Simple Statics Equilbrium [SOLVED]

[kevinr]

[SOLVED] Simple Statics Equilbrium

Homework Statement

A diving board of length 3m is supported at a point 1m from the end and a diver weighing 450N stands at the free end. The diving board is of uniform cross section and weighs 285N

Pic:
http://img408.imageshack.us/my.php?image=yf1124no0.jpg

Q) Find the magnitude of the force at the support point.

Homework Equations

Statics Equilbrium?

The Attempt at a Solution

Ok i am kind of confused on this one. I know that there are 2 forces down which is 450N + Force of the push down from supported end + the weight of the board = support force. I hope that's right.

Im not sure about the exact equation to use. Would it be Torque = 0? But even than i get 2 unknowns (Support Force and force of supported end)

But how could i do this without given the supported force down?

Thanks!

 

[Doc Al]

Ok i am kind of confused on this one. I know that there are 2 forces down which is 450N + Force of the push down from supported end + the weight of the board = support force. I hope that's right.

Unfortunately, the pic link doesn't seem to be working. But don't forget to consider the weight of the board itself.

Im not sure about the exact equation to use. Would it be Torque = 0?

Yes, the net torque about any point must be zero. That's one of the conditions for equilibrium.

But even than i get 2 unknowns (Support Force and force of supported end)

What's another condition for equilibrium? What must the sum of the forces equal?

You can also apply the torque = 0 condition at two different pivot points.

 

[kevinr]

Hello,
I reuploaded image for you:
http://allyoucanupload.webshots.com/v/2003386143771893898

Now i am still getting this wrong. I figured Torque in this way:
T = TF(sp) - TFg - weight of board (set starting point where Fn acts so this is 0)
This means:
F(sp) = 3*Fg + weight of board
so:
F(sp) = 3 * 450 + 285 = 1635 N but this is wrong and i don't understand why.

 

[Doc Al]

Now i am still getting this wrong. I figured Torque in this way:
T = TF(sp) - TFg - weight of board (set starting point where Fn acts so this is 0)

I'm not quite understanding what you are doing. First thing to do is identify all the forces acting on the board and where they act:
F(sp) = I assume this is the downward support force that acts at the left end of the board?
Fg = The weight of the girl, which acts at the right end of the board?
W = weight of the board, which acts at the middle of the board.
Fn = The upward force located 1 m from the left end.

OK, If you chose the location of Fn as your pivot, then you must set clockwise and counterclockwise torques equal to each other. And you must measure their distances from the pivot point in order to calculate torque.

This means:
F(sp) = 3*Fg + weight of board
so:
F(sp) = 3 * 450 + 285 = 1635 N but this is wrong and i don't understand why.

Redo this with the correct distances.

 

[kevinr]

Ok still seems to come out to the same thing. If i use the variables you have, i have to set my pivot at left end if i want to find Fn.

Fn creates counter clockwise direction (+)
W creates clockwise (-)
Fg creates clockwise (-) at radius of 3.

So here's my equation:
Torque = 1 * Fn - 1 * W -3 * Fg (looking for Fn)
Fn = W + 3 * Fg
Fn = 285 + 3 * 450 = 1685 N

But this seems to be incorrect.

Thanks for your time!

 

[Doc Al]

So here's my equation:
Torque = 1 * Fn - 1 * W -3 * Fg (looking for Fn)

The weight of the board acts at its center of mass, which is not 1 m from the left end. Fix this.

 

[kevinr]

Ah! Thank you!