# Determine the magnitude of force F

Determine the magnitude of force F so that the resultant $F_R$ of the three forces is as small as possible. What is the minimum magnitude of $F_R$? Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Let’s first draw the vector components as follows: The dashed orange arrows represent the x and y components of the force, F.

We can write down these x and y components.

$F_x= F\text{sin30}^0$

$F_y= F\text{cos30}^0$

The next step is to add all the x components together, and to add all the y components together. To do this, we will first establish which sides we consider to be positive. We will choose forces acting up and forces acting to the right as positive.

$+\rightarrow\sum(F_R)_x=\sum(F_x)$

$(F_R)_x=5-F\text{ sin30}^0$

$(F_R)_x=5-0.5F$

(we simplified the equation as sin $30^0$ = 0.5)

Notice that $F\text{ sin30}^0$ is negative. This is because the x component is acting to the left and we said forces acting to the right is positive.

$+\uparrow\sum(F_R)_y=\sum(F_y)$

$(F_R)_y=F\text{ cos30}^0-4$

$(F_R)_y=0.8660F-4$

(Again, we simplified the equation as cos $30^0$ = 0.8660)

Now, we will use the Pythagorean theorem to find the magnitude of the resultant force, $F_R$.

$F_R=\sqrt{(F_R)_x^2+(F_R)_y^2}$

$F_R=\sqrt{(5-0.50F)^2+(0.8660F-4)^2}$

(Expand the brackets inside the square root and simplify)

$F_R=\sqrt{F^2-11.93F+41}$

(square both sides to get rid of the square root)

$F_R^2=F^2-11.93F+41$

Because we need to find the minimum magnitude of $F_R$ we must take the derivative.

(Take the derivative of both sides)

$2F_R\frac{\text{d}F_R}{\text{d}F}=2F-11.93$

We can now use this to find the minimum resultant force. If we equate $\frac{\text{d}F_R}{\text{d}F}$ to 0, we will find the minimums.

$2F_R\dfrac{\text{d}F_R}{\text{d}F}=2F-11.93$

(Set $\dfrac{\text{d}F_R}{\text{d}F}$=0)

$0=2F-11.93$

Solving for F gives us:

F=5.964 kN

Now, we can substitute this value back into our square root equation (look above). Our equation was the following:

$F_R=\sqrt{F^2-11.93F+41}$

(Now that we know that F=5.964 kN, we can substitute it in)

$F_R=\sqrt{5.964^2-11.93(5.964)+41}$

Solving for $F_R$ gives us: $F_R=2.330\,kN$

And so, we have our answers. If F=5.96 kN, it will produce the minimum resultant force. The minimum resultant force, $F_R=2.330\,kN$.