Torque Application: Simple Q&A Diagram

[WildEnergy]

in this diagram i made:

[PLAIN]http://img201.imageshack.us/img201/1856/physicsqa.png

left: simple application of linear force at center of mass
middle: application of linear force offset from center of mass gives rise to torque around center of mass
right: how to explain the forces and torques here?

most books just say that the contact point with the fixed object becomes the fulcrum
but i want to understand how that happens

i have drawn about 20 diagrams - and none of them make sense

the closest to an explanation i have got:
as the torque is a force applied to all points in the mass
the contact force is 1/2 the torque
an equal and opposite torque reaction somehow adds this 1/2 to the original ...

can anybody explain this?

 

[WildEnergy]

shameless bump
can anybody help with this?

 

[Dickfore]

most of the books are wrong then.

 

[WildEnergy]

most of the books are wrong then.

I doubt they are wrong :-)

But they don't explain the physics behind the pivot

I am trying to understand the hidden force diagram "behind" the typical force diagram in books

 

[WildEnergy]

with a force of 10N and a distance from CoM of 2m the torque would be 20Nm
but with a pivot 2m from the center of mass (4m from the force) the torque is 14Nm
that is what I don't understand

 

[kuruman]

The right diagram, if I understand it correctly, shows basically what you do when you open a door. If you understand how a door opens, you should understand this. The fixed point marked by "X" is the hinge (call it fulcrum if you wish). As you apply a force away from the hinge, you generate a torque that rotates the center of mass about the hinge. Note that there is just the right force exerted at the hinge to keep the CM in a circular path centered at the hinge. Did I answer your question?

 

[kuruman]

with a force of 10N and a distance from CoM of 2m the torque would be 20Nm
but with a pivot 2m from the center of mass (4m from the force) the torque is 14Nm
that is what I don't understand

You need to understand that, because torques generally depend on the point about which you calculate them. A notable exception is a "couple", i.e. equal and opposite forces, in that the torque they generate is independent of the reference point. Here you have a single force so it matters about what point you calculate the torque. It is best to calculate the torque about the fixed point, then it would be 40 Nm (not 14 Nm). Why is is the best point? Because then you will be able to calculate the angular acceleration of the object using Newton's 2nd Law for rotations and from this the linear acceleration of the CM.

 

[WildEnergy]

It is best to calculate the torque about the fixed point, then it would be 40 Nm (not 14 Nm)

[PLAIN]http://img441.imageshack.us/img441/3527/physicsqc.png

so in my physics simulator - when it says Torque = 14.054054 Nm
it is measuring the torque around the center of mass

whereas the torque around the pivot will be 40 Nm

the problem here is that I know how to work out the answer according to the book
but I don't want to use that approach

I want to use forces only - so a pivot would be modeled as a reaction force etc.

that is what I am struggling with - that is why I drew a diagram with 3 rectangles
I want to understand the way forces and torques work when adding a pivot

not how to calculate things given a pivot

Am I making any sense?

 

[kuruman]

[PLAIN]http://img441.imageshack.us/img441/3527/physicsqc.png

so in my physics simulator - when it says Torque = 14.054054 Nm
it is measuring the torque around the center of mass

whereas the torque around the pivot will be 40 Nm

the problem here is that I know how to work out the answer according to the book
but I don't want to use that approach

I want to use forces only - so a pivot would be modeled as a reaction force etc.

that is what I am struggling with - that is why I drew a diagram with 3 rectangles
I want to understand the way forces and torques work when adding a pivot

not how to calculate things given a pivot

Am I making any sense?

So you need to replace the pivot with a pair of forces that will do the same job as adding a pivot in the simulation. Is that it? Can you post exactly what is given in this problem, i.e. dimensions of object, forces applied etc. I can only guess what these are.

 

[Dickfore]

I doubt they are wrong :-)

But they don't explain the physics behind the pivot

I am trying to understand the hidden force diagram "behind" the typical force diagram in books

Please provide references where the statement you claimed had been made.

