(^660) „„„„„ A Textbook of Engineering Mechanics
We know that RF + RR=Mg = 800 × 9·8 = 7840 N ...(i)
Since the car is moving with an acceleration (a), therefore accelerating force acting on it
=Ma = 800 a ...(ii)
We know that as the car has rear wheel drive, therefore the force of friction will act on the rear
pair of wheels only as shown in Fig. 32.4. Now the force of friction on the rear pair of wheels,
FR=μRR = 0·6 RR
Now taking moments about the centre of gravity (G) of the car and equating the same,
RF × 1·15 = (RR × 1·6) + (FR × 0·85)
1·15 RF= 1·6 RR + (0·6 RR × 0·85)
or 1·15 RF= 1·6 RR + 0·51 RR = 2.11 RR
∴ RF=
2·11
1.835
1·1 5
R
R
R
= R
Substituting this value of RF in equation (i),
1·835 RR + RR= 7840 or 2·835 RR = 7840
∴ RR=
7840
2765 N
2·835

and FR= 0·6 RR = 0·6 × 2765 = 1659 N ...(iii)
We know that as the force of friction is equal to the accelerating force, therefore equating
equations (ii) and (iii)
800 a= 1659
∴ a=^2
1659
2·07 m/s
800
= Ans.
(ii) Maximum possible acceleration when the car has front wheel drive
We know that as the car has front wheel drive, therefore the force of friction will act on the
front pair of wheels only as shown in Fig. 32.5. Now the force of friction on the front pair of wheels.
FF=μRF = 0·6 RF
Fig. 32.5.
Now taking moments about the centre of gravity (G) of the car and equating the same,
RF × 1·15 = (RR × 1·6) + (FF × 0·85)
1·15 RF= 1·6 RR + (0·6 RF × 0·85)