - - - - - ||

2. Relevant equations:
τ = Fd
F = ma
Fup = Fdown / Fleft = Fright

3. I get these equations:

Fup = Fdown
F1 + F2 = Mg + mg (where mg = downward force of diver, Mg = downward force of diving board at center of board)
F1 + F2 = M(9.8) + 686

τcw = τccw
Mg(1.5) + mg(3) = F2(1) (assuming F2 is the upward force for the middle support?)
Mg(1.5) + 70(9.8)(3) = F2(1)
Mg(1.5) + 2058 = F2

I assume the pivot point to be on the far left of the board. The main problem problem with both of these are that I end up with equations that have two variables. Also, I don't have M, which is the mass of the diving board - how would I figure this out?

What's wrong in my process to get the answer?

 

[tiny-tim]

Welcome to PF!

Hi fp252! Welcome to PF!

The question is: calculate the torque about the front support on the right of a diving board exerted by a 70 kg person 3.0 m from that support.

I don't think they're asking you for to look at the whole problem …

I think they literally only want the torque of the diver at the front support … you can ignore everything else.

 

[fp252]

Thank you for replying.

Unfortunately, I'm still stuck. Here's what I have now:
τcw = τccw
mg(4) = F2(1)
686(4) = F2(1)
2744 N = F2

That would seem like an answer, but the force is designated upwards, so it can't be right.

Every other diving board problem I've seen has a setup where I either cannot base it off of it to get the answer of the question above. So far I haven't seen any questions that have a downward force on the front support of the diving board either.

 

[tiny-tim]

Hi fp252!

Thank you for replying.

Unfortunately, I'm still stuck. Here's what I have now:
τcw = τccw
mg(4) = F2(1)
686(4) = F2(1)
2744 N = F2

Yes, that's the right formula and the right answer, but for the wrong question …

you've calculated the normal force at the front support, which is 2774 N CCW, or upwards.

But the question is only asking you for the torque of the diver about the front support, which is (70g)3 J clockwise.

That would seem like an answer, but the force is designated upwards, so it can't be right.

Every other diving board problem I've seen has a setup where I either cannot base it off of it to get the answer of the question above. So far I haven't seen any questions that have a downward force on the front support of the diving board either.

I'm not following you … your own answer gives a positive CCW value (I assume that stands for "counter-clockwise").

The correct way of writing the answer is:
Torque = force times perpendicular distance, clockwise = (70g)3 J clockwise.

 

[fp252]

Sorry, I'm only half awake while trying to do this, so excuse me if I don't get something obvious.

If Torque = force times perpendicular distance, clockwise, then why is it 70g(3), when it should be in Newtons, not grams?

 

[tiny-tim]

Sorry, I'm only half awake while trying to do this, so excuse me if I don't get something obvious.

Get some sleep! :zzz:

If Torque = force times perpendicular distance, clockwise, then why is it 70g(3), when it should be in Newtons, not grams?

My g was gravity!

And torque is force x distance, so it's in joules.