Fig. 3.8 Two unequal masses are att-
ached to the ends of a string passing
over a pulley in such a way that the
body A moves vertically and body
B moves on a smooth horizontal plane.
Answers
ANSWER
The given system of two masses and a pulley can be represented as shown in the following figure:
Smaller mass, m
1
=8 kg
Larger mass,m
2
=12kg
Tension in the string=T
Mass m
2
, owing to its weight, moves downward with acceleration, and mass m
1
moves upward.
Applying Newton’s second law of motion to the system of each mass:
For mass m
1
:
The equation of motion can be written as:
T–m
1
g=ma… (i)
For mass m2:
The equation of motion can be written as:
m
2
g–T=m2a … (ii)
Adding equations (i) and (ii), we get:
(m2−m1)g=(m1+m2)a
a=
(m1+m2)
(m2−m1)g
....(iii)
=
(12+8)
(12−8)
×10=
20
4×10
=2ms−2
Therefore, the acceleration of the masses is 2 m/s2.
Substituting the value of a in equation (ii), we get:
m
2
g−T=m
2
(m
1
+m
2
)
(m
2
−m
1
)g
T=(m2−
(m1+m2)
(m
2
2
−m1m2)
g
=
(m
1
+m
2
)
2m
1
m
2
g
T=2×12×8×10/(12+8)
T=96N
Therefore, the tension in the string is 96 N.
solution
3×8 two unequal masses are attached to the ends