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.

Each of the four forces acting at E

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.


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We will first write the locations of points A, B, C, D, and E in Cartesian vector form.

Each of the four forces acting at E

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.





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}\,=\,\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


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


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


Final Answers:

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


This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-102.

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