Stepwise
Start practisingPractise

WAEC 2010 · Paper 2 · Q8

A body of mass 6 kg is hung from a fixed point by a light inextensible string. A horizontal force is applied to the body such that the body is in equilibrium when the string is inclined at 35∘35^\circ to the vertical. Find, correct to one decimal place, the: [Take g=10g = 10 m s−2^{-2}]

  1. (a)

    horizontal force;

  2. (b)

    tension in the string.

Worked solution (try it first)
  1. The weight is 6×10=606 \times 10 = 60 N downwards.
  2. The forces are the weight, the horizontal force FF and the tension TT along the string, 35∘35^\circ to the vertical.

(a)

  1. Resolve vertically: Tcos⁡35∘=60T\cos35^\circ = 60.
  2. Resolve horizontally: Tsin⁡35∘=FT\sin35^\circ = F.
  3. Divide the two equations: F=60tan⁡35∘=42.0F = 60\tan35^\circ = 42.0 N.

(b)

  1. From the vertical equation: T=60cos⁡35∘=73.2T = \dfrac{60}{\cos35^\circ} = 73.2 N.

Report a problem with this question