## The Attempt at a Solution

Why can't we take the moment on the left wheel?
In this case,
2d(N2) - mgd
= mgd + m(v^2)h/r - mgd
=m(v^2)h/r ≠ 0
Shouldn't there be an anticlockwise moment about the left wheel?

Last edited:

## Homework Statement

View attachment 110777

## The Attempt at a Solution

Why can't we take the moment on the left wheel?
In this case,
2d(N2) - mgd
= mgd + m(v^2)h/r - mgd
=m(v^2)h/r ≠ 0
Shouldn't there be an anticlockwise moment about the left wheel?

In a real car there is a suspension and the torque about the left wheel is part of what causes the right suspension to compress. However, for the purpose of this problem the car and its wheels are a rigid body. There is no way to rotate clockwise about the left wheels because the right wheels can't go below the ground.

As long as the car is not rolling over you are welcome to calculate the torque about any axis and it must be 0. However for determining the moment it will stop being in equilibrium you want to check the torque about the point that will become the one and only actual axis when the car starts rolling.

Doc Al
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Why can't we take the moment on the left wheel?
The problem with using the left wheel as your pivot point is that it is accelerating. The center of mass has special properties that allow you to take torques about it even though it's accelerating. But not other points. (You can use that left wheel as a pivot, but you'll have to incorporate inertial forces.)

Clara Chung
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Why can't we take the moment on the left wheel?
In this case,
2d(N2) - mgd
= mgd + m(v^2)h/r - mgd
=m(v^2)h/r ≠ 0
Yes, that's all fine, and has given you the same equation as in the text that starts N2=.
Shouldn't there be an anticlockwise moment about the left wheel?
There is. What makes you think there is not?

haruspex
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The problem with using the left wheel as your pivot point is that it is accelerating
The wheel is, but it is ok to use a fixed point on the road where the wheel happens to be at some point in time. It seems to me that is what Clara has done.

haruspex
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As long as the car is not rolling over you are welcome to calculate the torque about any axis and it must be 0.
No, to negotiate the bend there must be a net horizontal force acting to the left through the mass centre. That has an anticlockwise moment about points on the ground.
for determining the moment it will stop being in equilibrium you want to check the torque about the point that will become the one and only actual axis when the car starts rolling.
It is generally simpler that way, but not essential. The equation Clara obtained was correct.
There is no way to rotate clockwise about the left wheels because the right wheels can't go below the ground.
I do not see where that is being suggested.

Doc Al
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It seems to me that is what Clara has done.
That wasn't clear to me.

haruspex
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That wasn't clear to me.
Let me put it this way... for the purpose of calculating a moment, it might as well be a point fixed in space. Using an accelerating point of reference is only at risk of going wrong when trying to determine angular accelerations using that point as axis.

Doc Al
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No, to negotiate the bend there must be a net horizontal force acting to the left through the mass centre.
Yet, from an inertial frame, the only forces acting horizontally are the friction forces from the ground.

haruspex
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Yet, from an inertial frame, the only forces acting horizontally are the friction forces from the ground.
The net force is horizontal. The actual forces leading to that are at ground level. But the effective line of action of the net force depends also on torques. The torque provided by the displacement between the line of action of gravity and those of the normal forces on the wheels provides such a torque.

Clara Chung
No, to negotiate the bend there must be a net horizontal force acting to the left through the mass centre. That has an anticlockwise moment about points on the ground.

Ah yes, too broad a statement. The torque about any axis parallel to the tangent of the curve must be zero and you are welcome to calculate that at the left tire or right

It is generally simpler that way, but not essential. The equation Clara obtained was correct.

If you want to know when gravity is insufficient to keep the car from flipping over the right tire is the axis to use. It's not just simpler. It eliminates forces you don't know (the constraint of the normal force at the right tire.) and leaves only force you do know: the torque from gravity about the right tire vs the torque from the centripetal force (friction) being applied off center (below) the center of gravity. The question was why not the left tire. This is the answer.

. I do not see where that is being suggested.

Let me rephrase. Calculating the torque about the left tire includes a constraint force: you can't rotate clockwise about that axis because the right tire can't go below the ground. Calculating torques about the left tire you will always get zero because the constraint force will always prevent clockwise rotation about the left tire. Clockwise rotation about the left tire contact point is not possible so you will never be able to determine when rolling over is possible using that axis. If you want to know when the car flips over you have to look at the right tire about which the care is actually free to rotate. That is why they calculate with the right tire.

Clara Chung
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The torque about any axis parallel to the tangent of the curve must be zero
No. Consider the critical speed, where it is in danger of overturning. Taking moments about the right tyre's point of contact with the ground, there is only mg. The net torque is not zero, yet that is an axis parallel to the tangent to the curve.
The net torque about the mass centre must be zero.
As I posted before, the forces and torques must be equivalent to a single horizontal force (centripteal) acting through the mass centre. That has a torque about any axis not in the same horizontal plane as the mass centre.
It's not just simpler. It eliminates forces you don't know
It is simpler because it eliminates a force that you do not care about. Whatever axis you use, you can combine the moments equation with the usual two linear force equations and solve.

No. Consider the critical speed, where it is in danger of overturning. Taking moments about the right tyre's point of contact with the ground, there is only mg. The net torque is not zero, yet that is an axis parallel to the tangent to the curve.
The net torque about the mass centre must be zero.
As I posted before, the forces and torques must be equivalent to a single horizontal force (centripteal) acting through the mass centre. That has a torque about any axis not in the same horizontal plane as the mass centre.

It is simpler because it eliminates a force that you do not care about. Whatever axis you use, you can combine the moments equation with the usual two linear force equations and solve.

Oops. I stand corrected. I was thinking in the accelerated reference frame with the resulting fictitious centrifugal force.

Big thanks for all of your replies. Here is my comprehension from above discussion.
So when I take the left wheel as an accelerating frame, I have to include the fictitious force(centrifugal force).
i.e.
mgd-2d(N2)+ mv^2(h)/r which is equal to zero, therefore, from the driver's perspective, he won't topple.

When I take the road as an inertia frame,
There is an anticlockwise torque which provides the torque needed to carry out circular motion.
Am I right?

haruspex