If each one of the ropes will break 6

 in Mechanics: Statics  tagged Engineering Mechanics: Statics / equilibrium / Forces

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.

If each one of the ropes will break

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

Solution:

Show me the final answer↓

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.

If each one of the ropes will break

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.

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-49.

Leave a comment

Your email address will not be published. Required fields are marked *

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