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 Amg2cotθ 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.