Engineering Mechanics

Chapter 32 : Motion of Vehicles „„„„„ 653

A little consideration will show that the horizontal forces acting on the vehicle may be
considered as :

1. A force equal to 2F passing through the c.g. of the vehicle (but acting opposite to the
   tractive force).


2. A clockwise couple having a moment equal to 2F × h.
   First of all, consider the motion of the vehicle, which is moving towards right. We know that
   the forces acting on it are tractive force P kN (towards right) and 2F kN (opposite to P ). As the body
   is moving towards right, therefore resultant force.

=P – 2F ...(i)

Since the vehicle is moving towards right with an acceleration (a), therefore force acting on it

=Ma ...(ii)

Equating equations (i) and (ii),

P – 2F=Ma ...(iii)

Now consider the motion of the wheels. We know that linear acceleration of the wheel,

a=rα or α =

a

r

...(iv)

and torque, T=F × r

We know that torque of the wheel,

T=

2

I

×α ...(where

2

I

is the mass moment

of inertia of^ each wheel)

=

22

IaIa

rr

×= ...(v)

Equating equations (iv) and (v),

F × r=

2

Ia

r

∴ F= 2

2

Ia

r

...(vi)

Substituting the value of F in equation (iii),

–2 22

Ia

P

r

× =Ma

∴ P= 22

Ia I

Ma a M

rr

+= +⎛⎞

⎜⎟

⎝⎠

and a=

2

P

I

M

r

⎛⎞

⎜⎟+

⎝⎠

...(vii)

Since the vehicle has motion of translation only, therefore the couple (having moment equal to
2 F.h) and the remaining forces balance among themselves. Now considering the vertical forces only.

RF + RR=Mg ...(viii)