Laws of Motion

JEE Main level, single correct, medium

91 of 210

Question 91 of 210

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

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Answer: (A) 2N,45∘\sqrt{2}N,45^{\circ}

Let additional force required be F⃗\vec{F}.
Vector sum of the four forces and additional force.
F⃗+5i^−6i^+7j^−8j^=0\vec{F}+5\hat{i}-6\hat{i}+7\hat{j}-8\hat{j}=0
Force required to make net acceleration zero
F⃗=i^+j^\vec{F}=\hat{i}+\hat{j}
Therefore, the required force is ∣F⃗∣=12+12=2N\left|\vec{F}\right|=\sqrt{{1}^{2}+{1}^{2}}=\sqrt{2}N
Angle with xx-axis: tan⁡θ=ycomponentxcomponent=11\tan \theta =\dfrac{ycomponent}{xcomponent}=\dfrac{1}{1}
Thus, θ=45∘\theta =45^{\circ}

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