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.

Solution:

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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:

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

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

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

 

We can now write our position vectors:

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_{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_{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. 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}.:

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

F_{AD}\,=\,\dfrac{F_{AB}}{2}\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}\,=\,(F_{AB})_xi+0j+(F_{AB})_zk

F_{AC}\,=\,-0.429F_{AB}i-0.143F_{AB}j+0.214F_{AB}k

F_{AD}\,=\,-0.429F_{AB}i+0.143F_{AB}j+0.214F_{AB}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:

(F_{AB})_x-0.429F_{AB}-0.429F_{AB}\,=\,0

y-components:

-0.143F_{AB}+0.143F_{AB}\,=\,0

z-components:

(F_{AB})_z+0.214F_{AB}+0.214F_{AB}-490.5\,=\,0

 

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.

F^2_{AB}\,=\,(F_{AB})^2_x+(F_{AB})^2_z

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

F^2_{AB}\,=\,(0.858F_{AB})^2+(490.5-0.428F_{AB})^2

(simplify)

0.080652F^2_{AB}+419.868F_{AB}-240590\,=\,0

(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.

\theta\,=\,\tan^{-1}\left(\dfrac{268}{446.2}\right)

\theta\,=\,31^0

 

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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