Four coplanar forces of magnitudes 40 N, 20 N, 50 N and F N act at a point in the directions shown in the diagram. The four forces are in equilibrium.
Find F and \(\alpha\).
Solution
Since the forces are in equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.
Resolving horizontally:
\(F \cos \alpha = 40 \sin 30^\circ + 20 \sin 60^\circ - 50 \sin 45^\circ\)
\(F \cos \alpha = 1.965 \ldots\)
Resolving vertically:
\(F \sin \alpha = 50 \cos 45^\circ + 20 \cos 60^\circ - 40 \cos 30^\circ\)
\(F \sin \alpha = 10.714 \ldots\)
Using Pythagoras' theorem to find \(F\):
\(F = \sqrt{(1.965 \ldots)^2 + (10.714 \ldots)^2} = 10.9 \text{ N}\)
To find \(\alpha\):
\(\alpha = \arctan\left(\frac{10.714 \ldots}{1.965 \ldots}\right) = 79.6^\circ\)
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