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)

Solution

Physics sir · 4 steps

Draw the free body diagram of the midpoint, not the block, and the angle does the rest.

1. Lower rope first. The 1 kg block hangs at rest, so the lower part of the rope pulls up on it with T₂ = mg = 1 × 10 = 10 N. That same rope pulls the midpoint straight down with 10 N.
2. Forces on the midpoint. Three forces act at the midpoint: T₁ along the upper rope at 45° to the vertical, T₂ = 10 N downward, and F horizontally. The rope length of 4 m does not enter the calculation.

   T₁cos 45° = T₂, T₁sin 45° = F ⇒ F = T₂tan 45°

3. Put in numbers. tan 45° = 1, so F = 10 × 1 = 10 N. As a check, T₁ = 10√2 N, and its two components are each 10 N.
4. So. The horizontal force is F = 10 N, option (b).

The idea to keep. When a rope is pulled sideways at a point, take that point as your body. Balance vertical and horizontal parts separately, and the horizontal force comes out as weight times tan θ.

Answer from the NTA final answer key. Solution by PrepWiser.