If the mass of the flowerpot

If the mass of the flowerpot is 50 kg, determine the tension developed in each wire for equilibrium. Set x = 1.5 m and z = 2 m.

If the mass of the flowerpot

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


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We will first find the locations of the points A, C, and D. We do not need to find the location of point B as the force in cable AB would lie along the y-axis.

If the mass of the flowerpot

Using the diagram, the locations of the points in Cartesian vector form is:




(remember x = 1.5 m, and z = 2 m)


We can now write position vectors for cables AC and AD.



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


Next, we will find the magnitude of each position vector.

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

magnitude of r_{AD}\,=\,\sqrt{(-1.5)^2+(-6)^2+(2)^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.


Now, we can find the unit vectors for each position vector.



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)}}


Our next step is to express each force in the cables.



(simplify by expanding the brackets and writing the fractions in decimal form)




(Remember, F_{AB} is the force that lies along the y-axis, it only has the y-component)


(Force W is the weight of the pot, which has only a z-component. The weight is equal to (50)(9.81)=490.5 N)


As the system is in equilibrium, all forces added together must equal zero.

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



Since all the forces added together must equal zero, then the individual components added together must also equal zero.








Solving the three equations gives us:

F_{AB}\,=\,1210 N

F_{AC}\,=\,605 N

F_{AD}\,=\,749 N

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-54.

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