Engineering Mechanics

(^658) „„„„„ A Textbook of Engineering Mechanics
Find from first principles, the (i) acceleration of the vehicle and (ii) normal reaction at the
front and rear pair of wheels. Neglect the rotational inertia of the wheel.
Solution. Given: Mass of vehicle (M) = 1200 kg; Tractive force = 1800 N; Height of the c.g.
above road surface (h) = 1 m; Distance between the centre of gravity of the vehicle and the point
through which tractive force act (y) = 1 – 0·75 = 0·25 m; Horizontal distance between centre of
wheels (2x) = 24 m or x = 1·2 m; Resistance to motion on front wheel (FF) =
1
12 F
×R and restistance
to motion on rear wheel (FR) =
1
12 R
×R.
(i) Acceleration of the vehicle
Fig. 32.3.
We know that RF + RR= Mg = 1200 × 9·8 = 11 760 N
Let a= Acceleration of the vehicle
We know that total force of resistance,
FF + FR=
11
( ) 11760 kg
12 FR 12
RR+=× = 980 N
∴ Net tractive force, P = 1800 – 980 = 820 N
We also know that the net accelerating force (P),
820 =Ma = 1200 a
or a=^2
820
0·68 m/s
1200
= Ans.
(ii) Normal reactions at the front and rear pair of wheels ...(i)
Taking moments about the centre of gravity (G) of the vehicle and equating the same,
(RF × 1·2) + (1800 × 0.25) = (RR × 1·2) + (FF + FR) × 1
(RF – RR) 1·2 = 980 – 450 = 530
∴ RF – RR=
530
442 N
1· 2
= ...(ii)
Adding equations (i) and (ii),
2 RF= 11 760 + 442 = 12 202
∴ RF= 6101 N
Substituting the value of RF in equation (i),
RF + RR= 11 760 N
∴ RR= 11 760 – RF = 11 760 – 6101 = 5649 N Ans.