 

[WildEnergy]

thanks for taking a look

I = 10.8333 (at center of mass)
mass = 5 kg
force applied 2 m away from CoM
pivot 2 m away from CoM
force, CoM and pivot are colinear
force applied is 10 N (perpendicular to force-CoM)

I am planning how to build a physics simulator
but there is no concept of a pivot - only rigid bodies (pos, vel, mass, moi, friction)
so a pivot could be modeled as (vel = 0, mass = moi = 9e99)

 

[kuruman]

I = 10.8333 (at center of mass)

You mean "about" the center of mass. I assume the length of the rectangle is 4.0 m, correct? If the rectangle represents a uniform rod, then its moment of inertia about the CM should be
I = ML²/12 = 5*4²/12 = 6.67 kg m², not 10.833. So what is this I = 10.833?
Also, as the rectangle rotates, does the force stay perpendicular to the rectangle (i.e. rotates with it) or is its direction fixed?

 

[WildEnergy]

You mean "about" the center of mass. I assume the length of the rectangle is 4.0 m, correct? If the rectangle represents a uniform rod, then its moment of inertia about the CM should be
I = ML²/12 = 5*4²/12 = 6.67 kg m², not 10.833. So what is this I = 10.833?
Also, as the rectangle rotates, does the force stay perpendicular to the rectangle (i.e. rotates with it) or is its direction fixed?

the rectangle is 5x1
the force and pivot are inset 0.5
10.833 is the default I_cm calculated by my (2D) simulator
maybe the simulator treats it as 5x1x1 I am not sure

 

[WildEnergy]

[PLAIN]http://img826.imageshack.us/img826/2307/physicsqd.png

 

[kuruman]

OK, so the length of the rectangle is 5.0 m not 4.0 m as I originally thought. Now I agree with the moment of inertia given by the simulation. What about the force? What is its direction as the object rotates? Also, is gravity in the picture or does this happen on a horizontal table? Finally, if I were you, I would turn friction off, understand the simpler problem, then turn friction back on.

 

[WildEnergy]

OK, so the length of the rectangle is 5.0 m not 4.0 m as I originally thought. Now I agree with the moment of inertia given by the simulation. What about the force? What is its direction as the object rotates? Also, is gravity in the picture or does this happen on a horizontal table? Finally, if I were you, I would turn friction off, understand the simpler problem, then turn friction back on.

there is no gravity, no friction and the force *will* rotate with the object
(but I only care about this single point in time so that shouldn't be a factor?)

 

[kuruman]

there is no gravity, no friction and the force *will* rotate with the object
(but I only care about this single point in time so that shouldn't be a factor?)

Good. This is what you do.

1. Use the parallel axis theorem to find the moment of inertia about the pivot.
2. Use Newton's Second Law for rotations to find the angular acceleration α.
3. Use α to find the linear acceleration of the center of mass, a_CM.
4. Use a_CM in Newton's Second Law for linear motion to find the horizontal force at the pivot. Note that the pivot exerts no vertical force because all the forces are horizontal.

 

[WildEnergy]

Good. This is what you do.

1. Use the parallel axis theorem to find the moment of inertia about the pivot.
2. Use Newton's Second Law for rotations to find the angular acceleration α.
3. Use α to find the linear acceleration of the center of mass, a_CM.
4. Use a_CM in Newton's Second Law for linear motion to find the horizontal force at the pivot. Note that the pivot exerts no vertical force because all the forces are horizontal.

yes - as per the book - and i can do that already - i did it to check the sim calculations

but try and derive the solution without the concept of a pivot ...

or to put it another way - imagine going back in time - you know basic Newtonian physics
and I say "hey look I just invented a pivot" and we have to work out the physics of the pivot
when we only have torques and forces

like you said - we need to work out the answer by modelling the pivot as forces - not as a pivot

the pivot could be something with very large mass or very large coefficient of static friction

either way is the same - it won't budge - but we can't consider it a pivot

 

[kuruman]

My suggestion is to replace the pivot with its effect on the rectangle, namely an additional force. You can find what it is by the method I described to you. So don't use a pivot to find that force, use the idea of "how much force must be exerted at point P located 0.5 m in from the end opposite to the one where F is applied such that point P is at rest while the rest of the rectangle rotates about it." Don't call it a pivot if you don't want to, but whatever it is, it must simulate the motion of the rectangle which requires that point P behave like a pivot, i.e. it is at rest while the rest of the body rotates about it. To do that, a force must be applied at P that you can find as I indicated to you.

