A) 0.5 N
B) 1.5 N
C) \[\frac{\sqrt{3}}{4}N\]
D) \[\sqrt{3}\,N\]
Correct Answer: A
Solution :
Minimum additional force needed \[F=-{{({{F}_{resul\tan t}})}_{x}}\] \[{{F}_{resul\tan t}}=[(4-2)(\cos 30\hat{j}-\sin 30\hat{i})\] \[+1(\cos 60\hat{i}+\sin 60\hat{j})\] \[=\left[ 2\left( \frac{\sqrt{3}}{2}\hat{j}-\frac{1}{3}\hat{i} \right)+\left( \frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j} \right) \right]\] \[=\left[ \left( \sqrt{3}+\frac{\sqrt{3}}{2} \right)\hat{j}+\left( -\hat{i}+\frac{{\hat{i}}}{2} \right) \right]\] \[=\left[ -\frac{1}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j} \right]\] \[=-\frac{1}{2}+\frac{3\sqrt{3}}{2}\hat{j}\] \[\therefore \] \[F=-\left( -\frac{{\hat{i}}}{2} \right)=\frac{1}{2}\hat{i}\] Hence,\[|F|=0.5N\]
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