How is Weight Distributed on a Man's Feet on a Turning Railroad Car?

[CGandC]

Homework Statement

The problem relates to the subject of torque and angular momentum:
A man of mass $M$ stands on a railroad car that is rounding an unbanked turn of radius R at speed v.
His center of mass is height $L$ above the car, and his feet are distance $d$ apart.
The man is facing the direction of motion. How much weight is on each of his feet?

Answers:

Homework Equations

Torque and angular momentum equation.
Conservation of angular momentum.
Newton's second law.

The Attempt at a Solution

What they did is they calculated the torque with respect to the center of mass of the man ( this torque equals zero because he doesn't rotate with respect to his own center of mass ) . If I solve the problem this I get the right answer, but I don't want to do the problem this way, so I decided to calculate the torque with respect to the middle point $\frac{d}{2}$ on the cart:
( the red cross in the picture below )

So, calculating the torque of the person with respect to the red-point I get that :
$-N_1*\frac{d}{2} +N_2*\frac{d}{2} = 0$ ( equals to zero since the cart doesn't rotate with respect to itself )
Thereby , having : $N_1 = N_2$
Which is the wrong answer.
( But the Newtons second law equations remain the same as in the answers, so the only problem is with the torque)

What I did wrong?

 

[BvU]

These two are not the only forces working on the cart (or else it would accelerate downwards rather rapidly...)

 

[CGandC]

These two are not the only forces working on the cart (or else it would accelerate downwards rather rapidly...)

Well, there are also friction forces whose directions are parallel to the red cross, therefore they do not contribute anything to the torque.
I also have the gravitational force on the person , but the direction of this force from the red cross is also parrallel , therefore it does not contribute anything to the torque also.
Only the Normal forces directions are not parallel to the red cross , hence , their contribution to the torque does not cancel.

 

[BvU]

You do a moment balance on the cart and I am telling you that you forget a bunch of forces. As Illustrated by the fact that the two forces you do mention point in the same direction, yet an acceleration in that direction is not what you expect.

(ps parallel to the red cross is rather weird, but I understand what you mean -- however, in one paragraph you have the friction parallel and in the next gravity is also parallel...? )

Only the Normal forces directions are not parallel to the red cross

is correct

Humor me and draw a free body diagram of the cart...

Consider force balance and moment balance

 

[jbriggs444]

So, calculating the torque of the person with respect to the red-point I get that :
$-N_1*\frac{d}{2} +N_2*\frac{d}{2} = 0$ ( equals to zero since the cart doesn't rotate with respect to itself )

Stop for a moment and re-examine that equation. You are trying to claim that the rate of change of angular momentum of the person is the same as the net torque on the person. Such a claim would be correct.

But what is the rate of change of angular momentum of the person about an inertial axis momentarily positioned between his feet?

Hint: Angular momentum can exist even without rotation.

Edit: I assume that you are intending to analyze in the tangent inertial frame and not the accelerating cart frame. In the cart frame, the relevant effect is accounted for by the fictitious centrifugal force.

 

[haruspex]

with respect to the middle point $\frac{d}{2}$ on the cart:

You need to be careful what you mean by that.

If you mean that as a point fixed to the cart then it is accelerating. As jbriggs444 points out, that means you have centrifugal force, and this will have a torque about your reference point.

If you mean it as a point fixed in space, but which happens to be the midpoint of his feet at some instant, then the man has angular velocity wrt that point by virtue of his linear velocity. That angular velocity is changing, and a torque is required to produce that precession.

 

[CGandC]

You do a moment balance on the cart and I am telling you that you forget a bunch of forces. As Illustrated by the fact that the two forces you do mention point in the same direction, yet an acceleration in that direction is not what you expect.

(ps parallel to the red cross is rather weird, but I understand what you mean -- however, in one paragraph you have the friction parallel and in the next gravity is also parallel...? )

is correct

Humor me and draw a free body diagram of the cart...

Consider force balance and moment balance

Stop for a moment and re-examine that equation. You are trying to claim that the rate of change of angular momentum of the person is the same as the net torque on the person. Such a claim would be correct.

But what is the rate of change of angular momentum of the person about an inertial axis momentarily positioned between his feet?

Hint: Angular momentum can exist even without rotation.

Edit: I assume that you are intending to analyze in the tangent inertial frame and not the accelerating cart frame. In the cart frame, the relevant effect is accounted for by the fictitious centrifugal force.

You need to be careful what you mean by that.

If you mean that as a point fixed to the cart then it is accelerating. As jbriggs444 points out, that means you have centrifugal force, and this will have a torque about your reference point.

If you mean it as a point fixed in space, but which happens to be the midpoint of his feet at some instant, then the man has angular velocity wrt that point by virtue of his linear velocity. That angular velocity is changing, and a torque is required to produce that precession.

