Question

N7M.10 Consider a car turning a corner to the left. Using the methods of chapter N4, show that in order to balance the torques on the car that seek to rotate the car around an axis going through its center of mass along the direction of its motion, the roadbed has to exert a greater normal force on the right wheels of the car than on the left wheels. Since this normal force will be transmitted to the car's body through the springs of the car's suspension, this means the suspen- sion springs on the right will have to compress more than those on the left, and thus the car will lean to the right, as I asserted in section N7.4.

Answer 1

When the car takes turn to the left, the car will make a circular path about a center located to the left of the car.

Let R be the radius of the circle.

Centrifugal force on the COM of the car = Mv² /R

there will be friction between the road surface and the tires.

The force applies a torque to the left CCW. For the car not to skid ,there must be a torque to the right CW.

The right side torque is provided by the frictional forces at the tires. For a net right side torque. Frictional force on the right side must be more than that on the left side.

Co-efficient of friction for all the 4 tires is same

F_f = $\mu$ N - N - normal force

F_fR > F_fL

N_R > N_L - normal reaction on the right ties must be more than than that on the left.

reaction on left is less, the springs are compressed less and the springs on the right are compressed more.

Mv2/R