3D force vector summation involves combining multiple force vectors acting on an object in three-dimensional space to determine the resultant force vector. Each force vector can be represented in terms of its components along the x, y, and z axes.
Solution:
Force Matrix
The force vectors acting on the body are given as column matrices in pounds (lb):
$$ \begin{aligned} & F _1=\left[\begin{array}{l} 3 \\ 0 \\ 0 \end{array}\right] lb \\ & F _2=\left[\begin{array}{l} 0 \\ 4 \\ 0 \end{array}\right] lb \\ & F _3=\left[\begin{array}{l} 0 \\ 0 \\ 5 \end{array}\right] lb \end{aligned}
$$
The force matrix can be represented as a collection of these individual force vectors. However, the question likely intends for us to find the resultant force vector, which is the vector sum of all forces acting on the body.
Resultant Force Vector
To find the resultant force vector $R$, we add the individual force vectors:
$$ R = F _1+ F _2+ F _3=\left[\begin{array}{l} 3 \\ 0 \\ 0 \end{array}\right]+\left[\begin{array}{l} 0 \\ 4 \\ 0 \end{array}\right]+\left[\begin{array}{l} 0 \\ 0 \\ 5 \end{array}\right]=\left[\begin{array}{l} 3+0+0 \\ 0+4+0 \\ 0+0+5 \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \\ 5 \end{array}\right] lb $$
So, the resultant force vector in US units is $\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right] lb$. Now, let's convert this to SI units (Newtons). The conversion factor is $1 lb \approx 4.44822 N$.
$$ R _{S I}=\left[\begin{array}{l} 3 \times 4.44822 \\ 4 \times 4.44822 \\ 5 \times 4.44822 \end{array}\right] N=\left[\begin{array}{l} 13.34466 \\ 17.79288 \\ 22.24110 \end{array}\right] N $$
The resultant force vector in SI units is approximately $\left[\begin{array}{l}13.34 \\ 17.79 \\ 22.24\end{array}\right] N$ (rounded to two decimal places).
Magnitude of the Resultant Force
The magnitude of a vector $R =\left[\begin{array}{l}R_x \\ R_y \\ R_z\end{array}\right]$ is given by $| R |=\sqrt{R_x^2+R_y^2+R_z^2}$. In US Units:
$$ \begin{aligned} & | R |=\sqrt{(3)^2+(4)^2+(5)^2}=\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2} lb \\ & | R | \approx 5 \times 1.414=7.07 lb \end{aligned} $$
In SI Units:
$$ \left| R _{S I}\right|=\sqrt{(13.34466)^2+(17.79288)^2+(22.24110)^2} N $$
Alternatively, we can convert the magnitude from US units to SI units:
$$ \left| R _{S I}\right|=5 \sqrt{2} lb \times 4.44822 \frac{N}{lb}=7.07106781 \times 4.44822 N \approx 31.46 N $$
Let's verify this with the components in SI units:
$$ \begin{aligned} & \left| R _{S I}\right|=\sqrt{(13.34)^2+(17.79)^2+(22.24)^2}=\sqrt{177.9556+316.4841+494.6176}= \\ & \sqrt{989.0573} \approx 31.45 N \end{aligned} $$
The slight difference is due to rounding $R =\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]$lb and its magnitude $| R |=5 \sqrt{2}$ lb :
$$ \begin{aligned} & \cos \alpha=\frac{3}{5 \sqrt{2}}=\frac{3 \sqrt{2}}{10} \approx \frac{3 \times 1.414}{10}=\frac{4.242}{10}=0.4242 \\ & \cos \beta=\frac{4}{5 \sqrt{2}}=\frac{4 \sqrt{2}}{10}=\frac{2 \sqrt{2}}{5} \approx \frac{4 \times 1.414}{10}=\frac{5.656}{10}=0.5656 \\ & \cos \gamma=\frac{5}{5 \sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \approx \frac{1.414}{2}=0.7071\end{aligned} $$
The direction cosines are approximately $0.4242,0.5656,0.7071$. Note that these are dimensionless and the same for both US and SI units since they depend on the ratio of the components to the magnitude.
Let's double-check if $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$ :
$$ (0.4242)^2+(0.5656)^2+(0.7071)^2 \approx 0.1799+0.3200+0.4999 \approx 0.9998 \approx 1 $$
The slight deviation from 1 is due to rounding. Using the exact values:
$$ \left(\frac{3}{5 \sqrt{2}}\right)^2+\left(\frac{4}{5 \sqrt{2}}\right)^2+\left(\frac{5}{5 \sqrt{2}}\right)^2=\frac{9}{50}+\frac{16}{50}+\frac{25}{50}=\frac{9+16+25}{50}=\frac{50}{50}=1 $$
So, the direction cosines are indeed correct.
Force Matrix: While not explicitly asked for in a standard format, the individual force vectors are:
$$ \begin{aligned} & F _1=\left[\begin{array}{l} 3 \\ 0 \\ 0 \end{array}\right] lb=\left[\begin{array}{c} 13.34 \\ 0 \\ 0 \\ 0 \\ 0 \\ 4 \\ 0 \end{array}\right] N \\ & F _2=\left[\begin{array}{l} 0 \\ 17.79 \\ 0 \\ 0 \\ 5 \end{array}\right] lb=\left[\begin{array}{c} 0 \\ 0 \\ 22.24 \end{array}\right] N \end{aligned}
$$
Resultant Force Vector: In US units: $R =\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right] lb$ In SI units: $R _{S I} \approx\left[\begin{array}{l}13.34 \\ 17.79 \\ 22.24\end{array}\right] N$ Magnitude of the Resultant Force: In US units: $| R |=5 \sqrt{2} \approx 7.07 lb$ In SI units: $\left| R _{S I}\right|=5 \sqrt{2} \times 4.44822 \approx 31.46 N$
Direction Cosines of the Resultant:
$$ \begin{aligned} & \cos \alpha=\frac{3 \sqrt{2}}{10} \approx 0.4242 \\ & \cos \beta=\frac{4 \sqrt{2}}{10} \approx 0.5656 \\ & \cos \gamma=\frac{\sqrt{2}}{2} \approx 0.7071 \end{aligned} $$
https://gist.github.com/viadean/b66d920e936d8636873af251482d1568