Three forces, F\(_1\) (8N, 030°), F\(_2\) (10N, 150° ) and F\(_3\) ( KN, 240° )are in equilibrium. Find the value of N
F\(_1\) (8N, 030°), F\(_2\) (10N, 150° ) and F\(_3\) ( KN, 240° )
Resolving the forces into x and y component
F\(_x\) = 8 cos 30 - 10cos 150 - N cos 240
F\(_y\) = 8 sin 30 + 10 sin 150 - N sin 240
For equilibrium, the sum of the forces in both x and y directions must be zero
F\(_x\) = 4\(\sqrt{3}\) - 5\(\sqrt{3}\) - \(\frac{N}{2}\) = 0, - \(\sqrt{3}\) = \(\frac{N}{2}\), so, N = - 2\(\sqrt{3}\)
F\(_y\) = 4 + 5 - \(\frac{N\sqrt{3}}{2}\) = 0, so, 9 = \(\frac{N\sqrt{3}}{2}\), 18 = N\(\sqrt{3}\), N = 6\(\sqrt{3}\)
Thus, the value of N = 6\(\sqrt{3}\).
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