Now assume that the frictional force f is not at its maximum value. What is the relation between the torque Ï„ applied to each wheel by the axles and the acceleration a of the car? Once you have the exact expression for the acceleration, make the approximation that the wheels are much lighter than the car as a whole.'

a)The direction the frictional force will acts is in the positive x direction.

Explanation:

a)The direction the frictional force will acts is in the positive x direction

b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.

F(total)=4f

''f'' is taken as the frictional force.

c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0

=4N(wheel)-mg=0

d) torque is the force that tends to bend rotation

ζ=rf

but acceleration=4×frictional force

cross multiply

f=ζ/r

f=ma/4

ma/4=ζ/r

a=4ζ/r

The question was dealt with as an incomplete question.

A. The frictional force is in the positive X direction

B. The net force is in the horizontal direction, f is the friction force in each wheel. so the answer is Fnet=4f

C. The weight is distributed equally on each wheel in the Y direction, gives us the formula 4Nwheel-mg=0 with a final answer of N=mg/4

D. torque force at each wheel is tau=rf, linear acceleration is obtained from the sum of the forces 4f=ma, f=tau/r and f=ma/4, (ma/4=tau/r) giving us the final answer of a=4tau/mr (please note that the tau symbol is in the drop down list and looks like a cross between a lowercase r and an upper case T)