Find the maximum length of x that will maintain equilibrium

[Richie Smash]

https://www.physicsforums.com/attachments/220647

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

 

[BvU]

Paint me a picture ...

 

[Richie Smash]

I have attached an image

 

[BvU]

Yep. Do they really want the minimum value ?
And: did you already perform the experiment ? two books will also do the trick... especially hardcovers

 

[Richie Smash]

Sorry typo maximum

 

[BvU]

the toppling force formula

sounds nice. No idea what it is.

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).

 

[Richie Smash]

https://physics.stackexchange.com/questions/49926/what-amount-of-force-is-needed-to-topple-a-person

That is where I found the formula for toppling force.

And to make a start I'm going to assume gravity is 10m/s² and then say the force of each plank is 100N

So in my mind it would look something like this 100N *x= 100N *(some multiple)x

 

[BvU]

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

something like this 100N *x= 100N *(some multiple)x

only leads to multiple = 1 I am afraid.

 

[Richie Smash]

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.

 

[jbriggs444]

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?

 

[Richie Smash]

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?

 

[jbriggs444]

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?

 

[Richie Smash]

Each mass is 10 kg

 

[jbriggs444]

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.

 

[Richie Smash]

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.

 

[BvU]

What should I do

do the experiment !

 

[Richie Smash]

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.

 

[jbriggs444]

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.

 

[Richie Smash]

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

 

[Richie Smash]

Might this be on the right track?

 

[tnich]

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.

 

[tnich]

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?

 

[Richie Smash]

with this equation perhaps? L being the length
x²/l²= x²-2lx+l²

 

[tnich]

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?

 

[Richie Smash]

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.

 

[tnich]

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?

 

[Richie Smash]

x*F= (2-x) *F

 

[tnich]

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?

 

[Richie Smash]

It would be gravity acting on it? at the edge of the board?

 

[tnich]

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?

 

[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