(^662) „„„„„ A Textbook of Engineering Mechanics
Example 32.5. An automobile of mass 4000 kg is travelling at 45 km.p.h. on a level road.
The height of centre of gravity of the vehicle is 1·2 m above the road surface and the distance
between the two axles is 2·8 m. The distance of centre of gravity from the front axle is 1·6 m.
Find the distance covered by the automobile in coming to stop,if brakes are applied on (i) rear
pairs of wheels only; (ii) fron pair of wheels only. and (iii) both the pair of wheels. Take resistance to
motion by application of brakes as 20% of the normal reactions.
Solution. Given: Mass of the automobile (M) = 400 kg; Speed of the vehicle (u) = 45 km.p.h.
= 12·5 m/sec; Height of the centre of gravity of the vehicle above road surface (h) = 1·2 m; Distance
between the two axles = 2·8 m; Distance of c.g. from the front axle (x 1 ) = 1·6 m; Distance of centre of
gravity from the rear axle (x 2 ) = 2·8 – 1·6 = 1·2 m and resistance to motion = 20% of normal reaction
(i) Distance covered by the automobile in coming to stop, it the brakes are applied on the rear pair
of wheels only.
Let s 1 = Distance covered by the automobile in coming to stop.
RF= Reaction at front pair of wheels, and
RR= Reaction at rear pair of wheels.
We know that RF + RR=Mg = 4000 × 9·8 = 39 200 N ...(i)
and kinetic energy of the automobile before the brakes are applied

(^22) 4000 (12·5)
312 500 N-m
22
Mu
== ...(ii)
Fig. 32.6.
We know that when the brakes are applied on the rear pair of wheels only, then resistance is
set up in the rear pairs of wheels as shown in Fig. 32.6. Therefore force of friction in the rear pair
of wheels,
FR= 0·2 RR ...(Given)
Now taking moments about the centre of gravity (G) of the automobile and equating the same,
RF × 1·6 = (RR × 1·2) + (FR × 1·2)
1.6 RF= 1·2 RR + 1·2 (0·2 RR ) = 1·44 RR ...(Q FR = 0·2 R)
∴ RF=
1· 4 4
0·9
1· 6
R
R
R
= R
Substituting the value of RF in equation (i),
0·9 RR + RR= 39 200 N
or RR=
39 200
20 630 N
1·9

and FR= 0·2 RR = 0·2 × 20 630 = 4126 N