Torque about different points gives different answers

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Homework Statement



A solid cube is placed on a horizontal surface. The coefficient of friction between them is μ, where μ < 1/2. A variable horizontal force is applied on the cube's upper face, perpendicular to one edge and passing through the mid-point of edge, as shown in figure. The maximum acceleration with which it can move without toppling is

[URL]http://203.196.176.41/VLEBT_RootRepository/Resources/d5595700-11ed-4e99-9708-9bc47ea580ba.gif[/URL]


The Attempt at a Solution



I am getting two different answers with different approaches:

1) Torque about the point of rotation when the cube is about to topple

mg*L/2 = P*L + ma*L/2
P = mg/2 - ma/2

ma = P - μmg
3ma/2 = mg/2 - μmg

a = g(1-2μ)/3

2) Torque about the centre of mass

P*l/2 + μmg*L/2 = mg*L/2
P = mg - μmg

a = g(1-2μ)

Why different answers?
 
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Answers and Replies

  • #2
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Hello Abdul,
it's a tricky problem, btw.
When you wrote
1) torque about the point of rotation when the cube is about to topple
mg*L/2 = P*L + ma*L/2

I think this the net sum of the torques.
Why ma*L/2 ?

It's already included in P.

I solved your exercise, but I got two different results, both for part 1) and part 2). I mean different from yours, both 1) and 2).
I'm really not sure about MY results, but they are the same in part 1) and 2) that is about the CoM and about the pivoting point. This is at least reassuring, but I'm not sure.

In any case, as a check, you may assume μ=0.
In this case, the torques about the pivoting point are:
PL and mgL/2, you'll agree.
so
PL-mgL/2 = 0
maL-mgL/2 = 0 (because μ=0 )
a = g/2

that I think makes sense at least intuitively.
You should obtain something that give a=g/2 when μ=0

If you like, I'll hand the solution, which is not obvious in part 2) I think.
Regards
 
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  • #3
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Hello Abdul,
it's a tricky problem, btw.
When you wrote


I think this the net sum of the torques.
Why ma*L/2 ?

It's already included in P.

How?

In any case, as a check, you may assume μ=0.

Why do you assume μ=0? The torque due to friction about the point of rotation is 0.
 
  • #4
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How?

Because P is already (μmg + ma)
The torque contribute of ma is already included in P.

Why do you assume μ=0?
To simplify things.
First let's make things work with μ = 0 , then we complicate them.
The final formula must work as well with μ = 0

The torque due to friction about the point of rotation is 0.
Yep.
But only when the center of rotation is the pivoting point.
 
  • #5
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So what is your final answer?
 
  • #6
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The answer I get is
[tex]a = g \left({1 \over 2} - \mu \right)[/tex]
 
  • #7
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@Abdul Quadeer:
Abdul Quadeer said:
mg*L/2 = P*L + ma*L/2

I hope I don't get you wrong; probably the ma part is the fictitious force (in the reference frame of the center of mass). If so, then it should be: mg*L/2 + ma*L/2 = P*L, as the fictitious force points to the left (if P points to the right). You should get the same answer as 2).

@Quinzio:
If mu = 0, there is no way the cube can move without toppling. Check your solution for that case. You forgot the normal force N by the ground :wink: Things will get tricky in the reference frame of the ground; consider the reference frame of center of mass.
 
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  • #8
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@Abdul Quadeer:


I hope I don't get you wrong; probably the ma part is the fictitious force (in the reference frame of the center of mass). If so, then it should be: mg*L/2 + ma*L/2 = P*L, as the fictitious force points to the left (if P points to the right). You should get the same answer as 2).

@Quinzio:
If mu = 0, there is no way the cube can move without toppling. Check your solution for that case. You forgot the normal force N by the ground :wink: Things will get tricky in the reference frame of the ground; consider the reference frame of center of mass.

Hello,
why do you say that if mu=0 you can't move it without toppling ?
You could suggest your idea.

My formula say that with mu=0 the maximum acceleration without toppling is g/2.
What you say happens if mu>=0.5 --> accel=0. That means that any force on the top of the cube will make it topple.

You can make this experiment.
Take a cube of any material (not too light, eg. wooden) and a sheet of rubber (likely mu>0.5). Cut a piece of rubber as wide as a face of the cube, than glue it to a face of the cube. Then put the cube on the rest of the rubber sheet and with your finger push one of the upper edges. See if you can make it slide or it will inveitably topple.
 
  • #9
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I'm sorry, wrong statement :uhh: It should be: If mu = 0, a cannot be g/2.

