Pot is supported from A by the three cables 2

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

The 50-kg pot is supported from A by the three cables. Determine the force acting in each cable for equilibrium. Take d = 2.5 m.

Pot is supported from A by the three cables

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

Solution:

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Let us first express each force in each rope in Cartesian vector form. To do so, the first step is to figure out the locations of points A, B, C, and D in cartesian vector notation.

Pot is supported from A by the three cables

From the diagram, the locations of the points are:

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

B:(6i+0j+2.5k) (Remember d=2.5m)

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

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

 

We can now write our position vectors:

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

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_{AB}\,=\,\sqrt{(6)^2+(0)^2+(2.5)^2}\,=\,6.5

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_{AB}\,=\,\left(\dfrac{6}{6.5}i+0j+\dfrac{2.5}{6.5}k\right)

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:

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

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

F_{AD}\,=\,F_{AD}\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}\,=\,0.923F_{AB}i+0j+0.385F_{AB}k

F_{AC}\,=\,-0.857F_{AC}i-0.286F_{AC}j+0.429F_{AC}k

F_{AD}\,=\,-0.857F_{AD}i+0.286F_{AD}j+0.429F_{AD}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:

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

y-components:

-0.286F_{AC}+0.286F_{AD}\,=\,0

z-components:

0.385F_{AB}+0.429F_{AC}+0.429F_{AD}-490.5\,=\,0

 

Solving the three equations simultaneously gives us:

F_{AB}\,=\,579 N

F_{AC}\,=\,312 N

F_{AD}\,=\,312 N

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-52.

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