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NEET 2008 (AIPMT) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from NEET 2008 (AIPMT)

Q22NEET 2008 · AIPMTMedium

Equilibrium, Third Law & Frames

Figure

Three forces acting on a body are shown in the figure. To have the resultant force only along the y- direction, the magnitude of the minimum additional force needed is

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Answer: (c)

  1. From the figure: 1 N1\text{ N} at 60∘60^\circ above +x+x; 4 N4\text{ N} at 30∘30^\circ from +y+y towards −x-x; 2 N2\text{ N} along the same line as the 4 N4\text{ N}, i.e. 60∘60^\circ below +x+x.
  2. ΣFx=1cos⁡60∘−4sin⁡30∘+2cos⁡60∘=0.5−2+1=−0.5 N\Sigma F_x = 1\cos 60^\circ - 4\sin 30^\circ + 2\cos 60^\circ = 0.5 - 2 + 1 = -0.5\text{ N}
  3. To leave only a yy-component, cancel FxF_x with a force along +x+x; this is the minimum force.
  4. So the answer is (c) 0.5 N.
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