contestada

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?

Respuesta :

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 Ï„ = 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 = Ï„ / r

                            = 295 Nm / 0.350 m

                            f_s = 842.8 N

onyebuchinnaji

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;

Ï„ = Fr

where;

  • Ï„ 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 = Ï„ /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

Otras preguntas