A mass M kg is suspended by a weightless string. The horizontal force required to hold the mass at 60° with the vertical is

1) Mg

2) Mg√3

3) Mg(√3 +1)

4) Mg/√3

Answer: 2) Mg√3

Solution:

The tension acting on the string at equilibrium is T and the applied force is F

In the horizontal direction: F = T sin600

In the vertical direction: Mg = T cos600

Dividing both the equations we get: F/Mg = tan 600

F = Mg√3

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