(^668) „„„„„ A Textbook of Engineering Mechanics
Taking moments about the centre of gravity (G) of the car and equating the same,
RF × 1·25 = ( RR × 1) + (FR × 0·5)
1·25 RF= RR + (0·35 RR × 0·5)
1·25 RF=RR + 0·175 RR = 1·175 RR
1·25 (9800 cos α – 28 000 sin α) = 1·175 × 28 000 sin α
9800 cos α – 28 000 sin α=
1·1 7 5
1· 2 5
× 28 000 sin α
0·35 cos α – sin α= 0·94 sin α
...(Dividing both sides by 28 000)
0·35 cos α= 0·94 sin α + sin α = 1·94 sin α
∴ tan α=
0·35
0·1804
1·9 4

α= 10·2° Ans.
(ii) Maximum inclination on which the car can climb when it is driven by the front pair of wheels only
Fig. 32.11.
We know that as the car is driven by the front pair of wheels only, therefore force of friction at
the front pair of wheels
FF=μ RF = 0·35 RF ...(iii)
Resolving the forces along the plane
FF=Mg sin α = 1000 × 9·8 sin α = 9800 sin α ...(iv)
Equating equations (iii) and (iv)
0·35 RF= 9800 sin α
∴ RF=
9800 sin
28 000 sin
0·35
α
=α
and RR= 9800 cos α – 28 000 sin α
Taking moments about the centre of gravity (G) of the car and equating the same,
RF × 1·25 + FF × 0·5 =RR × 1
1·25 RF + 0·35 RF × 0·5 =RR × 1
or 1·425 RF=RR × 1
1·425 × 28 000 sin α= 9800 cos α – 28 000 sin α
1·425 sin α= 0·35 cos α – sin α ...(Dividing both sides by 28 000)
2·425 sin α= 0·35 cos α