Question

Can I get some help on 3D vectors?

Answer 1

I give you some hint.

Answer 2

Make the ropes like a triangle comparing to the wall.

Answer 3

Then use Pythagoras theorem

Answer 4

a^2+b^2= C^2

Answer 5

Or use cosine, sine and tan if it is possible.

Answer 6

ok sowhat i did was i started with F(AB) and using the coordinates i made a triangle in the x-z plane and then found the hypotenuse which was the bottom of my second triangle in the y plane and found the angle using tan and the previous hypotenuse. i ended up with a angle of 43.6 degrees. then i separated the F(AB) into components like so,\[F_{AB}=-189.8i+63.1j+138k\]and then i did the same for F(AC) and got this,\[F_{AC}=98.1i+19.6j+50.7k\]
so then i added the two together and got this\[F_{AB+AC}=-91.7i+82.7j+188.7k\]and then i did this\[\sqrt{(-91.7i)^2+(82.7j)^2+(188.7k)^2}=225.5 ft*lbs\]please tell me i'm somewhat right...

Answer 7

Coordinates for Points A,B,C
A(6, 0, 10)
B(0, 6, 8)
C( 8, 6, 0)
-----------------

Vectors and Magnitudes
<AB> = < 0-6 , 6-0, 8-10> = <-6 , 6 , -2>
<AC> = <8-6, 6-0, 0-10> = < 2, 6, -10>

\[\left| \left| AB \right| \right| = \sqrt{(-6)^2+6^2+(-2)^2} = 2\sqrt{19}~ft\]
\[\left| \left| AC \right| \right|=\sqrt{2^2+6^2+(-10)^2}=2\sqrt{35}~ft\]
Unit Vector In direction of each rope:
\[\lambda _{AB} = \frac{ <AB> }{ \left| \left| <AB> \right| \right| }=\frac{ 1 }{ 2\sqrt{19} }<-6, 6, -2>\]
\[\lambda _{AC}=\frac{ <AC> }{ \left| \left| <AC> \right| \right| }=\frac{ 1 }{ 2\sqrt{35} }<2, 6, -10>\]

Answer 8

Since you are given the magnitude of each Force and the coordinates, you can find the unit vectors along each line of action, like above, then multiply those by the magnitude of each force.

\[F _{AB} = \left| \left| F _{AB} \right| \right|\lambda _{AB} = [200~lb]*\frac{ 1 }{ 2\sqrt{19} }*<-6, 6, -2>\]

\[F _{AC}=\left| \left| F _{AC} \right| \right|\lambda _{AC}= [100~lb]*\frac{ 1 }{ 2\sqrt{35}}*<2, 6 , -10> \]

Answer 9

The Resultant Force at point A is the sum of Force AB and Force AC.
\[R = \sum F = F _{AB}+F _{AC} = <(F _{ABx}+F _{ACx}), (F _{ABy}+F _{ACy}), (F _{ABz}+F _{ACz})>\]

Answer 10

Decimal approx...
\[F _{AB}=<-137.65,~ 137.65, -45.88>\]
\[F _{AC}=<16.90,~ 50.71, -84.52>\]

Answer 11

Add those to get the resultant Force at A, then use
\[\left| \left| R \right| \right|=\sqrt{R _{x}^{2}+R _{y}^{2}+R _{z}^{2}}\]

For the magnitude of the resultant force. If you want the angles, you can use the Direction Cosines.

Answer 12

Resultant R = < -120.75, 188.36, -130.4> lbf

Magnitude is about = 258.97 lbf

Give er a check, i went through it kinda fast.