Forces when car wheels "lay rubber"

[vparam]

Homework Statement

If the coefficient of static friction between tires and pavement is 0.65, calculate the minimum torque that must be applied to the 66 cm diameter tire of a 950 kg automobile in order to "lay rubber" (make the wheels spin, slipping as the car accelerates). Assume each wheel supports an equal share of the weight.

Relevant Equations

∑T = Iα

Suppose the car is moving to the right, so if the wheels roll without slipping, they are rolling clockwise. To get the wheel to slip, a counterclockwise torque would need to be applied to cause the wheel to have some angular acceleration. If the wheel was slipping, then the bottom of the wheel would move left with respect to the pavement, which means that friction would point to the right, providing a counterclockwise torque to offset the applied torque.

When the friction force maxes out, in other words F_f = μF_n, then an applied torque can exceed the frictional torque and cause net angular acceleration and the wheel to slip. In that sense, we set the two torques to be equal, and solve, which gets the correct answer according to the textbook.

However, I'm confused about why this works. If this is happening as the car accelerates as the problem suggests, wouldn't there need to be a net rightward force? Then, the translational velocity also increases, and can "keep up" with the increasing angular speed. Wouldn't this mean that the conditions laid out wouldn't work? In other words, how could this counterclockwise torque cause the car to accelerate to the right (more accurately, the force that causes the torque), while also getting the wheels to slip?

 

[haruspex]

To get the wheel to slip, a counterclockwise torque would need to be applied

If you mean applied from the axle then you mean clockwise, I hope.

wouldn't there need to be a net rightward force?

There is. The ##F_fs## on the tyres are the only horizontal forces on the car as a whole.

 

[vparam]

If you mean applied from the axle then you mean clockwise, I hope.

Yes, sorry that torque should be clockwise.

There is. The ##F_fs## on the tyres are the only horizontal forces on the car as a whole.

I think my main point of confusion is how would F_fs be the only force. Wouldn't there also need to be another horizontal force on the wheels/car to apply the clockwise torque?

Edit: After trying this out on paper, I realized that this torque generating force could also be purely vertical rather than horizontal. In this case, F_f would be the only force acting in the horizontal direction. Is this a valid interpretation of what you mean?

 

[haruspex]

this torque generating force could also be purely vertical

A torque is neither horizontal nor vertical. If the axle applies a clockwise torque to the wheel then the wheel exerts an anticlockwise torque on the axle. This pushes down the back of the car and lifts the front. So the normal forces on the wheels have a net clockwise torque on the car. As long as the front wheels stay in contact with the ground, this balances the anticlockwise torque on the car from the frictional forces.

 

[vparam]

A torque is neither horizontal nor vertical. If the axle applies a clockwise torque to the wheel then the wheel exerts an anticlockwise torque on the axle. This pushes down the back of the car and lifts the front. So the normal forces on the wheels have a net clockwise torque on the car. As long as the front wheels stay in contact with the ground, this balances the anticlockwise torque on the car from the frictional forces.

What about the force that the axle exerts on the wheel? Would that need to be vertical in order for the net force on the wheel to equal the friction force, while still being able to generate a clockwise torque?

 

[haruspex]

What about the force that the axle exerts on the wheel? Would that need to be vertical in order for the net force on the wheel to equal the friction force, while still being able to generate a clockwise torque?

The forces from the wheels on the axle will be vertical forces to carry the weight of the car and horizontal forces to accelerate the car (car excluding wheels in each case).
Forces from axle on wheels are necessarily equal and opposite to those.
The torques are in addition to the forces.