Question

Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.300 m. The engine of the car applies a torque of 222 N·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 countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

Answer 1

Answer:

740 N

Explanation:

We are given that

Radius,r=0.3 m

Torque,$\tau=222 Nm$

We have to find the magnitude of the static frictional force.

According to question

Torque by engine=Torque by static friction

$222=f\times r$

$f=\frac{222}{r}$

$f=\frac{222}{0.3}$

$f=740 N$

Hence, the magnitude of static frictional force=740 N