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