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WAEC 2022 · Paper 1 · Q37

A particle is acted upon by forces F=(10 N,060∘)F = (10\text{ N}, 060^\circ), P=(15 N,120∘)P = (15\text{ N}, 120^\circ) and Q=(12 N,200∘)Q = (12\text{ N}, 200^\circ). Express the force that will keep the particle in equilibrium in the form xi+yjx\mathbf{i} + y\mathbf{j}, where xx and yy are scalars.

Worked solution (try it first)
  1. The angles are bearings, so a force (R,θ)(R, \theta) is Rsin⁡θ i+Rcos⁡θ jR\sin\theta\,\mathbf{i} + R\cos\theta\,\mathbf{j}.
  2. The i\mathbf{i} parts: 10sin⁡60∘+15sin⁡120∘+12sin⁡200∘=8.660+12.990−4.10410\sin 60^\circ + 15\sin 120^\circ + 12\sin 200^\circ = 8.660 + 12.990 - 4.104
    =17.55= 17.55.
  3. The j\mathbf{j} parts: 10cos⁡60∘+15cos⁡120∘+12cos⁡200∘=5−7.5−11.27610\cos 60^\circ + 15\cos 120^\circ + 12\cos 200^\circ = 5 - 7.5 - 11.276
    =−13.78= -13.78.
  4. The resultant is 17.55i−13.78j17.55\mathbf{i} - 13.78\mathbf{j}.
  5. The force that keeps the particle in equilibrium is equal and opposite: −17.55i+13.78j-17.55\mathbf{i} + 13.78\mathbf{j}, option C.

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