Question 92 of 210

Four forces are acting at a point $P$ in equilibrium as shown in figure. The ratio of force ${F}_{1}$ to ${F}_{2}$ is $1:x$ where $x=$_____ .

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Answer: 3

Here, $\theta =45^{\circ}$
Applying equilibrium conditions,
Taking components along $x$ & $y$
Considering right side positive, $\sum {F}_{x}=0$
${F}_{1}+1\sin 45^{\circ}-2\sin 45^{\circ}=0 {F}_{1}=2\sin 45^{\circ}-1\sin 45^{\circ}$
${F}_{1}=\sqrt{2}-\dfrac{1}{\sqrt{2}}=\dfrac{2-1}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}$
Taking upward positive,$\sum {F}_{y}=0$
$1\cos 45^{\circ}+2\cos 45^{\circ}-{F}_{2}=0 {F}_{2}=1\cos 45^{\circ}+2\cos 45^{\circ}$
${F}_{2}=\sqrt{2}+\dfrac{1}{\sqrt{2}}=\dfrac{2+1}{\sqrt{2}}=\dfrac{3}{\sqrt{2}}$
${F}_{1}:{F}_{2}=1:3$
$x=3$

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