WAEC 2014 · Paper 2 · Q15

  1. (a)

    PQRSPQRS is a square. Calculate the magnitude and direction of the resultant of the forces 5 N5\text{ N} acting along PQPQ, 32 N3\sqrt2\text{ N} acting along PRPR and 3 N3\text{ N} acting along PSPS.

    Separate values with commas, e.g. 3, −2

  2. (b)

    A book of mass 0.8 kg0.8\text{ kg} is placed on a plane inclined at 30∘30^\circ to the horizontal. Find the frictional force if the book: (i) does not slide; (ii) slides with an acceleration of 1.2 m s−21.2\text{ m s}^{-2}. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. Take PQPQ and PSPS as the two directions.
  2. PRPR is the diagonal, at 45∘45^\circ to each.
  3. Along PQPQ: 5+32cos⁡45∘=5+35 + 3\sqrt2\cos45^\circ = 5 + 3
    =8 N= 8\text{ N}.
  4. Along PSPS: 3+32sin⁡45∘=3+33 + 3\sqrt2\sin45^\circ = 3 + 3
    =6 N= 6\text{ N}.
  5. R=82+62=10 NR = \sqrt{8^2 + 6^2} = 10\text{ N}, at tan⁡−168≈36.9∘\tan^{-1}\frac68 \approx 36.9^\circ to PQPQ.

(b)(i)

  1. The weight is 8 N8\text{ N}.
  2. Its part down the plane is 8sin⁡30∘=4 N8\sin30^\circ = 4\text{ N}.
  3. At rest, friction balances it: 4 N4\text{ N}.

(ii)

  1. Down the plane: 4−F=0.8×1.2=0.964 - F = 0.8 \times 1.2 = 0.96, so F=3.04 NF = 3.04\text{ N}.

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