Answer:
The unknown force will be 18.116 lb.
Explanation:
Given that,
Three forces act on a flange as shown in figure.
The net force acting on the flange is a minimum.
[tex]\dfrac{dF_{net}}{df}=0[/tex]
We need to calculate the unknown force
Using formula of net force
[tex]\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}[/tex]
Put the value into the formula
[tex]\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}[/tex]
[tex]\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}[/tex]
The magnitude of net force,
[tex]F_{net}=\sqrt{F_{x}^2+F_{y}^2}[/tex]
[tex]F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}[/tex]
[tex]F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}[/tex]
[tex]F_{net}=\sqrt{F^2+4899.78+36.232F}[/tex]
On differentiating w.r.to F
[tex](\dfrac{dF_{net}}{dF})^2=2F+36.232[/tex]
[tex]0=2F+36.232[/tex]
[tex]F=-\dfrac{36.232}{2}[/tex]
[tex]F=-18.116\ lb[/tex]
Negative sign shows the direction of force which is downward.
Hence, The unknown force will be 18.116 lb.
Following are the solution to the given question:
Calculating the Net force on flange:
[tex]\overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 70 \cos 30^{\circ} -40 ) \hat{i}+(70 \sin 30^{\circ} - F \sin 45^{\circ} )\hat{y} \\\\ \overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 20.62 ) \hat{i}+(35 -F\sin 45^{\circ})\hat{y} \\\\ [/tex]
Calculating the magnitude:
[tex]= \sqrt{f_{X}^2 +f_{y}^{2}}\\\\ = \sqt{( F \cos 45^{\circ} + 20.62 )^2+(35 -F\sin 45^{\circ})^2} \\\\ = \sqt{( F^2+20.62^2+ 35^2+ 41.24 F \cos 45^{\circ} -70 F\sin 45^{\circ})}\\\\ = \sqt{( F^2+20.33F +1650.1844)} \\\\[/tex]
Differentiate the value
[tex]\to \frac{F_{net}}{df}=0 [/tex]
[tex]\to 2F-20.33=0\\\\ \to 2F=20.33\\\\ \to F=\frac{20.33}{2}\\\\ \to F= 10.165\ lb [/tex]
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