Ok, so to sum up so far:

1. if the red-cross is a fixed point in space, then there is angular momentum on the person and the direction of this angular momentum changes but not the magnitude , therefore the derivative of this angular momentum will not be zero , hence , the torque on the person relative to the red-cross the instant the red-cross is exactly between his legs is not zero:

Calculating the angular momentum of the person with respect to the red-cross the instant it is exactly between the legs of the person:
$\vec{L} = \vec{R_c} \times m \vec{V} = R_c \hat{k} \times m \omega R \hat{\theta} = -R_c m \omega R \hat{r}$
$\frac{ \vec{dL} }{dt} = -R_c m \dot{\omega} R \hat{r} -R_c m {\omega}^2 R \hat{\theta}$
Therefore, at that instant of time :
$-N_1*\frac{d}{2}\hat{k} +N_2*\frac{d}{2}\hat{k} = \frac{ \vec{dL} }{dt} = -R_c m \dot{\omega} R \hat{r} -R_c m {\omega}^2 R \hat{\theta}$

where:
$R_c$ = distance from red-cross to the center-of-mass of the person , $\omega$ is the angular velocity of the person with respect to the center of circle about which he is rotating , $\dot{\omega}$ is the angular acceleration . $R$ = radius of circle

2. If the red-cross is attached to the cart ( therefore the cross is now a non-inertial system) then , the person is not moving relativly to the cart , therefore the angular momentum of the person is zero , this means the torque is zero and since I find no other new forces when I'm looking at the torque from this non-inertial point ( red-cross attached to moving cart ) , I still get: $-N_1*\frac{d}{2} +N_2*\frac{d}{2} = 0$

And the force diagram on the cart is:

where $M_c$ = mass of cart. $N_c$ = normal force on the cart from the ground . $f_1$ and $f_2$ are friction forces from the feet of the person.What do you think?

 

[BvU]

I think you are making a mess of this. Does it help at all if we try to guide you ? The original exercise

The man is facing the direction of motion. How much weight is on each of his feet?

seems to be forgotten and you are mixing up directions, leaving out crucial forces, etc.

if the red-cross is a fixed point in space,

It is not.

 

[jbriggs444]

2. If the red-cross is attached to the cart ( therefore the cross is now a non-inertial system) then , the person is not moving relativly to the cart , therefore the angular momentum of the person is zero , this means the torque is zero and since I find no other new forces when I'm looking at the torque from this non-inertial point ( red-cross attached to moving cart )

You forgot centrifugal force.

 

[CGandC]

You forgot centrifugal force.

I can't imagine the centrifugal force here and what makes it. I assume it is not parallel to the red cross so it doesn't cancel out.
I assume this centrifugal force is imaginary
Can you please provide image showing the force on the cart? It will help very much

 

[jbriggs444]

I can't imagine the centrifugal force here and what makes it. I assume it is not parallel to the red cross so it doesn't cancel out.

What does "parallel to the red cross" even mean? The red cross marks a point. What line do you think that it corresponds to?

I assume this centrifugal force is imaginary

When you adopt a non-inertial frame, you adopt fictitious forces. The centrifugal force points away from the center of the track's curvature.

 

[BvU]

I can't imagine the centrifugal force here and what makes it. I assume it is not parallel to the red cross so it doesn't cancel out.
I assume this centrifugal force is imaginary

It's not imaginary: if you are in a merry-go-round you feel it.
My urgent advice: forget about that kind of reference frames at least until you are able to work effectively with inertial reference frames (the normal kind). Instead of taking centrifugal and Coriolis forces into account, go on the lookout for the centripetal force that causes the curved trajectory

Can you please provide image showing the force on the cart? It will help very much

No. PF culture is not that we do your work for you. Not because we are lazy or unkind, but because it does not help you. I specifically asked you to set up a free body diagram and you did not do it so far. Post a start and we'll help you along.
[edit] OK, too quick and harsh on my part, sorry. Take the last picture in #7 as a start and ask:

what compensates for $f_1+f_2$ ?
what causes the cart to follow a curved trajectory ? Hint:

parallel to the red cross

what direction is that ? I have never heard of something "parallel to a cross".

 

[CGandC]

It's not imaginary: if you are in a merry-go-round you feel it.
My urgent advice: forget about that kind of reference frames at least until you are able to work effectively with inertial reference frames (the normal kind). Instead of taking centrifugal and Coriolis forces into account, go on the lookout for the centripetal force that causes the curved trajectory

No. PF cutture is not that we do your work for you. Not because we are lazy or unkind, but because it does not help you. I specifically asked you to set up a free body diagram and you did not do it so far. Post a start and we'll help you along.
[edit] OK, too quick and harsh on my part, sorry. Take the last picture in #7 as a start and ask:

what compensates for $f_1+f_2$ ?
what causes the cart to follow a curved trajectory ? Hint:View attachment 220853​

what direction is that ? I have never heard of something "parallel to a cross".

