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.

Three forces act on the bracket

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.

Three forces act on the bracket

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

Three forces act on the bracket

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 180^0to 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+\theta)}

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

F_2\,=\,88.8 lb

 

Final Answers:

F_2\,=\,88.8 lb

\theta\,=\,103^0

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-56.

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