Question

The wheel of a car has a radius of 0.350 m. The engine of the car applies a torque of 295N⋅m to this wheel , which does not slip against the road surface . Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a counter-torque. Moreover, the car has a constant velocity , so this counter-torque balances the applied torque . What is the magnitude of the static frictional force?

Answer 1

Answer:842.85 N

Step-by-step explanation:

Given

Engine of car applies a Torque of 295 N.m

radius of wheel r=0.350 m

since the car is moving with constant velocity therefore net torque on it must be zero

Let
`F_r` friction force force applies a counter torque given by

`T_f=F_r* r`

`295=F_r* 0.350`

`F_r=842.85 N`