AP EAMCET · PHYSICS · Laws of Motion
Three forces of magnitude \(F_1, F_2\) and \(F_3\) act on a body located at the origin as shown in the figure. The condition that gives zero net force is
- A \(F_2=-(2+\sqrt{3}) F_1: F_3=\frac{-4}{\sqrt{6}-\sqrt{2}} F_1\)
- B \(F_2=-(2-\sqrt{3}) F_1: F_3=\frac{-4}{\sqrt{6}+\sqrt{2}} F_1\)
- C F2 = [-F1 (2 + √3)]; F3= [√6-√2]F1
- D \(F_2=-(2+\sqrt{2}) F_1: F_3=\frac{-2}{\sqrt{6}-\sqrt{2}} F_1\)
Answer & Solution
Correct Answer
(C) F2 = [-F1 (2 + √3)]; F3= [√6-√2]F1
Step-by-step Solution
Detailed explanation
Forces given are as shown Resolving components we have, For equilibrium ; \(\Sigma F_y=0\) \(F_2 \sin 60^{\circ}+F_1 \sin 30^{\circ}=F_3 \cos 45^{\circ}\) \(\Rightarrow \quad F_2 \times \frac{\sqrt{3}}{2}+F_1 \times \frac{1}{2}=F_3 \times \frac{1}{\sqrt{2}}\)…
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