A 1 kg mass is suspended from the ceiling by a rope of length 4 m. A horizontal force F is applied at the mid point of the rope so that the rope makes an angle of 45° with respect to the vertical axis as shown in figure. The magnitude of F is (Assume that the system is in equilibrium and g=10 m/s²)

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(2) 10 N. Idea: take the knot as the body. Three forces meet there — the upper rope's tension, the applied F, and the lower rope's pull, which is vertical and equal to the weight. Let T be the tension in the upper half, at 45° to the vertical. Vertical: T cos 45°=mg=10 N. Horizontal: T sin 45°=F. Dividing one by the other, F=mg tan 45°=10×1=10 N. The rope's 4 m length and the fact that F acts at the midpoint are both irrelevant to the magnitude — they only decide where the bend appears. At 45° the horizontal push must equal the weight exactly, which is a useful thing to recognise on sight.

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