Consider your answer + your analysis when N (the normal force) is taken into account (see picture). In the reference frame of the center of mass, the cube stays at rest, so all the conditions about forces and torques in equilibrium can be applied (notice: in the reference frame of the ground, the cube is moving with acceleration, and that makes things tricky in this reference frame!).

We have 3 forces: P, F (which is mg + the fictitious force ma) and N. As you computed, a=g/2, so the forces should look like in the picture. The important point is, they do not converge into 1 point (the three forces P, F and N), which violates the equilibrium condition.
 

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  • #10
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I'm sorry, wrong statement :uhh: It should be: If mu = 0, a cannot be g/2.

Consider your answer + your analysis when N (the normal force) is taken into account (see picture). In the reference frame of the center of mass, the cube stays at rest, so all the conditions about forces and torques in equilibrium can be applied (notice: in the reference frame of the ground, the cube is moving with acceleration, and that makes things tricky in this reference frame!).

We have 3 forces: P, F (which is mg + the fictitious force ma) and N. As you computed, a=g/2, so the forces should look like in the picture. The important point is, they do not converge into 1 point (the three forces P, F and N), which violates the equilibrium condition.

I don't understand what is your opinion.
Are you saying that the correct formula is a = g(1-μ) ?
It's not important for me who's right or wrong, but just to understand the proble in the correct way.
In your diagram appear the force "ma", which I think is not there in the moment the cube is about to topple.
In my diagram there were of course only P and mg, and of course the toppling condition was g/2.
The force "ma" I think should not be there.
As this problem is intriguing I would like to make a real experiment if I found some suitable objects like a cube, and possibly I will record a video of the experiment.

Anyone reading that has an opinion on this problem ?
 
  • #11
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The ma force is the fictitious force, arising from the choice of reference frame (see my previous post, if you didn't read carefully). If your diagram contains only P and mg (and N presumably), then you probably chose the reference frame of the ground. In that reference frame, the cube is moving with acceleration, and this makes things more complicated. The simple formula torque = moment of inertia x angular acceleration doesn't always work. There are many subtleties behind it.

My answer is a = g(1-2mu).

P.S.:
To further elaborate on my points of the subtleties: The reason we can deduce the formula Torque = Moment of Inertia x Angular Acceleration is that we assume that the pivoting point is fixed and the body is a rigid body (so that the whole body rotates around the pivoting point with uniform angular acceleration). From this formula, we deduce the second equilibrium condition: Torque = 0 as Angular Acceleration must be 0 under equilibrium (the same for F = ma; as a must be 0 under equilibrium, F has to be 0).

However, in this problem, in the reference frame of the ground, the pivoting point (right lower edge of the cube) is moving and the cube doesn't simply rotate around any point. That is, the cube is not even in equilibrium. Incautiously applying the equilibrium condition here certainly messes things up. On the other hand, if we choose the reference frame of the center of mass, the cube would be under equilibrium, and things are way easier.
 
  • #12
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The ma force is the fictitious force, arising from the choice of reference frame (see my previous post, if you didn't read carefully). If your diagram contains only P and mg (and N presumably), then you probably chose the reference frame of the ground. In that reference frame, the cube is moving with acceleration, and this makes things more complicated. The simple formula torque = moment of inertia x angular acceleration doesn't always work. There are many subtleties behind it.

My answer is a = g(1-2mu).

P.S.:
To further elaborate on my points of the subtleties: The reason we can deduce the formula Torque = Moment of Inertia x Angular Acceleration is that we assume that the pivoting point is fixed and the body is a rigid body (so that the whole body rotates around the pivoting point with uniform angular acceleration). From this formula, we deduce the second equilibrium condition: Torque = 0 as Angular Acceleration must be 0 under equilibrium (the same for F = ma; as a must be 0 under equilibrium, F has to be 0).

However, in this problem, in the reference frame of the ground, the pivoting point (right lower edge of the cube) is moving and the cube doesn't simply rotate around any point. That is, the cube is not even in equilibrium. Incautiously applying the equilibrium condition here certainly messes things up. On the other hand, if we choose the reference frame of the center of mass, the cube would be under equilibrium, and things are way easier.
 
  • #13
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My answer is a = g(1-2mu).

Perfect! :smile:

I hope I don't get you wrong; probably the ma part is the fictitious force (in the reference frame of the center of mass). If so, then it should be: mg*L/2 + ma*L/2 = P*L, as the fictitious force points to the left (if P points to the right). You should get the same answer as 2).

Yeah I made that mistake. Thanks for clearing it!!!
 

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