# Cable AB exerts a force of 80 N 4

Cable AB exerts a force of 80 N on the end of the 3-m-long boom OA. Determine the magnitude of the projection of this force along the boom.

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

#### Solution:

We will write two position vectors, from A to O, and A to B. We will then find the unit vector of the position vector extending from A to B. Once we find the unit vector, we can express the 80 N in Cartesian vector form. We can then calculate the projection using the dot product. To do so, we will first find the locations of points A, O and B in Cartesian vector form.

Using the diagram, the locations of the points are:

$A:(3\cos60^0i+3\sin60^0j+0k)\,=\,(1.5i+2.6j+0k)$ m

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

$O:(0i+0j+0k)$ m

Let us now write our position vectors.

$r_{AO}\,=\,\left\{(0-1.5)i+(0-2.6)j+(0-0)k\right\}$

$r_{AO}\,=\,\left\{-1.5i-2.6j+0k\right\}$ m

$r_{AB}\,=\,\left\{(0-1.5)i+(0-2.6)j+(4-0)k\right\}$

$r_{AB}\,=\,\left\{-1.5i-2.6j+4k\right\}$ 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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

We will now find the magnitude of the position vectors.

magnitude of $r_{AB}\,=\,\sqrt{(-1.5)^2+(-2.6)^2+(4)^2}\,=\,5$

magnitude of $r_{AO}\,=\,\sqrt{(-1.5)^2+(-2.6)^2+(0)^2}\,=\,3$

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 will now find the unit vectors of each position vector.

$u_{AB}\,=\,\left(-\dfrac{1.5}{5}i-\dfrac{2.6}{5}j+\dfrac{4}{5}k\right)$

$u_{AO}\,=\,\left(-\dfrac{1.5}{3}i-\dfrac{2.6}{3}j+0k\right)$

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

We can now express the 80 N force that is along the position vector $r_{AB}$ in Cartesian vector form .

$F\,=\,80\left(-\dfrac{1.5}{5}i-\dfrac{2.6}{5}j+\dfrac{4}{5}k\right)$

$F\,=\,\left\{-24i-42j+64k\right\}$ N

We can now find the projection of this force along the boom AO. We will use the dot product to do so.

$\text{Proj}\,F\,=\,F\cdot U_{AO}$

$\text{Proj}\,F\,=\,\left[(-24)(-\dfrac{1.5}{3})+(-42)(-\dfrac{2.6}{3})+(-64)(0)\right]$

$\text{Proj}\,F\,=\,48.4$ N

The dot product is the product of the magnitudes of two vectors and the cosine angle $\theta$ between them. For example, if we had two vectors, $\mathbf{A}$ and $\mathbf{B}$, the dot product of the two is: $A\cdot B\,=\,AB\cos \theta$. In cartesian vector form, $A\cdot B\,=\,A_xB_x+A_yB_y+A_zB_z$. The angle formed between two vectors can be found by: $\theta\,=\,\cos^{-1}\left(\dfrac{A_xB_x+A_yB_y+A_zB_z}{(\text{magnitude of}\,A)(\text{magnitude of}\,B)}\right)$

The magnitude of the projection of the 80 N force along the boom AO is 48.4 N.

## 4 thoughts on “Cable AB exerts a force of 80 N”

• Wahteverm8

Doesnt change the reslut but r_AB should be:
r_AB = {(0−1.5)i+(0−2.6)j+(4-0)k}
r_AB = {-1.5i – 2.6j + 4k}

• Bumpkineer

Is it possible to get the answer with just 80cos(cos^1(3/5)) since the force vector creates a 3 4 5 right triangle? I tried calculating it this way but got a less precise answer of 48 instead of your 48.4. Why is this?