# Determine the height d of cable AB 2

Determine the height d of cable AB so that the force in cables AD and AC is one-half as great as the force in cable AB.What is the force in each cable for this case? The flower pot has a mass of 50 kg.

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

#### Solution:

We will figure out the position vectors for points from A to C, and A to D. We will not find the position vector for the rope AB since we don’t know the value of d.

The locations of the points are:

$A:(0i+0j+0k)$

$C:(-6i-2j+3k)$

$D:(-6i+2j+3k)$

We can now write our position vectors:

$r_{AC}\,=\,\left\{(-6-0)i+(-2-0)j+(3-0)k\right\}\,=\,\left\{-6i-2j+3k\right\}$

$r_{AD}\,=\,\left\{(-6-0)i+(2-0)j+3-0)k\right\}\,=\,\left\{-6i+2j+3k\right\}$

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

The magnitude of each position vector is:

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

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

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.

The unit vectors for each position vector is:

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

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

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

We can now express the force in each cable in Cartesian vector form. Remember, the question says “force in cables AD and AC is one-half as great as the force in cable AB,” thus, we will multiply each force by half of $F_{AB}$.:

$F_{AC}\,=\,\dfrac{F_{AB}}{2}\left(-\dfrac{6}{7}i-\dfrac{2}{7}j+\dfrac{3}{7}k\right)$

$F_{AD}\,=\,\dfrac{F_{AB}}{2}\left(-\dfrac{6}{7}i+\dfrac{2}{7}j+\dfrac{3}{7}k\right)$

(simplify by expanding brackets and writing fractions are decimals)

$F_{AB}\,=\,(F_{AB})_xi+0j+(F_{AB})_zk$

$F_{AC}\,=\,-0.429F_{AB}i-0.143F_{AB}j+0.214F_{AB}k$

$F_{AD}\,=\,-0.429F_{AB}i+0.143F_{AB}j+0.214F_{AB}k$

$F\,=\,\left\{0i+0j-490.5\right\}$

(Force F is the weight of the pot, which has a mass of 50 kg, $\therefore$ the weight is (50)(9.81) = 490.5 N)

Since the system is in equilibrium, all forces added together must equal zero.

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

$F_{AB}+F_{AC}+F_{AD}+F\,=\,0$

Furthermore, since each force added together must equal zero, then each component (x, y, z-component) added together must also equal zero.

x-components:

$(F_{AB})_x-0.429F_{AB}-0.429F_{AB}\,=\,0$

y-components:

$-0.143F_{AB}+0.143F_{AB}\,=\,0$

z-components:

$(F_{AB})_z+0.214F_{AB}+0.214F_{AB}-490.5\,=\,0$

Simplifying these equations leads us to the following:

$(F_{AB})_x\,=\,0.858F_{AB}$ (eq.1)

$(F_{AB})_z\,=\,490.5-0.428F_{AB}$ (eq.2)

We will now turn to the Pythagorean theorem to find the value of $F_{AB}$. The $(F_{AB})_x$ and $(F_{AB})_z$ values we found are the two sides of a right angle triangle. The value of $F_{AB}$ is the hypotenuse.

Thus, we can write the Pythagorean theorem for this triangle.

$F^2_{AB}\,=\,(F_{AB})^2_x+(F_{AB})^2_z$

Substitute the values of eq.1 and eq.2 we found.

$F^2_{AB}\,=\,(0.858F_{AB})^2+(490.5-0.428F_{AB})^2$

(simplify)

$0.080652F^2_{AB}+419.868F_{AB}-240590\,=\,0$

(Solve for the positive root)

$F_{AB}\,=\,520$ N

Since $F_{AD}$ and $F_{AC}$ are half of $F_{AB}$, then:

$F_{AD}\,=\,F_{AC}\,=\,\dfrac{520}{2}\,=\,260$ N

We can now figure out $(F_{AB})_x$ and $(F_{AB})_z$ by substituting the value of $F_{AB}$ we found into eq.1 and eq.2.

$(F_{AB})_x\,=\,(0.858)(520)\,=\,446.2$ N

$(F_{AB})_z\,=\,490.5-0.428(520)\,=\,268$ N

Referring to the triangle we formed before, we now have the values of all sides. Thus, we can use trigonometry to figure out the angle $\theta$.

$\theta\,=\,\tan^{-1}\left(\dfrac{268}{446.2}\right)$

$\theta\,=\,31^0$

Again, referring to the triangle we formed, remember that the base is 6 m in length. Thus, using trigonometry, we can figure out the opposite height. Using tangent, which is opposite over adjacent, we can write the following:

$\tan 31^0\,=\,\dfrac{d}{6}$

$d\,=\,6\tan 31^0$

$d\,=\,3.6$ m

$F_{AB}\,=\,520$ N

$F_{AC}\,=\,260$ N

$F_{AD}\,=\,260$ N

$d\,=\,3.6$ m

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