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JEE Advanced 2014 Paper 1
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© examsnet.com
Question : 10
Total: 60
In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle
θ
with the horizontal floor. The coefficient of friction between the wall and the ladder is
μ
1
and that between the floor and the ladder is
μ
2
.The normal reaction of the wall on the ladder is
N
1
and that of the floor is
N
2
.If the ladder is about to slip, then
μ
1
=
0
μ
2
≠
0
and
N
2
tan
θ
=
m
g
2
μ
1
≠
0
μ
2
=
0
and
N
1
tan
θ
=
m
g
2
μ
1
≠
0
μ
2
≠
0
and
N
2
=
m
g
1
+
μ
1
μ
2
μ
1
=
0
μ
2
≠
and
N
1
tan
θ
=
m
g
2
Validate
Solution:
Condition of transnational equilibrium
N
1
=
μ
2
N
2
N
2
+
μ
1
N
1
=
M
g
Solving
N
2
=
m
g
1
+
μ
1
μ
2
N
1
=
μ
2
m
g
1
+
μ
1
μ
2
Applying torque equation about corner(left) point on the floor
m
g
l
2
cos θ =
N
1
l
s
i
n
θ
+
μ
1
N
1
l
cos θ
Solving tan θ
=
1
−
μ
1
μ
2
2
μ
2
© examsnet.com
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