Stepwise
Start practisingPractise

WAEC 2019 · Paper 2 · Q14

  1. (a)

    Two forces T(8 N,030∘)T(8\text{ N}, 030^\circ) and Q(10 N,150∘)Q(10\text{ N}, 150^\circ) act on a body. Find the: (i) component of the resultant force along the xx-axis; (ii) magnitude of the acceleration if the body has a mass of 4 kg4\text{ kg}.

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

  2. (b)

    A uniform beam of mass 150 kg150\text{ kg} and length 4 m4\text{ m} is supported at a point 1 m1\text{ m} from one end, and a second support has a reaction of 588 N588\text{ N}. What is the distance between the two supports if the system is in equilibrium? [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

Worked solution (try it first)

(a)(i)

  1. TT: 8sin⁡30∘=48\sin30^\circ = 4 east.
  2. QQ: 10sin⁡150∘=510\sin150^\circ = 5 east.
  3. The xx-component of the resultant is 4+5=9 N4 + 5 = 9\text{ N}.

(ii)

  1. North parts: 8cos⁡30∘+10cos⁡150∘=6.928−8.6608\cos30^\circ + 10\cos150^\circ = 6.928 - 8.660
    =−1.732= -1.732.
  2. ∣R∣=81+3|\mathbf R| = \sqrt{81 + 3}
    =84= \sqrt{84}
    ≈9.165 N\approx 9.165\text{ N}.
  3. a=Fma = \dfrac{F}{m}
    =9.1654= \dfrac{9.165}{4}
    ≈2.29 m s−2\approx 2.29\text{ m s}^{-2}.

(b)

  1. The weight, 150×10=1500 N150 \times 10 = 1500\text{ N}, acts at the middle of the beam, 2 m from the end.
  2. That is 1 m1\text{ m} from the first support.
  3. Let the second support be xx m from the first.
  4. Moments about the first support: 588x=1500×1588x = 1500 \times 1, so x≈2.55 mx \approx 2.55\text{ m}.

Report a problem with this question