 

[WildEnergy]

My suggestion is to replace the pivot with its effect on the rectangle, namely an additional force

and this is where the problem lays

I have spent days working out solutions to this and got nowhere

it seems to come down to that weirdness whereby:

force X applied at CoM - result = linear force N
force X applied away from CoM - result = linear force N + rotational force N (huh?)

i can't even decide what the force against the pivot is because:

the linear force is directed to the right
the rotational force is directed to the left

what is an "equal and opposite torque" ?

and how to relate these when they are no longer equal?

F = linear force + torque / d
F = a / (a+b) N + b / (a+b) N / d

there are some alternative approaches along the lines of:

Ax = Ay * |x-p| / |c-p|
alpha = Ax - Ay / |x-c|
etc. which might lead to a solution

but I cannot determine a force diagram that explains how a pivot works!
and that is the "hole" in the physics books I have

 

[kuruman]

and this is where the problem lays

I have spent days working out solutions to this and got nowhere

it seems to come down to that weirdness whereby:

force X applied at CoM - result = linear force N

There is no force applied at the center of mass. The only force is F and that is applied at 0.5 from the top end of the rectangle. That's what your drawing shows, unless we are talking about a different problem.

force X applied away from CoM - result = linear force N + rotational force N (huh?)

Huh? is right. The above does not make sense.

i can't even decide what the force against the pivot is because:

the linear force is directed to the right

That is correct.

the rotational force is directed to the left

What is this rotational force? What piece of the Universe exerts it on the rectangle?

what is an "equal and opposite torque" ?

Why do you ask? There is only one torque about the pivot and that is exerted by force F.

and how to relate these when they are no longer equal?

I gave you a detailed plan on how to do this. You say you did it already, then you should know what is going on here.

F = linear force + torque / d
F = a / (a+b) N + b / (a+b) N / d

there are some alternative approaches along the lines of:

Ax = Ay * |x-p| / |c-p|
alpha = Ax - Ay / |x-c|
etc. which might lead to a solution

The above makes no sense.

but I cannot determine a force diagram that explains how a pivot works!
and that is the "hole" in the physics books I have

A pivot works by exerting contact forces, i.e. it exerts whatever forces are necessary to provide the observed acceleration. So first you answer the question "what is the observed acceleration?", then work from there to find what forces are needed to be exerted by the pivot so you can get that acceleration. That's what I said you should do, you said you did it, so what's the problem?

 

[vela]

and this is where the problem lays

"lies" not "lays" (pet peeve of mine)

I have spent days working out solutions to this and got nowhere

it seems to come down to that weirdness whereby:

force X applied at CoM - result = linear force N
force X applied away from CoM - result = linear force N + rotational force N (huh?)

"Rotational force" is another way of saying "torque," so you don't have a linear force N and a rotational force N. You have a linear force N and a rotational force rN, where r is the length of the lever arm.

i can't even decide what the force against the pivot is because:

the linear force is directed to the right
the rotational force is directed to the left

what is an "equal and opposite torque" ?

and how to relate these when they are no longer equal?

F = linear force + torque / d
F = a / (a+b) N + b / (a+b) N / d


there are some alternative approaches along the lines of:

Ax = Ay * |x-p| / |c-p|
alpha = Ax - Ay / |x-c|
etc. which might lead to a solution

but I cannot determine a force diagram that explains how a pivot works!
and that is the "hole" in the physics books I have

As kuruman explained, the effect of the pivot is a force on the object and, if there's friction at the pivot, a torque. The force will be whatever is necessary to produce the resulting linear and angular accelerations. So what you do is replace the pivot with the unknown force and then apply F=ma and τ=Iα to the object. You'll get a system of equation which you can hopefully solve for the unknown force.

 

[vela]

but try and derive the solution without the concept of a pivot ...

or to put it another way - imagine going back in time - you know basic Newtonian physics
and I say "hey look I just invented a pivot" and we have to work out the physics of the pivot
when we only have torques and forces

like you said - we need to work out the answer by modelling the pivot as forces - not as a pivot

the pivot could be something with very large mass or very large coefficient of static friction

either way is the same - it won't budge - but we can't consider it a pivot

I'm not sure I get your point. You say you want to understand the pivot in terms of forces and torques, and that is exactly the approach kuruman has told you to use, which you said you know how to do. So what exactly is your objection? Are you just saying you don't want to use the location of the pivot as the axis you're taking moments about?

