question_answer

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:

A) \[0.5\,N\]

B)  \[1.5\,N\]

C)  \[\frac{\sqrt{3}}{4}N\]

D) \[\sqrt{3}\,N\]

Correct Answer: A

Solution :

 The component of 1 N and 2 N forces along \[+x\] axis \[=(1cos{{60}^{o}}=2sin{{30}^{o}})\] \[=1\times \frac{1}{2}+2\times \frac{1}{2}=\frac{1}{2}+1=\frac{3}{2}=1.5\,N\] The component of 4 N forces along \[-x\]axis, \[\text{=4}\,\text{sin 3}{{\text{0}}^{\text{o}}}\text{=4 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{2}}\text{=2N}\] Hence net \[\text{x-}\]component of resultant \[{{\text{f}}_{x}}=-2+1.5=-0.5\,N\]along \[-x\]direction Therefore, if a force of 0.5 N is applied along \[+x-\] axis, the resultant force along \[x-\]axis will become zero and the resultant force will be obtained only along y- axis.