Question

Two equal forces are acting at a point with an angle of $60{}^{\circ }$ between them. If the resultant force is equal to $40\sqrt{3}N,$ the magnitude of each force is

• A. 40 N
• B. 20 N
• C. 80 N
• D. 30 N

Answer 1

Answer: (A)

Let equal forces $F_1=F_2=FN.$ Angle between the forces $(\theta )=60{}^{\circ }$ Resultant force $(R)=40\sqrt{3}N.$ Resultant force $R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$ $\therefore$ $40\sqrt{3}=\sqrt{F^2+F^2+2F.F\cos 60{}^{\circ }}$ or $40\sqrt{3}=\sqrt{2F^2+2F^2\times \frac{1}{2}}$ or $40\sqrt{3}=\sqrt{2F^2+F^2}$ or $40\sqrt{3}=\sqrt{3F^2}$ or $40\sqrt{3}=F\sqrt{3}$ or $F=40N$