A) 40 N
B) 20 N
C) 80 N
D) 30 N
E) 10 N
Correct Answer: A
Solution :
Let equal forces\[{{F}_{1}}={{F}_{2}}=F\,N.\] Angle between the forces \[(\theta )=60{}^\circ \] Resultant force\[(R)=40\sqrt{3}N.\] Resultant force \[R=\sqrt{F_{1}^{2}+F_{2}^{2}+2{{F}_{1}}\,{{F}_{2}}\cos \theta }\] \[\therefore \] \[40\sqrt{3}=\sqrt{{{F}^{2}}+{{F}^{2}}+2F.\,F\cos 60{}^\circ }\] or \[40\sqrt{3}=\sqrt{2{{F}^{2}}+2{{F}^{2}}\times \frac{1}{2}}\] or \[40\sqrt{3}=\sqrt{2{{F}^{2}}+{{F}^{2}}}\] or \[40\sqrt{3}=\sqrt{3{{F}^{2}}}\] or \[40\sqrt{3}=F\sqrt{3}\] or \[F=40\,N\]
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