Force on String:
A string ABCD, attached to the two fixed points A and D has two weight of 1000N each attached to it at points B and C. The weights rest with portions AB and CD inclined at an angle of 30° and 60° respectively, to the vertical as shown in the figure given below. Find out the tension in portion AB, BC, CD
Sol.: First the strings ABCD is split in to two parts, and take the joints B and C separately
Let,
T_{1} = Tension in the String AB
T_{2 } = Tension in the String BC
T_{3} = Tension in the String CD
As at joint B three forces are acting. Therefore apply lamis theorem at joint B,
T_{1}/sin60° = T_{2}/sin150° = 1000/sin150°
T_{1} = {sin60° × 1000}/sin150°
= 1732N .......ANS
T_{2} = {sin150° × 1000}/sin150°
= 1000N .......ANS
Apply lamis theorem at joint C,
T_{2}/sin120° = T_{3}/sin120° = 1000/sin120°
T_{3} = {sin120° × 1000}/sin120°
= 1000N .......ANS