 

[WildEnergy]

I am saying that I am not trying to solve a single physics puzzle ... but work out an approach to writing a physics simulator - this means I am looking for a generic approach, not one that requires a human brain to recognize pivots - but an approach that takes a set of rigid bodies and forces and calculates resulting accelerations

Elsewhere I have been told "just add up all the forces and torques" and that will work
but I can't see any way to do that ...

Here is an idea that might lead somewhere:

Thanks for your efforts - I am off to bed

 

[vela]

I am saying that I am not trying to solve a single physics puzzle ... but work out an approach to writing a physics simulator - this means I am looking for a generic approach, not one that requires a human brain to recognize pivots - but an approach that takes a set of rigid bodies and forces and calculates resulting accelerations

I guess what I'm asking is what you mean by "recognize pivots." We're just saying you replace it with the force it exerts on the body. As far as the equations go, you wouldn't be able to tell that the force is from a pivot or caused by some other agent.

If you're saying you want to solve the problem without having some sort of constraint that says the body pivots about a point, you can't do it because it's a different problem. Somehow the information that one point on the body doesn't move has to enter the problem somehow, whether it's in your assumption about the accelerations or about the force exerted at the point the pivot acts.

 

[WildEnergy]

[PLAIN]http://img43.imageshack.us/img43/7647/physicsqe.png

distance from CoM to Contact Force is: d

I don't know how to calculate the Contact Force

 

[kuruman]

distance from CoM to Contact Force is: d

I don't know how to calculate the Contact Force

Read post #17. It gives you the specific step by step strategy for calculating it. If you get stuck somewhere, let us know. I will do the first step for you.

Let
a,b,m = length, width and mass of rectangle
x₁=location of pivot from end of rectangle
x₂=location of application of force from same end
F = applied force

The moment of inertia about the pivot is

I_pivot = m(a²+b²)/12 + m(a/2 - x₁)²

Now go on to the next step. You will end up with an expression that, given any set of parameters as listed above, will return the required force.

 

[WildEnergy]

(and that matches the output in my simulator)

I think the acceleration at the contact point is 1.69 m/s²

(btw the latex support in this editor is very buggy)

 

[kuruman]

First off, if you want to be able to simulate the pivot regardless of input parameters, you need to find the expression in terms of symbols. Then the user will put any parameters (s)he wishes and the pivot will be simulated correctly. That is why I started you out by writing the I_pivot symbolically.

Secondly, I see that you have trouble applying Newton's Second Law for linear motion and I have the sneaking suspicion that you don't really understand it. I say this because you have repeatedly expressed your inability to find "the resultant force at the CoM."

There is no such thing.

Newton's Second Law says "the sum of all the external forces acting on an extended body is equal to the mass of the body multiplied by its acceleration." So, there are two external forces acting on the rectangle, the applied force F and the horizontal force at the pivot F_x. Following Newton's Second Law, we write

F + F_x = ma_CM.

What else is there to write?

Finally, to keep your sanity, I recommend that you use the available LateX for composing only equations, not text. That's what everybody else seems to do.

 

[WildEnergy]

First off, if you want to be able to simulate the pivot regardless of input parameters, you need to find the expression in terms of symbols. Then the user will put any parameters (s)he wishes and the pivot will be simulated correctly. That is why I started you out by writing the I_pivot symbolically.

understood - I was doing a check to match the output of my simulator
can I take it that my workings are correct?

Secondly, I see that you have trouble applying Newton's Second Law for linear motion and I have the sneaking suspicion that you don't really understand it. I say this because you have repeatedly expressed your inability to find "the resultant force at the CoM."There is no such thing.

Newton's Second Law says "the sum of all the external forces acting on an extended body is equal to the mass of the body multiplied by its acceleration." So, there are two external forces acting on the rectangle, the applied force F and the horizontal force at the pivot F_x. Following Newton's Second Law, we write

F + F_x = ma_CM.

What else is there to write?

that confuses me because ... we started with a single external force of 10N
and we end up with a resultant linear force of 12.974N and some torque

Finally, to keep your sanity, I recommend that you use the available LateX for composing only equations, not text. That's what everybody else seems to do.

LaTeX editing on this forum just doesn't work for me - the results are always corrupted - i have forced a new session etc. but didn't fix anything