Question

The wheel of a car has a radius of 0.350 m. The engine of the car applies a torque of 272 N·m to this wheel, which does not slip against the road surface. Since thewheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, sothis countertorque balances the applied torque. What is the magnitude of the static friction force?

Answer 1

F = tau/r
tau=295 N-m; r=0.350 m;


F = 842.9 Newtons

Answer 2

Torque balance:
r*F = tau

solve for F:
F = tau/r

b
tau =295 N-m; r:=0.350 m;


F = 842.9 Newtons

Answer 3

Better called the road traction force for easier understanding.


Forces acting on the wheel (forget about those in the vertical direction):
Axle reaction force, A (reverse)
Traction force, F (forward)

Engine torque is tau, acting to the left of the car.

Torque due to F about the axle must therefore point right.

Torque balance:
r*F = tau

solve for F:
F = tau/r

Data:
tau:=295 N-m; r:=0.350 m;

Result:
F = 842.9 Newtons