Page 161 - Tribology in Machine Design

Friction, lubrication and wear in lower kinematic pairs 147

Adding eqns (4.142) and (4.143) and eliminating (F l-F 2) from eqn (4.144),
then

Also, from eqns (4.143) and (4.144)

and eliminating a

This equation gives the least value of F t if wheel spin is to be avoided. For
2
2
example, suppose M = 1350 kg, 7 1 = 12.3kgm ; / 2 =8.1kgm and
a=0.33m, then
or

so that, if L exceeds this value, wheel spin will occur.

The maximum forward acceleration
Equation (4.145) gives the forward acceleration in terms of the driving
couple L, which in turn depends upon the limiting friction force F t on the
rear wheels. The friction force F 2 on the front wheels will be less than the
limiting value. Thus, if R t and R 2 are the vertical reactions at the rear and
front axles, then

To determine /? t and R 2, suppose that the wheel base is b and that the
centre of gravity of the car is x, behind the front axle and y, above ground
level. Since the car is under the action of acceleration forces, motion, for the
system as a whole, must be referred to the centre of gravity G. Thus the
forces F! and F 2 are equivalent to:
(i) equal and parallel forces FI and F 2 at G (Fig. 4.46, case (b)
(ii) couples of moment F {y and F 2y which modify the distribution of the
weight on the springs.
Treating the forces RI and R 2 in a similar manner, and denoting the weight
of the car by W, we have