# Determine the magnitude and direction

Determine the magnitude and direction of the force P required to keep the concurrent force system in equilibrium. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first express each force in Cartesian vector form, including force P. Using the diagram, we can write the forces in Cartesian vector form. We will start force force $F_1$

$F_1\,=\,(2\cos45^0i+2\cos60^0j+2\cos120^0k)\,=\,\left\{1.41i+1j-1k\right\}$ N

$F_1\,=\,\left\{1.41i+1j-1k\right\}$ kN

To find force $F_2$, we will first find the position vector, then find the unit vector of that position vector, and finally multiply the magnitude of the force by the unit vector to write the force in Cartesian vector notation.

$r_2\,=\,\left\{-1.5i+3j+3k\right\}$

A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

magnitude of $r_{2}\,=\,\sqrt{(-1.5)^2+(3)^2+(3)^2}\,=\,4.5$

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

$u_2\,=\,\left(-\dfrac{1.5}{4.5}i+\dfrac{3}{4.5}j+\dfrac{3}{4.5}k\right)$

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was $r\,=\,ai+bj+ck$, then unit vector, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

$F_2\,=\,0.75\left(-\dfrac{1.5}{4.5}i+\dfrac{3}{4.5}j+\dfrac{3}{4.5}k\right)$

$F_2\,=\,\left\{-0.25i+0.5j+0.5k\right\}$ kN

From the diagram, we see that force $F_3$ lies on the negative y-axis. Thus, we can write:

$F_3\,=\,\left\{-0.5j\right\}$ kN

We will also write force P in Cartesian notation.

$P\,=\,P_xi+P_yj+P_zk$

We can now write our equation of equilibrium. All forces added together must equal zero because the system is in equilibrium.

$\sum \text{F}\,=\,0$

$F_1+F_2+F_3+P\,=\,0$

Since the system is in equilibrium, the x, y, z (i, j, k) components must also equal to zero when added together.

$1.41-0.25+p_x\,=\,0$

$p_x\,=\,-1.16$ kN

$1+0.5-0.5+p_y\,=\,0$

$p_y\,=\,-1$ kN

$-1+0.5+p_z\,=\,0$

$p_z\,=\,0.5$ kN

Now that we found each component of force P, we can write it as:

$p\,=\,\left\{-1.16i-1j+0.5k\right\}$ kN

The magnitude of force P is:

magnitude of $p\,=\,\sqrt{(-1.16)^2+(-1)^2+(0.5)^2}\,=\,1.61$ kN

The coordinate direction angles can be found by taking the cosine inverse of each component of force P divided by it’s magnitude.

$\alpha\,=\,\cos^{-1}\left(\dfrac{-1.16}{1.61}\right)\,=\,136^0$

$\beta\,=\,\cos^{-1}\left(\dfrac{-1}{1.61}\right)\,=\,128^0$

$\gamma\,=\,\cos^{-1}\left(\dfrac{0.5}{1.61}\right)\,=\,72^0$

magnitude of $p\,=\,1.61$ kN
$\alpha\,=\,136^0$
$\beta\,=\,128^0$
$\gamma\,=\,72^0$