Determine the height d of cable AB 2

Determine the height d of cable AB so that the force in cables AD and AC is one-half as great as the force in cable AB.What is the force in each cable for this case? The flower pot has a mass of 50 kg.

Determine the height d of cable AB

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 figure out the position vectors for points from A to C, and A to D. We will not find the position vector for the rope AB since we don’t know the value of d.

Determine the height d of cable AB

The locations of the points are:





We can now write our position vectors:



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_{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:



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. Remember, the question says “force in cables AD and AC is one-half as great as the force in cable AB,” thus, we will multiply each force by half of F_{AB}.:



(simplify by expanding brackets and writing fractions are decimals)





(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



Furthermore, since each force added together must equal zero, then each component (x, y, z-component) added together must also equal zero.








Simplifying these equations leads us to the following:

(F_{AB})_x\,=\,0.858F_{AB} (eq.1)

(F_{AB})_z\,=\,490.5-0.428F_{AB} (eq.2)


We will now turn to the Pythagorean theorem to find the value of F_{AB}. The (F_{AB})_x and (F_{AB})_z values we found are the two sides of a right angle triangle. The value of F_{AB} is the hypotenuse.

Determine the height d of cable AB

Thus, we can write the Pythagorean theorem for this triangle.


Substitute the values of eq.1 and eq.2 we found.




(Solve for the positive root)

F_{AB}\,=\,520 N


Since F_{AD} and F_{AC} are half of F_{AB}, then:

F_{AD}\,=\,F_{AC}\,=\,\dfrac{520}{2}\,=\,260 N


We can now figure out (F_{AB})_x and (F_{AB})_z by substituting the value of F_{AB} we found into eq.1 and eq.2.

(F_{AB})_x\,=\,(0.858)(520)\,=\,446.2 N

(F_{AB})_z\,=\,490.5-0.428(520)\,=\,268 N


Referring to the triangle we formed before, we now have the values of all sides. Thus, we can use trigonometry to figure out the angle \theta.




Again, referring to the triangle we formed, remember that the base is 6 m in length. Thus, using trigonometry, we can figure out the opposite height. Using tangent, which is opposite over adjacent, we can write the following:

\tan 31^0\,=\,\dfrac{d}{6}

d\,=\,6\tan 31^0

d\,=\,3.6 m


Final Answers:

F_{AB}\,=\,520 N

F_{AC}\,=\,260 N

F_{AD}\,=\,260 N

d\,=\,3.6 m


This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-53.

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