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:
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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