A weight W is suspended by using two strings. One of the strings makes an angle of 30^(@) with the vertical. What should be the direction of the other string so that the tension in it becomes minimum? Find the tension in each string at this position.

A weight W is suspended by using two strings. One of the strings makes

1.	A weight W is suspended by using two strings. One of the strings makes an angle of 30^(@) with the vertical. What should be the direction of the other string so that the tension in it becomes minimum? Find the tension in each string at this position.
Answer» Solution :Let the angle between the two strings be `alpha` Fig. Given that the angle between the WEIGHT W and the first string =`180^(@)-30^(@)= 150^(@)` Hence, angle between the weight and the second string `=360^(@)-150^(@)-alpha` =`210^(@)-alpha` For equilibrium using Lami.s theorem, `T_(2)/(sin150^(@))=(W)/(sinalpha)or,(T_(2))/(0.5)=(W)/(sinalpha) or,T_(2)=(W)/(2sinalpha)` For `T_(2)` to be minimum, sin `alpha` should be maximum i.e., sin`alpha` =1 or `alpha=90^(@)` . Hence the tension in the second string will be minimum when it is at right angles to the first string. At this setting, `T_(2) = (W)/(2)` Also , `(W)/(sin90^(@)) = (T_(1))/(sin(210-90^(@)))= (T_(1))/(sin120^(@))` or, `W = (T_(1))/(sin60^(@))` `:. T_(1)= W sin 60^(@)= Wxx (SQRT(3))/(2)` `:." " T_(1) = (sqrt(3)W)/(2)" ""and"" "T_(2)=(W)/(2)`.