# Each of the four forces acting at E

Each of the four forces acting at E has a magnitude of 28 kN. Express each force as a Cartesian vector and determine the resultant force.

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, C, D, and E in Cartesian vector form.

Using the image, the locations of the points are:

$A:(6i-4j+0k)$ m

$B:(6i+4j+0k)$ m

$C:(-6i+4j+0k)$ m

$D:(-6i-4j+0k)$ m

$E:(0i+0j+12k)$ m

We will now write position vectors for points from E to A, E to B, E to C, and E to D.

$r_{EA}\,=\,\left\{(6-0)i+(-4-0)j+(0-12)k\right\}$ m

$r_{EA}\,=\,\left\{6i-4j-12k\right\}$ m

$r_{EB}\,=\,\left\{(6-0)i+(4-0)j+(0-12)k\right\}$ m

$r_{EB}\,=\,\left\{6i+4j-12k\right\}$ m

$r_{EC}\,=\,\left\{(-6-0)i+(4-0)j+(0-12)k\right\}$ m

$r_{EC}\,=\,\left\{-6i+4j-12k\right\}$ m

$r_{ED}\,=\,\left\{(-6-0)i+(-4-0)j+(0-12)k\right\}$ m

$r_{ED}\,=\,\left\{-6i-4j-12k\right\}$ m

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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Next we will find the magnitude of each position vector.

magnitude of $r_{EA}\,=\,\sqrt{(6)^2+(-4)^2+(-12)^2}\,=\,14$ m

magnitude of $r_{EB}\,=\,\sqrt{(6)^2+(4)^2+(-12)^2}\,=\,14$ m

magnitude of $r_{EC}\,=\,\sqrt{(-6)^2+(4)^2+(-12)^2}\,=\,14$ m

magnitude of $r_{ED}\,=\,\sqrt{(-6)^2+(-4)^2+(-12)^2}\,=\,14$ m

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.

We can now write the unit vector for each position vector.

$u_{EA}\,=\,\left(\dfrac{6}{14}i\,-\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

$u_{EB}\,=\,\left(\dfrac{6}{14}i\,+\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

$u_{EC}\,=\,\left(-\dfrac{6}{14}i\,+\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

$u_{ED}\,=\,\left(-\dfrac{6}{14}i\,-\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

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, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

Let us now express each force in Cartesian vector form. Remember that each force has a magnitude of 28 kN.

$F_{EA}\,=\,28\left(\dfrac{6}{14}i\,-\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

$F_{EA}\,=\,\left\{12i-8j-24k\right\}$ kN

$F_{EB}\,=\,28\left(\dfrac{6}{14}i\,+\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

$F_{EB}\,=\,\left\{12i+8j-24k\right\}$ kN

$F_{EC}\,=\,28\left(-\dfrac{6}{14}i\,+\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

$F_{EC}\,=\,\left\{-12i+8j-24k\right\}$ kN

$F_{ED}\,=\,28\left(-\dfrac{6}{14}i\,-\,\dfrac{4}{14}j\,-\,\dfrac{12}{14}k\right)$

$F_{ED}\,=\,\left\{-12i-8j-24k\right\}$ kN

The resultant force is equal to each corresponding coordinate of the forces added together.

$F_R\,=\,F_{EA}\,+\,F_{EB}\,+\,F_{EC}\,+\,F_{ED}$

$F_R\,=\,\left\{(12+12-12-12)i+(-8+8+8-8)j+(-24-24-24-24)k\right\}$ kN

$F_R\,=\,\left\{0i+0j-96k\right\}$ kN

$F_{EA}\,=\,\left\{12i-8j-24k\right\}$ kN

$F_{EB}\,=\,\left\{12i+8j-24k\right\}$ kN

$F_{EC}\,=\,\left\{-12i+8j-24k\right\}$ kN

$F_{ED}\,=\,\left\{-12i-8j-24k\right\}$ kN

$F_R\,=\,\left\{0i+0j-96k\right\}$ kN