Problem 1 : Two small spherical objects of mass “m” and “M” are attached with strings as shown in the figure. Find the angle “θ” such that the given system is in equilibrium.
Solution : Every point of the system is under action of balanced force system. We shall work through the points "A" and "B", where masses are attached.
Let “
T
1
T
1
” and “
T
2
T
2
” be the tensions in the two strings, meeting at point "A". The forces at "A" are shown in the figure above. Considering balancing of forces in "x" and "y" directions :
∑
F
x
=
T
1
cos
45
0

T
2
cos
45
0
=
0
∑
F
x
=
T
1
cos
45
0

T
2
cos
45
0
=
0
⇒
T
1
=
T
2
=
T
say
⇒
T
1
=
T
2
=
T
say
∑
F
y
=
T
1
sin
45
0
+
T
1
sin
45
0
=
m
g
∑
F
y
=
T
1
sin
45
0
+
T
1
sin
45
0
=
m
g
⇒
T
sin
45
0
+
T
sin
45
0
=
m
g
⇒
T
sin
45
0
+
T
sin
45
0
=
m
g
⇒
2
T
2
=
m
g
⇒
2
T
2
=
m
g
⇒
T
=
m
g
2
⇒
T
=
m
g
2
Let “
T
3
T
3
” be the tension in the upper section. The forces on the mass “M” is shown in the figure above. Considering balancing of forces in "x" and "y" directions :
∑
F
x
=
T
3
cos
θ

T
sin
45
0
=
0
∑
F
x
=
T
3
cos
θ

T
sin
45
0
=
0
Substituting for “T” and evaluating, we have :
⇒
T
3
cos
θ
=
m
g
2
X
1
2
=
m
g
2
⇒
T
3
cos
θ
=
m
g
2
X
1
2
=
m
g
2
∑
F
y
⇒
T
3
sin
θ
=
T
cos
45
0
+
M
g
∑
F
y
⇒
T
3
sin
θ
=
T
cos
45
0
+
M
g
Substituting for "T",
⇒
T
3
sin
θ
=
T
2
+
M
g
=
m
g
2
X
1
2
+
M
g
⇒
T
3
sin
θ
=
T
2
+
M
g
=
m
g
2
X
1
2
+
M
g
⇒
T
3
sin
θ
=
mg
2
+
M
g
⇒
T
3
sin
θ
=
mg
2
+
M
g
Taking ratio of the resulting equations in the analysis of forces on "M",
tan
θ
=
m
g
2
+
M
g
m
g
2
tan
θ
=
m
g
2
+
M
g
m
g
2
⇒
tan
θ
=
1
+
2
M
m
⇒
tan
θ
=
1
+
2
M
m
⇒
θ
=
tan
−
1
1
+
2
M
m
⇒
θ
=
tan
−
1
1
+
2
M
m