Find the maximum length of x that will maintain equilibrium

AI Thread Summary
The discussion revolves around determining the maximum length \( x \) of a plank that can extend over the edge of another plank while maintaining equilibrium. Participants explore concepts of moments and forces, emphasizing that the pivot point is at the edge of the lower plank. The calculations involve balancing the clockwise and counter-clockwise moments, leading to the conclusion that the maximum length \( x \) before toppling occurs is approximately 1.33 meters. The conversation highlights the importance of understanding the center of gravity and the distribution of mass in the planks. Ultimately, the participants arrive at a solution through collaborative problem-solving and experimentation.
Richie Smash
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Homework Statement


Two uniform planks each of mass 10Kg and length 2m, are arranged as shown in the picture.

Find the maximum value of x for which the top plank will remain in equlibrium.

Homework Equations


F= (mgl)/h

The Attempt at a Solution


I honestly have no clue, but I will try:

If the mass is equal, that must mean.. As the plank goes further and further over the edge, the mass over the edge is steadily getting higher...

Perhaps the edge of the bottom block acts as a pivot, and there must be a point where the clockwise moments will outweigh the anti-clockwise moments...

They want me to find the length of x that is just before toppling...
I was thinking perhaps I can use the toppling force formula.

I'm very unfamiliar with this topic, honestly.. I am trying but I don't have any concrete answers, just random guesses and thoughts
 

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Paint me a picture ...
 
I have attached an image
 
Yep. Do they really want the minimum value ?
And: did you already perform the experiment ? two books will also do the trick... especially hardcovers
 
Sorry typo maximum
 
Richie Smash said:
the toppling force formula
sounds nice. No idea what it is.
Richie Smash said:
Perhaps the edge of the bottom block acts as a pivot, and there must be a point where the clockwise moments will outweigh the anti-clockwise moments
sounds good too. Make a start and see where you stumble (topple... if you forgive me the pun).
 
I see. But in your exercise the thickness of the boards isn't given, so you are allowed to treat them as lines. And the only force in play is gravity, downward.

I take it we must assume the lower board is fixed ? In that case you can do the experiment with just one book, ruler or anything else rectangular !

Your equation
Richie Smash said:
something like this 100N *x= 100N *(some multiple)x
only leads to multiple = 1 I am afraid.
 
Last edited:
What should I do sir? I'm thoroughly confused, I believe it has something to do the the principle of moments and I'm pretty sure I have to assume g to be 10 m/s2 at this level, it's hard to figure out because no length is given except of the entire plank, and yes the bottom plank does remain fixed.
 
  • #10
Richie Smash said:
What should I do sir? I'm thoroughly confused, I believe it has something to do the the principle of moments and I'm pretty sure I have to assume g to be 10 m/s2 at this level, it's hard to figure out because no length is given except of the entire plank, and yes the bottom plank does remain fixed.
You should not have to know the value of g at all. When solving problems with algebra, a good rule of thumb is to NOT substitute in the values of any known constants until you have arrived at a symbolic formula for the desired result.

One way to proceed would be to compute the moment of force for just the portion of the upper board that extends beyond the end of the lower board. And then compute the moment of force for the remainder of the upper board, the portion that does not extend past the end. If the two moments are equal, the board is on the verge of tipping.

If the upper board has total length l and sticks out a distance x, what fraction of the board's mass extends beyond the lip of the lower board?
 
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  • #11
I think I'm onto something, it would be x/2 for the length that's sticking out, and x/10 for the mass yes?
 
  • #12
Richie Smash said:
I think I'm onto something, it would be x/2 for the length that's sticking out, and x/10 for the mass yes?
Where does the /10 come from?
 
  • #13
Each mass is 10 kg
 
  • #14
Richie Smash said:
Each mass is 10 kg
What did I tell you about plugging in known values into equations?

Don't do it!

It's simple. If the plank has half its length sticking out over the edge, it has half its mass sticking out over the edge. No factors of 10 involved.
 
  • #15
Yes you are right
I have a new equation

x/l * y/m = (m-y) (l-x)

where m is the mass and y is the fraction of the mass.
 
  • #16
Richie Smash said:
What should I do
do the experiment !
 
  • #17
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.
 
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  • #18
Richie Smash said:
Yes you are right
I have a new equation

x/l * y/m = (m-y) (l-x)

where m is the mass and y is the fraction of the mass.
If x/l is the fraction of the length that hangs over then x/l is also the fraction of the mass that hangs over.

y/m does not even make sense. It would have units of inverse mass.
m-y also does not make sense. You cannot subtract a mass from a pure number.
 
  • #19
Alright well if x/l is the fraction for both then my new expression is

x2/l2= x2-2lx+l2

Then I used l= 2 and have solved for x
And x = 4m, and x= 1.33m

And since the total length is 2m, it cannot be 4m, so the maximum length it can be at equilibrium is therefore 1.33 m
 
  • #20
Might this be on the right track?
 
  • #21
Richie Smash said:
https://www.physicsforums.com/attachments/220647
Perhaps the edge of the bottom block acts as a pivot, and there must be a point where the clockwise moments will outweigh the anti-clockwise moments...
I think you are on to something here.
 
  • #22
tnich said:
I think you are on to something here.
How would you set up an equation to find the value of x where the moments are equal?
 
  • #23
with this equation perhaps? L being the length
x2/l2= x2-2lx+l2
 
  • #24
Richie Smash said:
with this equation perhaps? L being the length
x2/l2= x2-2lx+l2
I don't know what you are trying to represent with that equation, but I don't see any moments in it. What is the moment of a force?
 
  • #25
a moment is the turning effect of a force, and it is calculated by the distance from the line of action of the force to the pivot times the size of the force.
 
  • #26
Richie Smash said:
a moment is the turning effect of a force, and it is calculated by the distance from the line of action of the force to the pivot times the size of the force.
OK. Now what is the moment acting on each side of the pivot, given x?
 
  • #27
x*F= (2-x) *F
 
  • #28
Richie Smash said:
x*F= (2-x) *F
You are getting closer - both sides now look something like moments. Let's just use L for the length of the board for now.
If the board has uniform density, and a length x is hanging over the edge, at what point would you say the force acts on the part hanging over?
What is the force acting on that part of the board?
 
  • #29
It would be gravity acting on it? at the edge of the board?
 
  • #30
Richie Smash said:
It would be gravity acting on it? at the edge of the board?
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?
 
  • #31
tnich said:
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
 
  • #32
Richie Smash said:
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.
 
  • #33
l-x *F= moment
 
  • #34
Richie Smash said:
l-x *F= moment
Write an equation for F.
 
  • #35
F=m*g

So

l-x *m*g = moment
 
  • #36
Richie Smash said:
F=m*g
Right. Now what is the mass of the part of the board hanging over?
 
  • #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]
 

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