Question

A uniform chain of mass "M" and "L" is attached with the vertical wall so that exactly half of its length is resting over rough horizontal surface and half of length is hanging.If tangent drawn at the point "P" in the "]chain makes an angle 37 with vertical as shown in figure,then frictional force acting on the chain,is

Answer 1

see free body diagram,

weight of centre of mass = Mg/2 [ shown in figure ]

now at equilibrium,

$T_A=T_Bsin37^{\circ}$.....(1)

and $\frac{Mg}{2}=T_Bcos37^{\circ}$.....(2)

from equations (1) and (2),

so, $\frac{T_A}{\frac{Mg}{2}}=tan37^{\circ}$

or, $T_A=\frac{Mg}{2}tan37^{\circ}$

= $\frac{3Mg}{8}$.......(3)

as frictional force only acting on point A.

so, at equilibrium, frictional force at A = tension at A

or, frictional force = $T_A=\frac{3Mg}{8}$