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.

#### Solution:

Show me the final answer↓

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:

(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:

(Simplify)

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

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

\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 180^0to 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+103^0)}

F_2\,=\,88.8 lb

#### Final Answers:

Why should add 180 at theta?

To get it in the bottom quadrant since that’s where the force lies.

Thaks