Question

4. Two equal weights of 20 N are attached to the ends of a thin string which passes over three smooth pegs in a wall arranged in the form of an equilateral triangle with one side horizontal. Find the tension on each peg.​

Answer 1

Tension in topmost peg = 40rt(5/3)

Tension in other pegs = 40/rt(3)

Step-by-step explanation:

Note point: I am taking T as a vector quantity and not scalar.

If the tension in the horizontal pegs is T, then the tension in the topmost peg will be T + T.

Lets find the magnitude of the vector T

|T| = 20/cos 30 can be easily seen by drawing FBD

= 40/rt(3)

Then, tension in topmost peg = rt{1600/3 + 1600/3 + (3200/3)*1/2}

= rt{(3200/3)*(5/2)} = rt(8000/3)