Question

Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acting on the flange is a minimum.

Answer 1

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.

$\dfrac{dF_{net}}{df}=0$

We need to calculate the unknown force

Using formula of net force

$\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}$

Put the value into the formula

$\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}$

$\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}$

The magnitude of net force,

$F_{net}=\sqrt{F_{x}^2+F_{y}^2}$

$F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}$

$F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}$

$F_{net}=\sqrt{F^2+4899.78+36.232F}$

On differentiating w.r.to F

$(\dfrac{dF_{net}}{dF})^2=2F+36.232$

$0=2F+36.232$

$F=-\dfrac{36.232}{2}$

$F=-18.116\ lb$

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.