Find the maximum length of x that will maintain equilibrium

[Richie Smash]

What would Isaac Newton have to say about the force acting on the part of the board hanging over?
If you were balancing something on your finger, would you say that gravity acted at the edge of it?

I would say that the force of the part hanging over is acting down, while the force of the pivot of the board is acting up

 

[tnich]

I would say that the force of the part hanging over is acting down, while the force of the pivot of the board is acting up

Trying writing an equation for the force on the left side of the board.

 

[Richie Smash]

l-x *F= moment

 

[tnich]

l-x *F= moment

Write an equation for F.

 

[Richie Smash]

F=m*g

So

l-x *m*g = moment

 

[tnich]

F=m*g

Right. Now what is the mass of the part of the board hanging over?

 

[Richie Smash]

The mass is... m?

 

[tnich]

The mass is... m?

Say the mass of the board is m. How much of that mass is hanging over the edge?

 

[Richie Smash]

m/10 Kg

 

[tnich]

m/10 Kg

What is the fraction of the board hanging over?

 

[Richie Smash]

Sorry that I can't figure this out,

I know the fraction of the board in length hanging over is x/l and I suppose for the mass it would be x/m?

 

[tnich]

Might this be on the right track?

No, that is still not right.

 

[tnich]

Sorry that I can't figure this out,

I know the fraction of the board in length hanging over is x/l and I suppose for the mass it would be x/m?

Let's try it this way. What is the mass per unit length of the board?

 

[Richie Smash]

That would be 5kg per metre

 

[tnich]

That would be 5kg per metre

OK. If x meters of board are hanging over, how many kilograms would that be?

 

[Richie Smash]

x Kg

 

[tnich]

x Kg

No. x is a measure of length in meters. You have the mass per length of the board as 5kg/m, so how would you find the mass of x meters of board?

 

[tnich]

Sorry, got to go.

 

[Richie Smash]

Hello, I have figured out that it would be 5x Kg.

I cannot figure out this problem, so I won't bother anyone, I apologize for so much posts.

 

[jbriggs444]

Hello, I have figured out that it would be 5x Kg.

Correct.

So we know that if x of the plank extends out, the mass of that portion of the plank is 5x.
The center of gravity of that portion of the plank is how far out from the edge?

Hint: that section starts at 0 from the edge and extends to x from the edge. Its average position is ___ from the edge?

[It would have been nice to keep this symbolic, so that the mass of the portion that extends over the edge is $m \cdot \frac{x}{l}$ but we can go with 5x instead]

 

[Richie Smash]

Would the average position be 5/x?

I hope I am not judged for having so many posts in one question, this concept is foreign to me.

 

[tnich]

Would the average position be 5/x?

I hope I am not judged for having so many posts in one question, this concept is foreign to me.

You seem to be just guessing. Try figuring it out.

 

[BvU]

I did it with two books just now and it seems to be about half the length of book with length of about 25cm is the point of max equilibrium.

Very good! It earned you a like. From old days you may remember the see-saw that is horizontal if both loads are the same and at the same distance. No toppling if moments clockwise and anticlockwise cancel. And the support is in the center (something else you can discover from the books experiment: the main contact force in the extreme situation is at the edge of the supporting book). You could draw a free body diagram of the forces that act for the top book and check force and moment balances.

 

[Richie Smash]

cg = ((m1 * x1) + (m2 * x2))/ (m1 + m2).

I found this on a website, but I was only asked to find the COG for the right side, so perhaps I would exclude m2 and x2

and write that the average position of the COG on the right side is (5x *x)/5x

 

[jbriggs444]

cg = ((m1 * x1) + (m2 * x2))/ (m1 + m2).

I found this on a website, but I was only asked to find the COG for the right side, so perhaps I would exclude m2 and x2

and write that the average position of the COG on the right side is (5x *x)/5x

That cannot be right. (5x*x)/5x simplifies to just x. That would put the center of gravity of the right half of the board exactly at the tip of the board.

Hint: The center of gravity of a uniform object is in the middle of the object.

 

[Richie Smash]

Oh I didnt know that!

Then it would simply be x/2 yes?

 

[jbriggs444]

Oh I didnt know that!

Then it would simply be x/2 yes?

Yes!

 

[Richie Smash]

Ah hold on mr J briggs, I'm on to something, the centre of gravity is a the point where the total weight is acting upon it, we know the mass.. we know the length..

The full equation is x *5x *x/2?

Or perhaps just x*x/2

This is for the left side only

 

[jbriggs444]

Ah hold on mr J briggs, I'm on to something, the centre of gravity is a the point where the total weight is acting upon it, we know the mass.. we know the length..

The full equation is x *5x *x/2?

The full equation for what? The torque due to gravity acting on the right half of the top board?

We have the position of the center of gravity: x/2.
We have the mass of the right half. 5x.
Where did that extra x come from?

 

[Richie Smash]

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

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