Question

A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals: (Take g = 10 ms^–2 and the rope to be massless)

(1) 100 N

(2) 90 N

(3) 75 N

(4) 70 N

Answer 1

Answer is (1) 100 N

For equilibrium,

T sin 45° = F ....(1)

and T cos 45° = 10g ....(2)

equation (1)/(2)

we get F = 10g

= 100 N

Answer 2

Answer is (1) 100 N

F = 100 N