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Question

Assuming frictionless contact everywhere, determine the magnitude of external horizontal force P applied at the lower end for equilibrium of the rod as shown in figure. The rod is uniform and its mass is m.


A
mg2cotθ
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B
mgtanθ
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C
mg2sinθ
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D
mgcosθ
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Solution

The correct option is A mg2cotθ
Let the length of the rod =l. Let N1 and N2 be the normal reaction forces acting on the end A and B in a direction to the contact surface.


For rod to be in complete equilibrium, Fnet=0 & τnet=0,
For translational equilibrium along horizontal & vertical direction
Fx=0
N2=P ...(i) for horizontal direction.
Similarly Fy=0 along vertical direction
N1=mg ...(ii)

For rotational equilibrium the net torque be zero about point B, τnet=0 ...(iii)
Taking anticlockwise sense of rotation for torque as +ve
& τ=F×r
Force N1 passes through reference point B, hence its torque about point B is zero.


Substituting the torquex with proper signs in Eq (iii),
τ=0
N2(lsinθ)+mg(l2cosθ)+0=0
N2=mg2cotθ

Substituting N2=P from equation (i)
P=mg2cotθ
Option (a) is correct.

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