JAMB - Mathematics (1978 - No. 17)
A force of 5 units acts on a particle in the direction to the east and another force of 4 units acts on the particle in the direction north-east. The resultants of the two forces is
\(\sqrt{3}\) units
3 units
\(\sqrt{41 + 20 \sqrt{2}}\) units
\(\sqrt{41 - 20 \sqrt{2}}\) units
Explanation
Force to the east = 5 units and angle to the east = 0º
force to the Northeast = 4 units and angle to northeast = 45º
Resolve these forces into x and y components
| FORCES | x - component | y - component |
| 5 | 5cos0 = 5 | 5sin 0 = 0 |
| 4 | 4cos45 = 4 x \(\frac{\sqrt2}{2}\) | 4sin45 = \(\frac{\sqrt2}{2}\) |
| Total force | 5 + 2\(\sqrt{2}\) | 0 + 2\(\sqrt{2}\) |
Resultant force = \(\sqrt{(F_x)^2 + (F_y)^2}\)
= \(\sqrt{ (5 + 2\sqrt{2})^2 + (2\sqrt{2})^2}\)
= \(\sqrt{(25 + 20\sqrt{2} + 8) + 8}\)
= \(\sqrt{33 + 20\sqrt{2} + 8}\) = \(\sqrt{41 + 20\sqrt{2}}\)
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