Problem 6 : The blocks A and B weighing 10 N and 20 N are connected by a string. The block B, in turn, is connected to block C with another string passing over a pulley. Friction forces at all interfaces is negligible. If the block system is in equilibrium, find the weight of C and tensions in the two strings.
Solution : There are two strings. Hence, the tensions in the strings will be different. Let
T
1
T
1
and
T
2
T
2
be the tension in strings AB and BC respectively.
Looking at the various body systems, we guess that the simplest force system is the one associated with block C. However, force analysis of block C will not yield anything as we do not know its weight or the tension
T
2
T
2
.
Thus, we begin with block A.
Free body diagram of A
Free body diagram of A
The external forces are (i) weight of A = 10 N (ii) Normal force applied by incline i.e.
N
1
N
1
(say) (iii) tension in AB,
T
1
T
1
.
∑
F
x
=
10
sin
30
0

T
1
=
0
⇒
T
1
=
10
sin
30
0
=
5
N
∑
F
x
=
10
sin
30
0

T
1
=
0
⇒
T
1
=
10
sin
30
0
=
5
N
We need not analyze forces in y – direction as we are not required to determine normal force
N
1
N
1
and it is not expected to be used for analyzing force on block B.
Free body diagram of B
Free body diagram of B
The external forces are (i) weight of B,
W
C
W
C
= 20 N (ii) Normal force applied by incline i.e.
N
2
N
2
(say) (iii) tension in AB,
T
1
T
1
and (iv) tension in BC,
T
2
T
2
.
∑
F
x
=
20
sin
30
0
+
T
1

T
2
=
0
⇒
T
2
=
20
x
1
2
+
5
=
15
N
∑
F
x
=
20
sin
30
0
+
T
1

T
2
=
0
⇒
T
2
=
20
x
1
2
+
5
=
15
N
Free body diagram of C
Free body diagram of C
The external forces are (i) weight of c = ? and (ii) tension in BC,
T
2
T
2
.
∑
F
y
=
T
2

W
C
=
0
⇒
W
C
=
T
2
=
15
N
∑
F
y
=
T
2

W
C
=
0
⇒
W
C
=
T
2
=
15
N