# Opposing force of the axle when a torque is applied to a wheel?

• alkaspeltzar
In summary, the axle has an opposing force to keep the wheel/axle assembly from accelerating linearly.
alkaspeltzar
Okay let's say we have a wheel attached to a fixed axle, free to spin, ignore friction and we apply a force on the edge of the wheel. That force creates a torque, does the axle have an opposing force to keep the wheel/axle assembly from accelerating linearly?

Ive learned on this forum that when a powered car wheel encounters friction from road, friction causes both counter torque(against engine) and forward force. My question is why doesn't the force of the piston on the crank do similar. We only ever look at the torque. Is it becuase internally any forces trying to move the crank linearly are canceled out, leaving only the torque ?

alkaspeltzar said:
Okay let's say we have a wheel attached to a fixed axle, free to spin, ignore friction and we apply a force on the edge of the wheel. That force creates a torque, does the axle have an opposing force to keep the wheel/axle assembly from accelerating linearly?
We're ignoring friction. So I assume we've hammered a spike into the wheel, giving us a place to push.

And since we're ignoring friction, we can stand the axle on end and pound it into the ground so it is firmly fixed in place. Now we do not have to concern ourselves with any road.

The axle acts as a constraint on the position of the center of the wheel. It will exert any linear force that is required to keep the center of the wheel fixed in place. If you apply a force one way at the rim of the wheel, the axle must provide an equal and opposite force the other way. Otherwise the wheel would move. And that would violate the constraint.

alkaspeltzar said:
Ive learned on this forum that when a powered car wheel encounters friction from road, friction causes both counter torque(against engine) and forward force. My question is why doesn't the force of the piston on the crank do similar.
It does. The piston (through the connecting rod) exerts both a force and a torque on the crank shaft. The bearings where the crank shaft fits into the engine block provide the relevant constraint that keeps the crank shaft right where it belongs rather than being shoved into the oil pan.

As you suggest, this negates the linear force, leaving nothing but a rotational torque. The bearings are designed not to prevent rotation.

[Of course, the cylinder head is also firmly attached to the engine block. So the expanding gasses do not act like a rocket exhaust and are contained instead].

Last edited:
JBriggs444, so needless to say is it becuase we know the axle or whatever the point of rotation is fixed, we many times don't even think of the linear portion of the force on the axle? I see many examples of torque where they just skip right over that.

Also then, can you just confirm that what I originally asked then was right. Like in the engine, the crank is locked down, so the only thing it can receive is torque. Now where friction on a tire pushes externally to the car, the tire, axle, car etc. are all free to move relative to the ground, so we have to take both torque and force on the wheel into consideration. Yes?

alkaspeltzar said:
JBriggs444, so needless to say is it becuase we know the axle or whatever the point of rotation is fixed, we many times don't even think of the linear portion of the force on the axle? I see many examples of torque where they just skip right over that.
Yes, just so.
alkaspeltzar said:
Also then, can you just confirm that what I originally asked then was right. Like in the engine, the crank is locked down, so the only thing it can receive is torque. Now where friction on a tire pushes externally to the car, the tire, axle, car etc. are all free to move relative to the ground, so we have to take both torque and force on the wheel into consideration. Yes?
I have highlighted the part that I think you want confirmed.

If we are concerned with the linear motion of the car, we are certainly concerned with the linear force of road on tire.

If we are unconcerned with weight transfer from front to rear wheels and are unconcerned with the possibility of the car doing a "wheely" then we may not be concerned about the torque associated with that force. On the other hand, if those things are important to us then we may need to be concerned with the torque after all.

There can be a problem with trying to 'see' what's going on throughout the system, all at once. Each link in the chain can either be studied individually or you can just look at the ends of the chain. It helps to bear in mind that the Physics of the system does actually work.
alkaspeltzar said:
friction causes both counter torque(against engine) and forward force.
This is a version of a Third Law Pair of torques / forces (just mixed up a bit). Inside the drive chain there is a series of third law pairs - at each gear , crank and piston. The 'counter torque' on the crank is equal to the torque due to the force on the con rod and the instantaneous radius of rotation (which changes all the time). The force will be due to the varying pressure in the combustion chamber. There is another torque in there, due to the rotating mass of the flywheel and that smooths things out a bit. At low enough revs (like in an old steam engine) the traction force on the ground will not be uniform. The mass of the car and any give on the transmission chain will act as another low pass filter and keep the speed / acceleration of the car uniform.

it could, perhaps, make things easier if we replace the tyres with a rack and pinion drive instead of a Friction force between the tyre and road.

## 1. What is the opposing force of the axle when a torque is applied to a wheel?

The opposing force of the axle when a torque is applied to a wheel is known as the reaction force. This force acts in the opposite direction of the applied torque and is necessary to maintain the equilibrium of the wheel.

## 2. How is the opposing force of the axle calculated?

The opposing force of the axle can be calculated using the formula F = T/r, where F is the reaction force, T is the applied torque, and r is the radius of the wheel.

## 3. Does the opposing force of the axle change if the torque is increased?

Yes, the opposing force of the axle will increase if the applied torque is increased. This is because the reaction force is directly proportional to the applied torque.

## 4. What factors affect the opposing force of the axle?

The opposing force of the axle is affected by the magnitude of the applied torque, the radius of the wheel, and the coefficient of friction between the wheel and the surface it is rolling on.

## 5. Why is the opposing force of the axle important in understanding the motion of a wheel?

The opposing force of the axle is important in understanding the motion of a wheel because it is necessary for the wheel to move in a circular motion. Without this force, the wheel would not be able to maintain its rotation and would simply slide or skid on the surface.

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