Engineering Mechanics

(^656) „„„„„ A Textbook of Engineering Mechanics

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.


3. An anticlockwise moment equal to P × y.
   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 motion of the wheels. We know that acceleration of the wheels,

a= rα or α =

a

r

and torque, T=F × r ...(iv)

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

× =M.a

∴ P= 2 2

Ia I

Ma a M

rr

+=⎛⎞+

⎜⎟

⎝⎠

or a=

2

P

I

M

r

⎛⎞

⎜⎟+

⎝⎠

...(vii)

Since the vehicle has motion of translation only, therefore the couple (having moment equal to

2 Fh) and moment (whose magnitude is equal to P × y) and the remaining forces balance among

themselves. Now considering the vertical forces only,

RF + RR=Mg (in kN) ...(viii)

Now taking moments about centre of gravity (G) of the vehicle, RF x – RR x = 2Fh – Py

RF – RR= 2

2

• 2

hPyIa

xxr

⎡⎤

⎢⎥⎣⎦ ... 22

Ia

F

r

⎛⎞

⎜⎟=

⎝⎠

Q