# If each one of the ropes will break 4

If each one of the ropes will break when it is subjected to a tensile force of 450 N, determine the maximum uplift force F the balloon can have before one of the ropes breaks. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

To solve this problem, we must first express the force in each rope in Cartesian vector form. To do so, we have to first write the locations of points A, B, C and D in Cartesian vector form. Using the diagram, we can locate the points are write them in Cartesian vector form:

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

$B:(-1.5i-2j+0k)$ m

$C:(2i-3j+0k)$ m

$D:(0i+2.5j+0k)$ m

The position vectors for each rope are:

$r_{AB}\,=\,\left\{(-1.5-0)i+(-2-0)j+(0-6)k\right\}\,=\,\left\{-1.5i-2j-6j\right\}$

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

$r_{AD}\,=\,\left\{(0-0)i+(2.5-0)j+(0-6)k\right\}\,=\,\left\{0i+2.5j-6j\right\}$

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 $r_{AB}\,=\,\sqrt{(-1.5)^2+(-2)^2+(-6)^2}\,=\,6.5$

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

magnitude of $r_{AD}\,=\,\sqrt{(0)^2+(2.5)^2+(-6)^2}\,=\,6.5$

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 are:

$u_{AB}\,=\,\left(-\dfrac{1.5}{6.5}i-\dfrac{2}{6.5}j-\dfrac{6}{6.5}k\right)$

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

$u_{AD}\,=\,\left(0i+\dfrac{2.5}{6.5}j-\dfrac{6}{6.5}k\right)$

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 each force in Cartesian vector form:

$F_{AB}\,=\,F_{AB}\left(-\dfrac{1.5}{6.5}i-\dfrac{2}{6.5}j-\dfrac{6}{6.5}k\right)$

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

$F_{AD}\,=\,F_{AD}\left(0i+\dfrac{2.5}{6.5}j-\dfrac{6}{6.5}k\right)$

(further simplify this by expanding the brackets using FOIL and simplifying the fractions into decimal values)

$F_{AB}\,=\,\left\{-0.231F_{AB}i-0.308F_{AB}j-0.923F_{AB}k\right\}$

$F_{AC}\,=\,\left\{0.286F_{AC}i-0.429F_{AC}j-0.857F_{AC}k\right\}$

$F_{AD}\,=\,\left\{0i+0.385F_{AD}j-0.923F_{AD}k\right\}$

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

(Force F is the net uplift force, which is applied directly upwards, thus it only has a z-component)

We can now write our equations of equilibrium. All forces added together must equal zero.

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

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

Since all forces added together must equal zero, then all individual components (x, y, z-components) added together must also equal zero.

x-components:

$-0.231F_{AB}+0.286F_{AC}\,=\,0$

y-components:

$-0.308F_{AB}-0.429F_{AC}+0.385F_{AD}\,=\,0$

z-components:

$-0.923F_{AB}-0.857F_{AC}-0.923F_{AD}+F\,=\,0$

The question states that the maximum tension a cable can withstand is 450 N, therefore, we will assume that rope AD, i.e $F_{AD}$ has a force value of 450 N. Substituting 450 N for $F_{AD}$ gives us the following equations:

x-components:

$-0.231F_{AB}+0.286F_{AC}\,=\,0$

y-components:

$-0.308F_{AB}-0.429F_{AC}+173.25\,=\,0$

z-components:

$-0.923F_{AB}-0.857F_{AC}-415.35+F\,=\,0$

Solving the three equations gives us:

$F_{AB}\,=\,264.7$ N

$F_{AC}\,=\,213.8$ N

$F\,=\,842.9$ N

Notice how both tensions in the cables are less than 450 N. That means the assumption we made is correct. If, however, we got tension values higher than 450 N for the other two cables, then we would have to change our assumption and assume another cable is 450 N.

## 4 thoughts on “If each one of the ropes will break”

• farhanah

why the coordinate for point A is 4 at z xis?

• questionsolutions Post author

Thanks for pointing out the mistake, we fixed it! It seems it was a typo since the correct value was used for the rest of the calculations.

• charles

why the coordinate for point C is not negative at y xis?

• questionsolutions Post author

This was a typo and it has been fixed. Thank you for letting us know 🙂