Force, Momentum and Newton’s Laws - Dynamics and Relativity - page 48

or

a = g(µ s cos θ − sin θ)

in case (a) where the car is going uphill. Under normal circumstances a will be
positive, but we can well imagine the situation where the road is icy; µ s is then very
small and we may easily get tan θ>µ s . Then the acceleration becomes negative
and the car slides downhill. Under these circumstances, the wheels will be spinning
and we no longer have a condition of static friction between the tyres and the road
and µ k replaces µ s to give

a = g(µ k cos θ − sin θ).

For case (b) where the car is heading downhill we simply reverse the sign of the
friction to give

a = g( − µ s cos θ − sin θ).

A couple of comments are perhaps in order following this example. Firstly, it may
seem strange that static friction is used to describe the behaviour of a wheel, which
after all is designed to roll. A little thought should enable you to visualise what is
happening. At any instant, there is some area of the wheel that makes contact with
the road and over this contact area the surfaces of the wheel and the road may look
something like those shown in Figure 2.18: that part of wheel that is in contact with
the road is instantaneously at rest. The wheel pushes against the road but does not
slip, rather the forward motion creates a new region of contact between the tyre and
the road, and again we have instantaneous static friction over this new area. The
continual making and breaking of these contact regions does have a cost, however,
producing what is known as rolling friction, characterised by the coefficient of
rolling friction µ r . Rolling friction accounts for the slowing down of a wheel
rolled on a surface in vacuum. In many dynamical problems µ r can be ignored,
since it is typically an order of magnitude smaller than µ s . Secondly, in the solution
to the previous example we did not explicitly consider the fact that there were four
wheels in contact with the road. This simplification is partially justified because the
final result is independent of the mass. Thus is doesn't matter how the weight is
distributed over the four wheels, the maximum value of a is the same. In practice,
uneven forces on the wheels might lead to other problems, such as the rotation of the
car about its centre of mass, but we shall not concern ourselves with such matters.

2.3.4 Momentum conservation
Consider a system of particles and assume that the i th particle experiences a
force F i and has momentum p i . Since each particle obeys Newton's Second Law
(Eq. (2.14)) we can write

d P
d t ,

F = (2.30)

= i p i is the total momentum
of, the system of particles. If the net force is zero then it follows that the total

= i F i is the net force acting on, and P

where F