Engineering Mechanics

Chapter 32 : Motion of Vehicles „„„„„ 655

and reaction on the rear pair of wheels,

RR= 2

1

• 2

h IP

Mg

x IMr

⎡⎤⎛⎞

⎢⎥⎜⎟+

⎣⎦⎝⎠

= 2

11·50·09 5·4

89·8– 38·98kN

21·20·09 8(0·4)

⎡ ⎛⎞× ⎤

⎢⎥×=⎜⎟

⎣⎦⎝⎠+^

Ans.

32.4.MOTION OF A VEHICLE ALONG A LEVEL TRACK WHEN THE TRACTIVE
FORCE PASSES THROUGH A POINT OTHER THAN ITS CENTRE OF GRAVITY.

Fig. 32.2. Force passing through a point other than centre of gravity.
Consider a four-wheeled vehicle moving along a lelvel track. Let the tractive force (P) tending
to move the vehicle to the right pass through a point C below the centre of gravity of vehicle as shown
in Fig. 32.2.

Let M= Total mass of the vehicle
m= Mass of the two pair of wheels and their axles,
k= Radius of gyration of the wheels,
I= Total mass moment of inertia of both the pair of wheels
and their axles (such that I = mk^2 )
r= Radius of the wheels,
h= Height of centres of gravity (G) of the vehicle above the
level track,
y= Distance between the centre of gravity (G) of the vehicle and
the point through which the tractive force (P) acts,
a= Linear acceleration of the vehicle due to tractive force,
α= Angular acceleration of the wheels,
RF and RR= Reactions at the pair of front and rear wheels respectively,
2 x= Horizontal distance between the reactions RA and RB,
F= Forces of friction acting at each wheel (such that total force of
friction is 2F).
For simplicity, let us consider the front and back pair of wheels to be symmetrical about the
axis dividing the wheels. Now let us apply two equal and opposite forces of magnitude 2F parallel to
the force of friction and through the centre of gravity (G) of the vehicle as shown in Fig. 32.2. A little
consideration will show that the horizontal forces acting on the vehicle may be considered as :