# If the balloon is subjected to a net uplift 5

If the balloon is subjected to a net uplift force of F = 800 N, determine the tension developed in ropes AB, AC, AD.

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

#### Solution:

We will first write each tension developed in Cartesian vector form. To do so, we need to write the locations of points A, B, C, and D in Cartesian vector form.

From the diagram, the locations of the points are:

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

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

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

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

$F\,=\,\left\{0i+0j+800k\right\}$

(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:

$-0.923F_{AB}-0.857F_{AC}-0.923F_{AD}+800\,=\,0$

Solving the three equations simultaneously gives us:

$F_{AB}\,=\,251.2$ N

$F_{AC}\,=\,202.9$ N

$F_{AD}\,=\,427.1$ N

## 5 thoughts on “If the balloon is subjected to a net uplift”

• questionsolutions Post author

Where is it wrong? Just saying solution is wrong doesn’t help anyone, please point out what value is wrong, it’s easy to miss when checking so it would be very helpful if you can point out the error. I checked through and none of the signs seem to be incorrect.

• Joris De Dier

I think that the Y component of C should be -3 instead of +3