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.

#### Solution:

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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:

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_{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\}

The magnitude of each position vector is:

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 unit vectors are:

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)

We can now express each force in Cartesian vector form:

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.

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.

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

y-components:

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

z-components:

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:

-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_{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.

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

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.

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

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

Thank you so much for the explanation. Really clear.

Really glad to hear that 🙂