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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Aicharya · Accepted Answer

Minimum additional force needed $\begin{aligned} & F=-\left(F_{	ext {resultant }}ight)_x \ & F_{	ext {resultant }}=[(4-2)(\cos 30 \hat{\mathbf{j}}-\sin 30 \hat{\mathbf{i}}) \ = & {\left[2\left(\frac{\sqrt{3}}{2} \hat{\mathbf{j}}-\frac{1}{2} \hat{\mathbf{i}}ight)+\left(\frac{1}{2} \hat{\mathbf{i}}+\frac{\sqrt{3}}{2} \hat{\mathbf{j}}ight)ight] } \ = & {\left[\left(\sqrt{3}+\frac{\sqrt{3}}{2}ight) \hat{\mathbf{j}}+\sin 60 \hat{\mathbf{j}}ight) } \ = & {\left.\left[-\frac{1}{2} \hat{\mathbf{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathbf{i}}+\frac{\hat{\mathbf{i}}}{2}ight)ight] } \ = & -\frac{\hat{\mathbf{i}}}{2}+\frac{3 \sqrt{3}}{2} \hat{\mathbf{j}} \quad F=-\left(-\frac{\hat{\mathbf{i}}}{2}ight)=\frac{1}{2} \hat{\mathbf{i}}\end{aligned}$ Hence, $|F|=0.5 \mathrm{~N}$

Three forces acting on a body are shown in the figure. To have the resultant force only along the…

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