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,

T₁  = Tension in the String AB

T₂  = Tension in the String BC 

T₃ = Tension in the String CD

As at joint B three forces are acting. Therefore apply lamis theorem at joint B,

T₁/sin60° = T₂/sin150° = 1000/sin150°

T₁ = {sin60° × 1000}/sin150°

= 1732N                                                   .......ANS

T₂ = {sin150° × 1000}/sin150°

= 1000N                                                   .......ANS

Apply lamis theorem at joint C,

T₂/sin120° = T₃/sin120° = 1000/sin120°

T₃ = {sin120° × 1000}/sin120°

= 1000N                                                   .......ANS

 

Posted Date: 10/16/2012 2:38:43 AM | Location : United States

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