Determine the coordinate direction angles of force F_1
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
We will start off by drawing a vector diagram to help us visualize the steps better like so:
Note that we only focus on force F_1 since the question only asks us to determine the coordinate direction angles of force F_1. Thus, we only drew the vector components of force F1. No need to draw unnecessary components since that will only make it more difficult for us to visualize the steps.
Let us now look at the components of force F_1.
We can write each component like so:
(F_1)_x=600(\frac{4}{5})\,\text{cos}\,30^0 N
(F_1)_y=600(\frac{4}{5})\,\text{sin}\,30^0 N
(F_1)_x=600(\frac{3}{5}) N
We can now write force F_1 in Cartesian vector form like so:
F_1=600\left\{\frac{4}{5}\,\text{cos}\,30^0(+i)\,+\,\frac{4}{5}\,\text{sin}\,30^0(-j)\,+\,\frac{3}{5}(+k)\right\}
F_1=600\left[0.6928{i}-0.4{j}+0.6{k}\right] N
(the signs for i,j, and k can be seen looking at the diagram. We can see that the j component is in the negative direction, thus, we can write it as -j)
Now, we need to find the unit vector. To do so, remember that we simply divide the Cartesian vector form by the value of F_1 which in this case is 600 N.
(u_F)_1=\frac{600(0.6928{i}-0.4{j}+0.6{k})\,\text{N}}{600\,\text{N}}
(u_F)_1=0.6928{i}-0.4{j}+0.6{k}
The last step is to find the coordinate direction angles.
\alpha=\text{cos}^{-1}((u_{F})_{1})_x
\alpha=\text{cos}^{-1}(0.6928) =46.1^0
\beta=\text{cos}^{-1}((u_{F})_{1})_y
\beta=\text{cos}^{-1}(-0.4) =114^0
\gamma=\text{cos}^{-1}((u_{F})_{1})_z
\gamma=\text{cos}^{-1}(0.6) =53.1^0
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