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Masja [62]
3 years ago
14

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 th

e 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?
Physics
1 answer:
Murrr4er [49]3 years ago
8 0

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

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