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WAEC 2009 · Paper 2 · Q8

A body of mass 4 kg hangs from a fixed point OO by a light inextensible string. It is pulled aside by a horizontal force FF N and rests in equilibrium with the string inclined at an angle of 60∘60^\circ to the downward vertical. [Take g=10 m s−2g = 10\text{ m s}^{-2}] Calculate, correct to two decimal places, the:

  1. (a)

    magnitude of FF;

  2. (b)

    tension in the string.

Worked solution (try it first)

(a)

  1. The weight is mg=4×10=40mg = 4 \times 10 = 40 N, acting downwards.
  2. Three forces act on the body: the weight, the horizontal force FF and the tension TT along the string, at 60∘60^\circ to the downward vertical.
  3. Resolve horizontally: Tsin⁡60∘=FT\sin 60^\circ = F.
  4. Resolve vertically: Tcos⁡60∘=40T\cos 60^\circ = 40.
  5. Divide the first equation by the second: tan⁡60∘=F40\tan 60^\circ = \dfrac{F}{40}, so F=40tan⁡60∘=403=69.28F = 40\tan 60^\circ = 40\sqrt3 = 69.28 N (2 d.p.).

(b)

  1. From the vertical equation: T=40cos⁡60∘T = \dfrac{40}{\cos 60^\circ}
    =400.5= \dfrac{40}{0.5}
    =80.00= 80.00 N.

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