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Two forces act at a point such that the sum of their magnitudes is 16 N. Their resultant is perpendicular to the smaller force and has magnitude 8 N. What is the magnitude of the smaller force?

1. 2 N
2. 4 N
3. 6 N
4. 7 N

Correct answer: 6 N

Solution

Let the smaller force be \(x\) and the larger force be \(16-x\). Since the resultant is perpendicular to the smaller force and has magnitude 8 N, the vector relation gives \(x(16-x)\cos\theta + x^2 = 0\) and \(R^2 = x^2 + (16-x)^2 + 2x(16-x)\cos\theta\); eliminating \(\cos\theta\) yields \(R^2 = (16-x)^2\), so \(16-x=8\). Hence \(x=6\,\text{N}\).

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