Engineering Mechanics

Chapter 32 : Motion of Vehicles „„„„„ 667

∴ a 2 =

• 100 –3.125m/s 2
  (2 16)

=

×

...(Minus sin due to retardation)

and final velocity of the vehicle (v),

0=u + a 2 t 2 = 10 – 3·125 t 2

∴ t=

10

3.2 s

3·125

= Ans.

Note: It will be interesting to know that when the vehicle is going up the plane then distance
covered and time taken in coming to stop is less than that when the vehicle is coming down the plane.

Example 32.7. A military car of 1000 kg mass has two pair of wheels 2·25 m apart. When
the car is standing on a level track, its centre of gravity is 1·25 m from the front wheel and 50 cm
above the road. If the coefficient of friction between the tyres and road is 0.35, find the maximum
inclination on which the car can climb with a constant velocity when
(i) It is driven by the rear pair of wheels only; and
(ii) It is driven by the front pair of wheels only.
Solution. Giben: Mass of the car (M) = 1000 kg; Distance between the centre of wheel (d) =
2·25 m; Distance of C.G. of the car from front wheel (x 1 ) = 1·25 m; Height of c.g. of the vehicle above
the road (h) = 50 cm = 0·5 m and coefficient of friction (μ) = 0·35

(i) Maximum inclination on which the car can climb when it is driven by the rear pair of wheels only

Fig. 32.10.
Let RF= Reaction at the front wheel,
RR= Reaction at the rear wheel, and
α= Maximum angle of inclination which the car can climb.
We know that RF + RR=Mg cos α = 1000 × 9·8 cos α = 9800 cos α
and as the car is driven by the rear pair of wheels only, therefore force of friction acts on the
rear pair of wheels only,

FR=μ RR = 0·35 RR ...(i)

Resolving the forces along the plane,

FR=Mg sin α = 1000 × 9·8 sin α = 9800 sin α ...(ii)

Equating equations (i) and (ii)

0·35 RR= 9800 sin α

∴ RR=

9800 sin

28 000 sin

0·35

α

=α

and RF= 9800 cos α – 28 000 sin α