Question

A fine light string ABCDE whose extremity A is fixed, has weights W₁ and W₂ attached to it at B and C. It passes round a small smooth peg at D carrying a weight of 40 N at the free end E as shown in fig. If in the position of equilibrium, BC is horizontal and AB and CD makes 150° and 120° with BC, find (i) Tension in the portion AB,BC and CD of the string and (ii) Magnitude of W₁ and W₂.

Answer 1

First string ABCD is split in to two parts, and consider the joints B and C separately Let

T₁ = Tension in String AB

T₂ = Tension in String BC

T₃ = Tension in String CD

T₄ = Tension in String DE 

T₄ = T₃ = 40 N

Since at joint B and C three forces are acting on both points. But at B all three forces are unknown and at point C only two forces are unknown 

SO Apply lamis theorem first at joint C,

T₂ /sin 150° = W₂/sin 120° = 40/sin 90°

= 20N

T₂ = {sin 150° × 40}/sin 90°

W₂ = {sin 120° × 40}/sin 90°

= 34.64N

Now for point B, We know the value of T₂ 

So, Again Apply lamis theorem at joint 

B, T₁/sin 90° = W₁/sin 150° = T₂/sin 120°

T₁ = {sin 90° × 20}/sin 120° 

= 23.1 N

W₁ = {sin 150° × 20}/sin 120°

= 11.55N