Drawing the forces on the picture you put:

where $f_{11}$ and $f_{22}$ are the friction forces

And on the cart:

where:
$f_{11} \: f_{22} \: f_{33} \: f_{44}$ are the centripetal friction forces acting on each wheel ( and these make the cart stay in circular motion )
$f_{11c} \: f_{22c} \: f_{33c} \: f_{44c}$ are the centrifugal friction forces acting on each wheel by the contact with the rails ( however, since there is a centrifugal force , the cart shouldn't move in a circular path, so perhaps there is a mistake in my assumptions here? )And when I said " parallel to the red-cross" I meant that the distance vectors from the red-cross to the friction forces $f_2$ and $f_1$ are in parallel direction. for example:

since the distance vectors are parallel to the forces , then the torque contribution from these forces must be zero by the cross-product.

 

[BvU]

That's quite a lot of forces now. Can we agree that $f_1$ and $f_2$ make the man describe a circle ? And $f_{11}+f_{22}+f_{33}+f_{44}$ make the cart go round and on top of that compensate for $f_1+f_2$ ?

Now, if $f_1$ and $f_2$ cause no torque around a certain axis ( through red cross and in the direction of motion), but then $f_{11}+f_{22}+f_{33}+f_{44}$ certainly do, right ? And who are available to compensate for that ?

(by the way, I was inclined to simplify a little and let the outer wheel flanges exert a single $f_{33} + f_{44}$ and assume the other two can be set to zero -- with no effect on the torque balance)

 

[CGandC]

Can we agree that $f_1$ and $f_2$ make the man describe a circle ?

Yes.

And $f_{11}+f_{22}+f_{33}+f_{44}$ make the cart go round and on top of that compensate for $f_1+f_2$ ?

Yes.

Now, if $f_1$ and $f_2$ cause no torque around a certain axis ( through red cross and in the direction of motion), but then $f_{11}+f_{22}+f_{33}+f_{44}$ certainly do, right ?

Yes

And who are available to compensate for that ?

Looking at the problem a little big longer, then I'd say the normal forces acted on each wheel by the rails compensate.
( ie: $N_{11} \: N_{22} \: N_{33} \: N_{44}$ )

Looking now at the problem, I realized that I have at least 5 new unknowns : $N_{11} \: N_{22} \: N_{33} \: N_{44}$ and the constant of friction between each wheel and the rail $\mu_c$
This makes the problem very difficult and I now realize why It is better to solve at the man's center of mass.
In addition, It doesn't matter if I solve the torque with respect to inertial or non-inertial center-of-mass frame, the torque remains the same.

I have another question: what do the centrifugal friction forces which act on each wheel: $f_{11c} \: f_{22c} \: f_{33c} \: f_{44c}$ contribute? , shouldn't they deviate the cart out of it's circular path?

 

[jbriggs444]

I have another question: what do the centrifugal friction forces which act on each wheel: $f_{11c} \: f_{22c} \: f_{33c} \: f_{44c}$ contribute? , shouldn't they deviate the cart out of it's circular path?

What centrifugal friction forces? The forces of friction on the wheels are centripetal.

 

[BvU]

This makes the problem very difficult and I now realize why It is better to solve at the man's center of mass.

My original agenda was as follows: you concentrate on the cart and gradually discover that for the forces $f_1$ and $f_2$, $N_1$ and $N_2$, you can't extract the necessary information that way. They simply depend on what happens to the man, whether he is on a cart, on a merry-go-round or in an airplane.

Your subscript c forces slow down the vehicle, which would have consequences for the sideways forces, yes. However, $v$ is a given quantity, so you can forget about all four.

 

[haruspex]

centripetal friction forces acting on each wheel

centrifugal friction forces acting on each wheel

Don't mix the two.

If you take an inertial frame of reference (see post #6) then there is no centrifugal force acting. There are only gravitation, normal forces and (radial) frictional forces. The centripetal force is the radial component of the resultant. (And in the present case there is no tangential component.)
$\vec {F_g}+\vec {F_n}+\vec {F_f}=\vec {F_r}$.

If you take a frame of reference moving and rotating with the cart, there is no centripetal resultant since there is no acceleration in this frame. Instead, the acting forces are gravitation, normal forces, frictional forces and centrifugal force.
$\vec {F_g}+\vec {F_n}+\vec {F_f}+\vec {F_{centrifugal}}=0$.

Since $\vec {F_{centrifugal}}=-\vec {F_r}$, these two equations are the same.

If you take a non-inertial frame in which the cart is nevertheless accelerating then you will have an acting centrifugal force and a centripetal resultant, but that smacks of masochism.