A dog of mass M sits in the bed of a pickup truck.

Question

The only thing keeping him in place is friction. The pickup is driving with constant speed v, on an unbanked circular road with radius R.

a. Draw a free body diagram of the dog.

b.Determine an expression for the minimum value of the coefficient of friction to keep the dog from sliding.

 

Let’s assume now the bed of the truck is frictionless. If the bank is at a proper angle, then the dog will not slide.

c. Draw a free body diagram of the dog when the truck is on this banked roadway.

d. Of the two forces acting on the dog, is one of them larger than the other?

e. Determine an expression for the minimum angle at which the curb must be banked at to keep the frictionless dog from sliding.

Answer 1

The coefficient of friction is µ so µMg is the friction opposing the outward force Mv²/R caused by circular motion. The dog will not slide if µMg>Mv²/R so µ>v²/(Rg). The minimum value is v²/(Rg).

If the angle of the bank is θ then the gravitational force must balance the outward force caused by rotation:

Mgsinθ=Mv²/R and sinθ=v²/(Rg) so θ=arcsin(v²/(Rg)).

As long as the gravitational force caused by the incline is greater than the force experienced by the circular motion, the dog will slide under gravity inwards to the side of the bed (down the incline). If the centrifugal force is greater than the force of gravity the dog will slide outwards. When the incline is at a particular angle the dog will stay in place, so neither force is bigger than the other. The mass of the dog is irrelevant.