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 m s⁻² and the rope to be massless)

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(2) 100 N. Idea: the knot at the midpoint is in equilibrium under three forces — the tension above, the applied F, and the weight hanging below on a vertical rope. With T the tension in the upper half at 45° to the vertical: Vertical: T cos 45°=mg=10×10=100 N. Horizontal: T sin 45°=F. Dividing, F=mg tan 45°=100 N. The rope's length plays no part. At 45° the horizontal force always equals the weight — and the upper half then carries T=100√2≈141 N, more than either.

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 m s⁻² and the rope to be massless)

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