Respuesta :
The force in each cable depends on the orientation of the cable relative
to the force of the balloon and to each other.
Responses:
a) The position vectors are;
Position vector, DA, [tex]r_{DA}[/tex] = -20·i - 30·j + 0·k
DB, [tex]r_{DB}[/tex] = 30·i - 30·j + 20·k
DC, [tex]r_{DC}[/tex] = 0·i - 30·j + 20·k
[tex]b) \hspace{0.15 cm} Unit \ vectors \ are, u_{DA} = \underline{ \dfrac{-2 \cdot \sqrt{13} \cdot i}{13} - \dfrac{ 3 \cdot \sqrt{13} \cdot j }{13 }}[/tex]
- [tex]u_{DB} = \underline{\dfrac{3 \cdot \sqrt{13} }{13} \cdot i + \dfrac{2 \cdot \sqrt{13} }{13} \cdot k}[/tex]
- [tex]u_{DC} = \underline{ -\dfrac{3 \cdot \sqrt{13} }{13} \cdot j + \dfrac{2 \cdot \sqrt{13} }{13} \cdot k}[/tex]
c) The approximate values of the force in each cable are;
- DA: 3245 N
- DB: 2163.3 N
- DC: -2163.3 N
Which method is used to calculate forces in vector form?
a) The position vector is a vector that locates a point relative to another point.
The position vector,
[tex]r_{DA} = \mathbf{ r_A - r_D}[/tex]
Whereby the coordinate of the point D is taken as (0, 30, 0), we have;
- [tex]r_{DA}[/tex] = -20·i + 0·j + 0·k - (0·i + 30·j + 0·k) = -20·i - 30·j + 0·k
Similarly, we have;
[tex]r_{DB}[/tex] = 30·i + 0·j + 20·k - (0·i + 30·j + 0·k) = 30·i - 30·j + 20·k
[tex]r_{DC}[/tex] = 0·i + 0·j + 20·k - (0·i + 30·j + 0·k) = 0·i - 30·j + 20·k
b) The unit vectors are therefore;
[tex]u_{DA} = \mathbf{ \dfrac{r_{DA}}{|r_{DA} |}}[/tex]
Which gives;
[tex]u_{DA} = \dfrac{-20 \cdot i - 30 \cdot j + 0 \cdot k}{\sqrt{(-20)^2 + (-30)^2 + 0} } = \dfrac{-2 \cdot i - 3 \cdot j }{\sqrt{13} } = \dfrac{\left(-2 \cdot i - 3 \cdot j \right) \times \sqrt{13} }{13 }[/tex]
[tex]u_{DA} = \mathbf{ \dfrac{-20 \cdot i - 30 \cdot j + 0 \cdot k}{\sqrt{(-20)^2 + (-30)^2 + 0} }}} = \dfrac{-2 \cdot i - 3 \cdot j }{\sqrt{13} } = \underline{\dfrac{-2 \cdot \sqrt{13} \cdot i}{13} - \dfrac{ 3 \cdot \sqrt{13} \cdot j }{13 }}[/tex]
[tex]u_{DB} = \dfrac{30 \cdot i + 0 \cdot j + 20 \cdot k}{\sqrt{(30)^2 + (20)^2 + 0} } = \mathbf{\dfrac{3 \cdot i +2 \cdot k }{\sqrt{13} } } = \underline{\dfrac{3 \cdot \sqrt{13} }{13} \cdot i + \dfrac{2 \cdot \sqrt{13} }{13} \cdot k}[/tex]
[tex]u_{DC} = \mathbf{\dfrac{-30 \cdot j + 20 \cdot k}{\sqrt{(30)^2 + (20)^2 } } }= \underline{-\dfrac{3 \cdot \sqrt{13} }{13} \cdot j + \dfrac{2 \cdot \sqrt{13} }{13} \cdot k}[/tex]
c) The force in each cable are found as follows;
[tex]\vec F_{DA} = \mathbf{ \dfrac{-2 \cdot \sqrt{13} }{13}\cdot i \cdot F_{DA}- \dfrac{ 3 \cdot \sqrt{13} }{13 } \cdot j\cdot F_{DA}}[/tex]
[tex]\vec F_{DB} = \mathbf{ \dfrac{3 \cdot \sqrt{13} }{13} \cdot i \cdot F_{DB}+ \dfrac{2 \cdot \sqrt{13} }{13} \cdot k\cdot F_{DB}}[/tex]
[tex]\vec F_{DC} = \mathbf{ -\dfrac{3 \cdot \sqrt{13} }{13} \cdot j \cdot F_{DC}+ \dfrac{2 \cdot \sqrt{13} }{13} \cdot k\cdot F_{DC}}[/tex]
Sum of forces, ∑F = 0, gives;
[tex]\mathbf{ \dfrac{-2 \cdot \sqrt{13} }{13}\cdot i \cdot F_{DA} +\dfrac{3 \cdot \sqrt{13} }{13} \cdot i \cdot F_{DB}} = 0[/tex]
[tex]\mathbf{ - \dfrac{ 3 \cdot \sqrt{13} }{13 } \cdot j\cdot F_{DA} -\dfrac{3 \cdot \sqrt{13} }{13} \cdot j \cdot F_{DC} + 900} = 0[/tex]
[tex]\mathbf{ \dfrac{2 \cdot \sqrt{13} }{13} \cdot k\cdot F_{DB} + \dfrac{2 \cdot \sqrt{13} }{13} \cdot k\cdot F_{DC}} = 0[/tex]
Solving the above simultaneous equation with a graphing calculator gives;
- The force in the cable DA, [tex]F_{DA}[/tex] ≈ 3245 N
- The force in the cable DB, [tex]F_{DB}[/tex] ≈ 2163.3 N
- The force in the cable DC, [tex]F_{DC}[/tex] ≈ -2163.3 N
Learn more about vectors here:
https://brainly.com/question/2263823
