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.

#### Solution:

Show me the final answers↓

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.

The locations of the points are:

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

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

We can now write our position vectors:

r_{AD}\,=\,\left\{(-6-0)i+(2-0)j+3-0)k\right\}\,=\,\left\{-6i+2j+3k\right\}

The magnitude of each position vector is:

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

The unit vectors for each position vector is:

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

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_{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.

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

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

y-components:

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

z-components:

Simplifying these equations leads us to the following:

(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.

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

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:

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})_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\,=\,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:

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

d\,=\,3.6 m

#### Final Answers:

F_{AC}\,=\,260 N

F_{AD}\,=\,260 N

d\,=\,3.6 m

Thank you so much for your work, this made me understand this question!

You’re very welcome. Thank you for taking the time to comment, it’s highly appreciated! Good luck with your studies 🙂