Find the maximum length of x that will maintain equilibrium

In summary: I'm sorry, I cannot reply to questions or provide help with problems. My purpose is to provide a summary of the conversation. In summary, the conversation discusses the problem of finding the maximum value of x for which the top plank will remain in equilibrium in a setup with two uniform planks of equal mass and length. The conversation includes the use of the toppling force formula, the principle of moments, and the idea of computing the moment of force for the portion of the plank that extends beyond the edge. The final conclusion is that, after performing the experiment, it seems that the maximum value of x is about half the length of the bottom plank.
  • #36
Richie Smash said:
F=m*g
Right. Now what is the mass of the part of the board hanging over?
 
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  • #37
The mass is... m?
 
  • #38
Richie Smash said:
The mass is... m?
Say the mass of the board is m. How much of that mass is hanging over the edge?
 
  • #39
m/10 Kg
 
  • #40
Richie Smash said:
m/10 Kg
What is the fraction of the board hanging over?
 
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  • #41
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?
 
  • #42
Richie Smash said:
Might this be on the right track?
No, that is still not right.
 
  • #43
Richie Smash said:
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?
 
  • #44
That would be 5kg per metre
 
  • #45
Richie Smash said:
That would be 5kg per metre
OK. If x meters of board are hanging over, how many kilograms would that be?
 
  • #46
x Kg
 
  • #47
Richie Smash said:
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?
 
  • #48
Sorry, got to go.
 
  • #49
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.
 
  • #50
Richie Smash said:
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]
 
  • #51
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.
 
  • #52
Richie Smash said:
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.
 
  • #53
Richie Smash said:
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.
 
  • #54
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
 
  • #55
Richie Smash said:
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.
 
  • #56
Oh I didnt know that!

Then it would simply be x/2 yes?
 
  • #57
Richie Smash said:
Oh I didnt know that!

Then it would simply be x/2 yes?
Yes!
 
  • #58
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
 
  • #59
Richie Smash said:
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?
 
  • #60
Ah yes I see you're right, I understand now finally... wow..

So would you confirm 5x2/2 is onto something?
 
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  • #61
Richie Smash said:
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 ___ ?
 
  • #62
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
 
  • #63
Richie Smash said:
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?
 
  • #64
tnich said:
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.
 
  • #65
Richie Smash said:
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?
 
  • #66
Yes
(5x2-10lx+5l2)/2=5x2/2

And i know l =2
So x ultimately is 1?
 
  • #67
Richie Smash said:
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.
 
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