Engineering Mechanics

Chapter 25 : Motion of Connected Bodies „„„„„ 507

Solution. Given : First mass (m 1 ) = 15 kg ; Second mass (m 2 ) = 6 kg and third mass (m 3 ) = 4 kg
From the system of pulleys and masses, we find that at pulley A, the 15 kg mass will come
down with some acceleration as the total mass on the other side of the string is less than 15 kg.
Similarly, at pulley B, the 6 kg mass will come down with some acceleration.

Let a 15 = Acceleration of 15 kg mass

a 6 = Acceleration of 6 kg mass, and

a 4 = Acceleration of 4 kg mass.

We know that acceleration of the 15 kg mass,

() ()

()

12 2

15

12

9.8 15 6 4

1.96 m/s

15 6 4

gm m

a

mm

− ⎡⎤⎣⎦−+

== =

+++

Similarly, a 6 =

() 2

4

9.8 6 4

1.96 m/s

64

a

−

==

+

Now if we look at all the three masses to move simultaneously at pulleys A and B, we find that

1. The mass 15 kg will come downwards with an acceleration of 1.96 m/s^2. Ans.


2. The pulley B will go up with an acceleration of 1.96 m/s^2.


3. The 6 and 4 kg masses will go up (because the pulley B is going up) with an acceleration
   of 1.96 m/s^2. Moreover, at pulley B, the 6 kg mass will come down (because 4 kg mass
   will go up) with an acceleration of 1.96 m/s^2. Thus the net acceleration of the 6 kg mass
   will be zero. Ans.


4. At pulley B the 4 kg mass will go up (because the 6 kg mass will come down) with an
   acceleration of 1.96 m/s^2. Thus the net acceleration of the 4 kg mass will be 1.96 + 1.96
   = 3.92 m/s^2. Ans.

25.3.MOTION OF TWO BODIES CONNECTED BY A STRING, ONE OF

WHICH IS HANGING FREE AND THE OTHER LYING ON A SMOOTH
HORIZONTAL PLANE
Consider two bodies of masses m 1 and m 2 kg
respectively connected by a light inextensible string as shown
in Fig. 25.4.

Let the body of mass m 1 hang free and the body of
mass m 2 be placed on a smooth (i.e., friction between the
body of mass m 2 and the horizontal plane is neglected) hori-
zontal plane. It may be noted that the body of mass m 1 will
move downwards and the body of mass m 2 along the surface
of the plane.

We know that velocity and acceleration of the body
of mass m 1 will be the same as that of the body of mass m 2.
Since the string is inextensible, therefore, tension in both the strings will also be equal,

Let a = Acceleration of the system and
T = Tension in the strings
First of all, consider the motion of body 1 of mass m 1 downwards. We know that forces acting
on it are m 1 .g (downwards) and T (upwards). As the body is moving downwards, therefore resultant
force

= m 1 g – T (downwards) ...(i)

Fig. 25.4 Body lying over

smooth horizontal plane.