Find the maximum length of x that will maintain equilibrium

[jbriggs444]

Ah yes I see you're right, I understand now finally... wow..

So would you confirm 5x²/2 is onto something?

Yes.

So that gives us the [clockwise] torque due to gravity acting on the overhanging right end of the top board. Now we have to figure out the counter-clockwise torque from the remaining left half of the top board.

The mass of the left half is ___ ?
The position of the center of gravity of the left half is ___ ?
The torque due to gravity acting on the left half is then ___ ?

 

[Richie Smash]

the mass of the left half would be 5(l-x) and the postion of the center of gravity would be (l-x)/2

So the Moment would be (5l²-10lx+5x²)/2

 

[tnich]

Ah yes I see you're right, I understand now finally... wow..

So would you confirm 5x²/2 is onto something?

Yes! Now you have the moment acting on the right side of the board, according to the diagram in terms of the length x of that side.
What is the length of the other side?

 

[tnich]

Yes! Now you have the moment acting on the right side of the board, according to the diagram in terms of the length x of that side.
What is the length of the other side?

Oops, sorry, missed the last two posts. Yes, you have the right expression for the moment on the other side.

 

[jbriggs444]

the mass of the left half would be 5(l-x) and the postion of the center of gravity would be (l-x)/2

So the Moment would be (5l²-10lx+5x²)/2

Good. That looks entirely correct.

The tipping point is when the torques from the two sides are equal. You have formulas for both torques. Can you write down an equation that states that they are equal?

 

[Richie Smash]

Yes
(5x²-10lx+5l²)/2=5x²/2

And i know l =2
So x ultimately is 1?

 

[jbriggs444]

Yes
(5x²-10lx+5l²)/2=5x²/2

And i know l =2
So x ultimately is 1?

Yes.

One could phrase it more generally: $x=\frac{l}{2}$. That is, regardless of what the board length is, the tipping point is when exactly half of the board hangs over.