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JEE Main 2022 (25 Jul, Shift 2) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from JEE Main 2022 (25 Jul, Shift 2)

Q13JEE Main 2022 · 25 JUL · SHIFT 2Medium

Equilibrium, Third Law & Frames

For a free body diagram shown in the figure, the four forces are applied in the 'x' and 'y' directions. What additional force must be applied and at what angle with positive x-axis so that the net acceleration of body is zero?

Figure

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Answer: (a)

  1. ΣFx=5−6=−1 N\Sigma F_x = 5 - 6 = -1\text{ N}, ΣFy=7−8=−1 N\Sigma F_y = 7 - 8 = -1\text{ N}
  2. Resultant =2 N= \sqrt{2}\text{ N}, pointing at 225∘225^\circ (third quadrant).
  3. The balancing force is equal and opposite: 2 N\sqrt{2}\text{ N} at 45∘45^\circ to +x+x.
  4. So the answer is (a) 2 N,45∘\sqrt{2}\text{ N}, 45^\circ.
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