# The man pulls on the rope at C

The man pulls on the rope at C with a force of 70 lb which causes the forces FA and FC at B to have this same magnitude. Express each of these two forces as Cartesian vectors.

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 figure out the locations of points A, B and C with respect to the origin.

Using the diagram, the locations of the points are:

$A: (5i-7j+5k)$ ft

$B: (-1i-5j+8k)$ ft

$C: (5i+7j+4k)$ ft

We can now write position vectors from points B to A and B to C.

$r_{BA}\,=\,\left\{(5-(-1))i+(-7-(-5))j+(5-8)k\right\}=\left\{6i-2j-3k\right\}$ ft

$r_{BC}\,=\,\left\{(5-(-1))i+(7-(-5))j+(4-8)k\right\}=\left\{6i+12j-4k\right\}$ ft

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 figure out the magnitude of each position vector.

magnitude of $r_{BA}\,=\,\sqrt{(6)^2+(-2)^2+(-3)^2}=7$ ft

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

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.

Next, we will write the unit vectors for each of the position vectors.

$u_{BA}\,=\,\left(\dfrac{6}{7}i-\dfrac{2}{7}j-\dfrac{3}{7}k\right)$

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

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 by multiplying the unit vector by the magnitude of the force. Remember, the question states that each rope has a force of 70 lb.

$F_{BA}=70\left(\dfrac{6}{7}i-\dfrac{2}{7}j-\dfrac{3}{7}k\right)$

$F_{BA}=\left\{60i-20j-30k\right\}$ lb

$F_{BC}=70\left(\dfrac{6}{14}i+\dfrac{12}{14}j-\dfrac{4}{14}k\right)$

$F_{BC}=\left\{30i+60j-20k\right\}$ lb

#### Final Answers:

$F_{BA}=\left\{60i-20j-30k\right\}$ lb

$F_{BC}=\left\{30i+60j-20k\right\}$ lb