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Two forces F1 and F2 are acting at point A as shown in. The angle between the two forces is 50°. It is found that the resultant R is 500 N and makes angles 20° with the force F1 as shown in the figure. Determine the forces F1 and F2. |
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Answer» Let ∆ ABC be the triangle of forces drawn to some scale. In this ∠BAC = α = 20° ∠ABC = 180 – 50 = 130° ∴ ∠ACB = 180 – (20 + 130) = 30° Applying sine rule to ∆ ABC, we get \(\cfrac{AB}{sin\,30^\circ}\) = \(\cfrac{BC}{sin\,20^\circ}\) = \(\cfrac{500}{sin\,130^\circ}\) AB = 326.35 N and BC = 223.24 N. Thus F1 = AB = 326.35 N and F2 = BC = 223.24 N |
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