If FB = 560 N and FC = 700 N determine 3


If F_B = 560 N and F_C = 700 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

If FB = 560 N and FC = 700 N

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 write the locations of points A, B and C are in Cartesian vector form.

If FB = 560 N and FC = 700 N

Looking at the diagram, we can write the following:

A:(0i+0j+6k)

B:(2i-3j+0k)

C:(3i+2j+0k)

 

We can now find the position vectors from A to B and A to C.

r_{AB}\,=\,\left\{(2-0)i+(-3-0)j+(0-6)k\right\}

r_{AB}\,=\,\left\{2i-3j-6k\right\} m

 

r_{AC}\,=\,\left\{(3-0)i+(2-0)j+(0-6)k\right\}

r_{AC}\,=\,\left\{3i+2j-6k\right\} m

 

We now need to find the magnitude of each position vector so that we can use it to find the unit vector in the next step.

Magnitude of r_{AB}\,=\,\sqrt{(2^2)+(-3)^2+(-6)^2}

Magnitude of r_{AB}\,=\,7 m

 

Magnitude of r_{AC}\,=\,\sqrt{(3^2)+(2)^2+(-6)^2}

Magnitude of r_{AC}\,=\,7 m

 

Let us now find the unit vector for each force. Remember, this is found by dividing each component of the position vector by it’s magnitude.

u_{AB}\,=\,\left(\dfrac{2}{7}i\,-\,\dfrac{3}{7}j\,-\,\dfrac{6}{7}k\right)

u_{AC}\,=\,\left(\dfrac{3}{7}i\,+\,\dfrac{2}{7}j\,-\,\dfrac{6}{7}k\right)

 

We can now write force F_B and F_C in Cartesian form.

F_B\,=\,560\left(\dfrac{2}{7}i\,-\,\dfrac{3}{7}j\,-\,\dfrac{6}{7}k\right)

F_B\,=\,\left\{160i\,-\,240j\,-\,480k\right\} N

 

F_C\,=\,700\left(\dfrac{3}{7}i\,+\,\dfrac{2}{7}j\,-\,\dfrac{6}{7}k\right)

F_C\,=\,\left\{300i\,+\,200j\,-\,600k\right\} N

(Here we multiplied each individual component by the force. In other words, expand the brackets using FOIL)

 

We can find the resultant force by adding both forces together like so:

F_R\,=\,F_B\,+\,F_C

F_R\,=\,\left\{160i\,-\,240j\,-\,480k\right\}\,+\,\left\{300i\,+\,200j\,-\,600k\right\}

F_R\,=\,\left\{460i-40j-1080k\right\} N

 

To figure out the coordinate direction angles, we need to find the magnitude of the resultant force we just found.

magnitude of F_R\,=\,\sqrt{(460^2)+(-40)^2+(-1080)^2}

magnitude of F_R\,=\,1174.56 N

 

Coordinate direction angles can be found taking the cosine inverse of each component of the resultant divided by the magnitude:

\alpha\,=\,\cos^{-1}\left(\dfrac{460}{1174.56}\right)\,=\,67^0

\beta\,=\,\cos^{-1}\left(\dfrac{-40}{1174.56}\right)\,=\,92^0

\gamma\,=\,\cos^{-1}\left(\dfrac{-1080}{1174.56}\right)\,=\,157^0

 

Final Answers:

magnitude of F_R\,=\,1174.56 N

\alpha\,=\,67^0

\beta\,=\,92^0

\gamma\,=\,157^0

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-92.

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3 thoughts on “If FB = 560 N and FC = 700 N determine

    • questionsolutions Post author

      I assume you mean lengths of each of the ropes? All you need to do is write a position vector from A to B, and then find the magnitude of it. That will give you the length. The same goes for rope AC, so you write a position vector from A to C, and then find the magnitude.