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JEE Main
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JEE Main
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Physics
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Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

Three forces acting on a body are shown in the figure. To have the...

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

Options

  1. A0.5 N0.5 \mathrm{~N}
  2. B1.5 N1.5 \mathrm{~N}
  3. C34 N\frac{\sqrt{3}}{4} \mathrm{~N}
  4. D3 N\sqrt{3} \mathrm{~N}

Correct answer

A. 0.5 N0.5 \mathrm{~N}

Step-by-step solution

Minimum additional force needed F=−(Fresultant )xFresultant =[(4−2)(cos⁡30j^−sin⁡30i^)=[2(32j^−12i^)+(12i^+32j^)]=[(3+32)j^+sin⁡60j^)=[−12i^+332i^+i^2)]=−i^2+332j^F=−(−i^2)=12i^\begin{aligned} & F=-\left(F_{\text {resultant }}\right)_x \\ & F_{\text {resultant }}=[(4-2)(\cos 30 \hat{\mathbf{j}}-\sin 30 \hat{\mathbf{i}}) \\ = & {\left[2\left(\frac{\sqrt{3}}{2} \hat{\mathbf{j}}-\frac{1}{2} \hat{\mathbf{i}}\right)+\left(\frac{1}{2} \hat{\mathbf{i}}+\frac{\sqrt{3}}{2} \hat{\mathbf{j}}\right)\right] } \\ = & {\left[\left(\sqrt{3}+\frac{\sqrt{3}}{2}\right) \hat{\mathbf{j}}+\sin 60 \hat{\mathbf{j}}\right) } \\ = & {\left.\left[-\frac{1}{2} \hat{\mathbf{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathbf{i}}+\frac{\hat{\mathbf{i}}}{2}\right)\right] } \\ = & -\frac{\hat{\mathbf{i}}}{2}+\frac{3 \sqrt{3}}{2} \hat{\mathbf{j}} \quad F=-\left(-\frac{\hat{\mathbf{i}}}{2}\right)=\frac{1}{2} \hat{\mathbf{i}}\end{aligned} Hence, ∣F∣=0.5 N|F|=0.5 \mathrm{~N}
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