Laws of Motion
Motion of bodies Connected by Strings
Two bodies in Contact
a
T m1 F
m2
a a
Free body diagrams :
T F T a
a
m1 m2
FBD for m1 FBD for m2 F m1 T T m2
Force on mass m1, F - T = m1a _____(1)
Force on mass m2, T = m2a _____(2)
F
a= ______(3)
m1+m2
Putting value of 'a' from equation (3) into (2), we get-
m2F
T=
m1+ m2
Three Connected bodies
T2 T2
T1 T1 m3
m1 m2
F
/////////////////////////////////////////////
a
a
a
T2 T2
T1 T1 m
m1 m2 3
F
/////////////////////////////////////////////
F – T1 = m1a _____(1)
T1 – T2 = m2a _____(2)
T2 = m3a _____(3)
F – T1 = m1a _____(1)
T1 – T2 = m2a _____(2)
T2 = m3a _____(3)
F ______(4)
a=
m1+ m2 + m3
Put the value of 'a' in equation (1)
F(m2 +m3)
T1 =
m1+m2 +m3
Now, put the value of 'a' in equation (3)
Fm3
T2 =
m1+m2 +m3
Question
Two bodies whose masses are m1 = 50 kg and m2 = 150 kg are
tied by a light string and are placed on a frictionless horizontal
surface. When m1 is pulled by a force F, an acceleration of
5 ms–2 is produced in both the bodies. Calculate the value of F.
What is the tension in the string connecting the two blocks?
(A) 1000 N, 750 N (B) 1250 N, 500 N
(C) 1500 N, 500N (D) 2000 N, 250 N
Question
Ans: (A) 1000 N, 750 N
Question
Two bodies of masses 10 kg and 20 kg respectively kept on a
smooth, horizontal surface are tied to the ends of a light string.
A horizontal force F = 600 N is applied to
(i) A,
(ii) B along the direction of string. What is the tension in the
string in each case?
Question
Ans: (i)400 N, (ii) 200 N
Question
Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each
other with light string and are then placed on a frictionless
surface as shown in the figure. The system is pulled by a force
F = 10 N, then tension T1 =
2 Kg 3 Kg 5 Kg
10N T1 T2
(A) 1N (B) 5 N (C) 8 N (D) 10 N
Ans: (C)
Body Hanged Vertically
//////// ////////
T3
m3
T1
T2
T1 = m1g _____(1) m1
m2
m1g
T1
m1
T2 //////// ////////
T3
m2
m3
m2g + T1
T2
T2 = ( m1 + m2 )g [From equation (1)]
T3 m2
m3 T1
m3g + T2 m1
T3 = (m1 + m2 + m3 )g [From equation (2)]
Bodies Accelerated Vertically Upwards
F
T1
T1 – m1 g = m1a T3
T1 = m1 (g + a) ____(1) m1
m3
m1g T2
a
T2
m2
T2 – m2 g – T1 = m2 a m2
T1
T2 = (m1 + m2 ) (g + a) ____(2)
m2g + T1
m1
F
T3 T3
m3
T3 – m3g – T2 = m3a
a m3
T2
a
m3g + T2
m2
T1
T3 = (m1 + m2 + m3 ) (g + a)
m1
Body Accelerated Vertically Downwards
F
T1
m1 g – T1 = m1a T3
T1 = m1 (g – a) ____(1) a m1
m3
m1g T2
a
T2
m2 g + T1 – T2 = m2 a m2
T2 = (m1 + m2 ) (g - a) ____(2) a m2
T1
m2g + T1
m1
F
T3
m3g + T2 – T3 = m3a T3
m3
a m3 T2
a
m3g + T2 m2
T1
T3 = (m1 + m2 + m3 ) (g – a)
m1
Question
Three blocks of masses m1, m2 and m3 are connected
by massless strings as shown in fig. They are pulled
with a force T3 = 40 N. If m1 = 10 kg, m2 = 6 kg and
m3 = 4 kg, the tension T2 will be
a
T1 T2 T3
m1 m2 m3
Question
Ans: 32 N
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Homework
Question
Two blocks, joined by a string, have masses of 6.0 and 9.0 kg.
They rest on a frictionless horizontal surface. A 2nd string,
attached only to the 9-kg block, has horizontal force = 30 N
applied to it. Both blocks accelerate. Find the tension in the
string between the blocks.
(A) 18 N (B) 28 N (C) 12 N (D) 15 N
Ans: (C)
Question
Three blocks are connected by string as shown in figure below,
and are pulled by a force T3 = 60N. If m1 = 3 kg, m2 = 4 kg and
m3 = 5 kg. Calculate the acceleration of the system, T1 and T2.
Ans: 35N
Question
When the frictionless system shown below is accelerated by an
applied force of magnitude F the tension in the string between
the blocks is? a
T 2 kg F
1kg
Ans: F/3N
Question T3
Three blocks A, B and C weighing 1, 8 and 27 kg
respectively are connected as shown in the figure
with an inextensible string and are moving upward.
T3 is equal to 36 N. Then T2 is m3
T2
a
m2
T1
m1
Ans: 9 N
Question 10 N
Three blocks of masses 2 kg, 3 kg and 5 kg are
connected to each other with light string and as
shown in the figure. The system is accelerated
downward then tension T1 = 2 kg
T1
a
3 kg
T2
5 kg
Ans: 8 N
