Question

the wheel of a car has a radius of .350 m. the engine of the car applies applies a torque of 295 N m to this wheel, which does not slip against the road surface. since the wheel does not slip the road must be appplying 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

The magnitude of static friction force is f_s = 842.8 N

Explanation:

Write down the values given in the question

The wheel of a car has radius r = 0.350 m

The car applies the torque is τ = 295 N m

It is said that the wheels does not slip against the road surface,

Here we apply a force of static friction,

It can be calculated as  

Frictional force f_s = τ / r

                            = 295 Nm / 0.350 m

                            f_s = 842.8 N

Answer 2

The magnitude of static frictional force is 842.86 N.

Magnitude of static frictional force

The magnitude of static frictional force is calculated as follows;

τ = Fr

where;

• τ is torque
• F is applied force which is equal to the static frictional force
• r is radius

Substitute the given parameters and solve for static frictional force.

F = τ /r

F = 295/0.35

F = 842.86 N

Thus, the magnitude of static frictional force is 842.86 N.

Learn more about torque here: https://brainly.com/question/14839816