# Three forces act on the bracket 3

Three forces act on the bracket. Determine the magnitude and direction ϴ of $F_2$ so that the resultant force is directed along the positive u axis and has a magnitude of 50 lb.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first draw a component diagram showing the x and y components of each force.

We will also draw a component diagram showing the components of $F_R$.

Using our first component diagram, we can add the x-components and y-components together as follows:

$\rightarrow ^+ F_{R_x} \,=\,\sum\text{F}_\text{x}$

(Remember, $F_R$ has a magnitude of 50 lb, refer to the second diagram)

$80+F_2\cos(25^0+\theta)+52\left(\dfrac{5}{13}\right)\,=\,50\cos 25^0$

(simplify)

$F_2\,=\,\dfrac{-54.684}{\cos(25^0+\theta)}$ (eq.1)

Now the y-components:

$+\uparrow F_{R_y}\,=\,\sum\text{F}_\text{y}$

$52\left(\dfrac{12}{13}\right)-F_2\sin(25^0+\theta)=-50\sin 25^0$

(Simplify)

$F_2\sin(25^0+\theta)=69.131$ (eq.2)

We can solve both equations to figure out $F_2$ and $\theta$.

Substitute eq.1 into eq.2.

$\dfrac{-54.684}{\cos(25^0+\theta)}\sin(25^0+\theta)=69.131$

(Remember, $\dfrac{\sin\theta}{\cos\theta}\,=\,\tan\theta$)

$-54.684\tan(25^0+\theta)\,=\,69.131$

$\tan(25^0+\theta)\,=\,\dfrac{69.131}{54.684}$

$25^0+\theta\,=\,tan^{-1}\left(\dfrac{69.131}{54.684}\right)$

$\theta\,=\,-76.65^0$

We must add $180^0$to this answer since $F_2$ lies on the bottom quadrant.

$\theta\,=\,-76.65^0+180^0\,=\,103^0$

Substitute the value of $\theta$ we found into eq.1 to solve for $F_2$.

$F_2\,=\,\dfrac{-54.684}{\cos(25^0+\theta)}$

$F_2\,=\,\dfrac{-54.684}{\cos(25^0+103^0)}$

$F_2\,=\,88.8$ lb

$F_2\,=\,88.8$ lb

$\theta\,=\,103^0$