Engineering Mechanics

Chapter 32 : Motion of Vehicles „„„„„ 663

This resisting force stops the automobile in a distance of s 1 metres. Therefore work done by
the force
= Force × Distance = 4126 s 1 ...(iii)
We know that the kinetic energy of 312 500 N-m [as per equations (ii)] of the automobile is
absorbed by the resisting force in travelling through a distance s 1. Therefore equating equations (ii)
and (iii),
312 500 = 4126 s 1

∴ s 1 =

312 500

75·7 m

4126

= Ans.

(ii) Distance covered by the automobile in coming to stop, if the brakes are applied on the front pair
of wheels only

Fig. 32.7.
Let s 2 = Distance covered by the automobile in coming to stop.
We know that when the brakes are applied on the front pair of wheels only, the resistance is
set up in the front pair of wheels, as shown in Fig. 32.7. Therefore force of friction in the front pair
of wheels,
FF= 0·2 RF ...(Given)
Now taking moments about the centre of gravity (G) of the automobile and equating the same,
RF × 1·6 = (RR × 1·2) + (FF × 1·2)
1·6 RF= 1.2 RR+ 1·2 (0·2 RF)
1·2 RR= 1·6 RF – 0·24 RF = 1·36 RF

∴ RR=

1·3 6

1·1 3

1· 2

F

F

R

= R

Substituting the value of RR in equation (i),

RF + 1·13 RF= 39 200 N

or RF=

39 200

18 400 N

2·13

=

and FF= 0·2 RF = 0·2 × 18 400 = 3680 N

This resisting force stops the automobile in a distance of s 2 meters. Therefore work done by
the force

= Force × Distance = 3680 s 2 N-m ...(iv)
We know that kinetic energy of 312 500 N-m [as per equation (ii)] of the automobile is
absorbed by the resisting force in travelling through a distance s 2. Therefore equating equation
(ii) and (iv),

312 500 = 3680 s 2

∴ s 2 =

312 500

84·9 m

3680

= Ans.