Stepwise
Start practisingPractise

WAEC 2009 · Paper 2 · Q17✱

Coplanar forces of 4 N, 8 N, 6 N, 4 N and 5 N act at a point as shown in the diagram (not drawn to scale). If the 6 N force acts in the direction 090∘090^\circ, calculate the:

8 N6 N4 N4 N5 N55°25°100°130°
  1. (a)

    magnitude of the resultant force;

  2. (b)

    direction of the resultant force.

Worked solution (try it first)

(a)

  1. Take east (the 6 N direction) as the xx-axis and north as the yy-axis.
  2. Measured anticlockwise from east, the forces act at: 8 N at 55∘55^\circ, 6 N at 0∘0^\circ, 4 N at −25∘-25^\circ, 4 N at 155∘155^\circ (100∘100^\circ beyond the 8 N) and 5 N at 205∘205^\circ (130∘130^\circ beyond the lower 4 N).
  3. East components: 8cos⁡55∘+6+4cos⁡25∘−4cos⁡25∘−5cos⁡25∘=4.5886+6−4.53158\cos 55^\circ + 6 + 4\cos 25^\circ - 4\cos 25^\circ - 5\cos 25^\circ = 4.5886 + 6 - 4.5315
    =6.0571= 6.0571.
  4. North components: 8sin⁡55∘+0−4sin⁡25∘+4sin⁡25∘−5sin⁡25∘=6.5532−2.11318\sin 55^\circ + 0 - 4\sin 25^\circ + 4\sin 25^\circ - 5\sin 25^\circ = 6.5532 - 2.1131
    =4.4401= 4.4401.
  5. Magnitude: ∣F∣=6.05712+4.44012|\mathbf F| = \sqrt{6.0571^2 + 4.4401^2}
    =56.403= \sqrt{56.403}
    =7.51= 7.51 N (2 d.p.).

(b)

  1. The resultant makes an angle tan⁡−14.44016.0571=36.2∘\tan^{-1}\dfrac{4.4401}{6.0571} = 36.2^\circ with east, towards north.
  2. As a bearing: 090∘−36.2∘=053.8∘090^\circ - 36.2^\circ = 053.8^\circ, which is 054∘054^\circ to the nearest degree.

Report a problem with this question