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WAEC 2020 · Paper 2 · Q14

A body of mass 10 kg10\text{ kg} rests on a rough plane inclined at an angle of tan⁡−1(512)\tan^{-1}\left(\frac{5}{12}\right) to the horizontal. The coefficient of friction between the body and the plane is 34\frac34. A force of magnitude P NP\text{ N} acts on the body along the inclined plane. Find the value of PP if the body is at the point of moving: [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

  1. (a)

    down the plane;

  2. (b)

    up the plane.

Worked solution (try it first)
  1. tan⁡θ=512\tan\theta = \frac{5}{12} gives a 5–12–13 triangle: sin⁡θ=513\sin\theta = \frac{5}{13} and cos⁡θ=1213\cos\theta = \frac{12}{13}.
  2. R=100cos⁡θR = 100\cos\theta
    =120013= \frac{1200}{13}
    ≈92.31 N\approx 92.31\text{ N}, so limiting friction is 34×92.31≈69.23 N\frac34 \times 92.31 \approx 69.23\text{ N}.
  3. The weight down the plane is 100sin⁡θ≈38.46 N100\sin\theta \approx 38.46\text{ N}.

(a)

  1. About to move down: friction acts up the plane, so P+38.46=69.23P + 38.46 = 69.23 and P≈30.77 NP \approx 30.77\text{ N}.

(b)

  1. About to move up: friction acts down the plane, so P=38.46+69.23≈107.69 NP = 38.46 + 69.23 \approx 107.69\text{ N}.

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