Stepwise
Start practisingPractise

WAEC 2023 · Paper 2 · Q7✱✱

Forces (8 N,080∘)(8\text{ N}, 080^\circ), (18 N,240∘)(18\text{ N}, 240^\circ) and (6 N,300∘)(6\text{ N}, 300^\circ) act on a particle. Find, correct to two decimal places, the magnitude of the resultant force.

  1. (a)

    Magnitude of the resultant

Worked solution (try it first)
  1. East parts: 8sin⁡80∘+18sin⁡240∘+6sin⁡300∘=7.878−15.588−5.1968\sin80^\circ + 18\sin240^\circ + 6\sin300^\circ = 7.878 - 15.588 - 5.196
    =−12.906= -12.906.
  2. North parts: 8cos⁡80∘+18cos⁡240∘+6cos⁡300∘=1.389−9+38\cos80^\circ + 18\cos240^\circ + 6\cos300^\circ = 1.389 - 9 + 3
    =−4.611= -4.611.
  3. ∣R∣=12.9062+4.6112|\mathbf R| = \sqrt{12.906^2 + 4.611^2}
    =187.83= \sqrt{187.83}
    ≈13.71 N\approx 13.71\text{ N}.

Report a problem with this question