Three forces of magnitude F_1, F_2 and F_3 ac | vedclass

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(A) For the body to be in equilibrium,the net force must be zero,i.e.,$\Sigma F_x = 0$ and $\Sigma F_y = 0$.Resolving the forces into $x$ and $y$ comp

Vedclass · Accepted Answer

$F_2=-(2+\sqrt{3}) F_1; F_3=\frac{-4}{\sqrt{6}-\sqrt{2}} F_1$

Vedclass · Answer

(A) For the body to be in equilibrium,the net force must be zero,i.e.,$\Sigma F_x = 0$ and $\Sigma F_y = 0$.Resolving the forces into $x$ and $y$ components:$F_{1x} = F_1 \cos 30^{\circ} = F_1 \frac{\sqrt{3}}{2}$,$F_{1y} = F_1 \sin 30^{\circ} = F_1 \frac{1}{2}$$F_{2x} = -F_2 \cos 60^{\circ} = -F_2 \frac{1}{2}$,$F_{2y} = F_2 \sin 60^{\circ} = F_2 \frac{\sqrt{3}}{2}$$F_{3x} = F_3 \sin 45^{\circ} = F_3 \frac{1}{\sqrt{2}}$,$F_{3y} = -F_3 \cos 45^{\circ} = -F_3 \frac{1}{\sqrt{2}}$For $\Sigma F_x = 0$:$F_1 \frac{\sqrt{3}}{2} - F_2 \frac{1}{2} + F_3 \frac{1}{\sqrt{2}} = 0 \Rightarrow \sqrt{3} F_1 - F_2 + \sqrt{2} F_3 = 0$ ... $(i)$For $\Sigma F_y = 0$:$F_1 \frac{1}{2} + F_2 \frac{\sqrt{3}}{2} - F_3 \frac{1}{\sqrt{2}} = 0 \Rightarrow F_1 + \sqrt{3} F_2 - \sqrt{2} F_3 = 0$ ... (ii)Adding $(i)$ and (ii):$(\sqrt{3}+1) F_1 + (\sqrt{3}-1) F_2 = 0 \Rightarrow F_2 = -F_1 \frac{\sqrt{3}+1}{\sqrt{3}-1} = -F_1 \frac{(\sqrt{3}+1)^2}{3-1} = -F_1 \frac{4+2\sqrt{3}}{2} = -F_1(2+\sqrt{3})$Substituting $F_2$ in $(i)$:$\sqrt{3} F_1 - [-F_1(2+\sqrt{3})] + \sqrt{2} F_3 = 0$$\sqrt{3} F_1 + 2 F_1 + \sqrt{3} F_1 + \sqrt{2} F_3 = 0$$\sqrt{2} F_3 = -F_1(2+2\sqrt{3}) \Rightarrow F_3 = -F_1 \frac{2(1+\sqrt{3})}{\sqrt{2}} = -\sqrt{2} F_1(1+\sqrt{3}) = -F_1(\sqrt{2}+\sqrt{6})$Note: The option $C$ is $F_3 = -F_1(\sqrt{2}+\sqrt{6})$,which is equivalent to $F_3 = \frac{-4 F_1}{\sqrt{6}-\sqrt{2}}$.

Three forces of magnitude $F_1, F_2$ and $F_3$ act on a body located at the origin as shown in the figure. The condition that gives zero net force is

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