Question

What are the tensions in the cords.

Answer 1

\[\sum_{}^{} F _{x}=0\]
\[F _{T _{1}} \cos(135) + F _{T _{2}} \cos(330)= 0\]

\[\sum_{}^{} F _{y}=0\]
\[F _{T _{1}} \sin(135) + F _{T _{2}} \sin (330) + -1\]


I need to solve for FT1 and FT2, But am having trouble figuring it out.... my professor said that FT1= 49N and FT2= 40N

Answer 2

So the second one you wrote is \(F _{T _{1}} \sin(135) + F _{T _{2}} \sin (330) + -1
\). Is that equal to anything? Is that equal to zero or something?

Answer 3

zero, yes sorry

Answer 4

Alright, that might help:\[F _{T _{1}} \sin(135) + F _{T _{2}} \sin (330) + -1=0\]Right?

Answer 5

yes

Answer 6

would you set them equal to each other?

Answer 7

Any diagram? I'm trying to get a concept of what's going on...

That could be one approach that might help!

Answer 8

okay 1 sec

Answer 9

I see! So we're talking about tension! Not torque. And I still have to find where the \(\+-1\) comes from.

Answer 10

\(+-1\)*

Answer 11

CCW is +
CW is -

Answer 12

sin 270 is -1

Answer 13

Oh! And so would that be \(m\ g\ sin(270^\circ)=m\ g\ (1)=m\ g\)?

Answer 14

it would become just -mg

Answer 15

Because it would be the force times that angle.. And I meant negative, sorry!

Answer 16

exactly, no it is okay

Answer 17

\(\left|m\ g\right|\ sin(270^\circ)=\left|m\ g\right|\ (-1)=-\left|m\ g\right|\)

Answer 18

why absolute?

Answer 19

oh nvm, I get it

Answer 20

That makes more sense! :)

So, now we have those two equations set up:\[\sum F_x=F _{T _{1}} \cos(135) + F _{T _{2}} \cos(330)= 0\]and\[\sum F_y=F _{T _{1}} \sin(135) + F _{T _{2}} \sin (330) - m\ g=0\]

Answer 21

Nice!

Answer 22

yes, okay. now what?

Answer 23

Oh sorry, I have to go! Here's my plan. Solve it like a "system of equations." You can look that up if you're unsure!

1. So, pick one equation and solve it for one force you pick.
2. Then, substitute that into the second equation.
3. Then solve that equation for the other force.
4. Then you'll know that force.
5. Substitute it's value back into the first equation, solved for the force you don't know (the one you solved for first)
6. Then you'll know both forces.

I'll be on later tonight, and I'd be happy to see if we arrive at the same result. Sorry I have to go! And best of luck :)

Answer 24

Those instructions were just quick, and some where actually just statements... But I'll use them here anyway. I hope this helps!

\[\sum F_x=F _{T _{1}} \cos(135) + F _{T _{2}} \cos(330)= 0\]\[\huge\sf 1. \]I'll solve for \(F_{T_1}\).\[F _{T _{1}} \cos(135) + F _{T _{2}} \cos(330)= 0\]\[F _{T _{1}} \cos(135) =- F _{T _{2}} \cos(330)\]\[F _{T _{1}} =\frac{- F _{T _{2}} \cos(330)}{\cos(135)}\]\[\\\huge\sf 2.\]Now I can substitute that into the other equation (the one for \(\sum \left[F_x\right]\)).\[\ \downarrow\]\[F _{T _{1}} \sin(135) + F _{T _{2}} \sin (330) + m\ g=0\]After substitution...
\[\frac{- F _{T _{2}} \cos(330)}{\cos(135)}\sin(135) + F _{T _{2}} \sin (330) + m\ g=0\]\[\\\huge\sf 3.\]And now we solve for \(F_{T_1}\).\[F _{T _{2}}\left(\frac{- \cos(330)}{\cos(135)}\sin(135) + \sin (330)\right) + m\ g=0\]\[F _{T _{2}}\left(\frac{- \cos(330)}{\cos(135)}\sin(135) + \sin (330)\right) = -m\ g\]\[F _{T _{2}} = \frac{-m\ g}{\frac{- \cos(330)}{\cos(135)}\sin(135) + \sin (330)}\]\[=\frac{-m\ g}{0.36602540378443804}=40.202127633376094\ [N]\]Where \(g=-9.81\ [m/s^2]\).
\[\\\huge\sf 4.\]So we have \(F_{T_1}\).

Answer 25

Correction: now we have \(F_{T_2}\).

\[\huge\sf 5.\]So we solved for \(F_{T_1}\) earlier and got\[F _{T _{1}} =\frac{- F _{T _{2}} \cos(330)}{\cos(135)}\]We put in \(F_{T_2}\)'s value in, and we have\[F _{T _{1}} =\frac{-40.202127633376094\ [N]\ \cos(330)}{\cos(135)}\]\[=49.237349638007452\ [N]\]

\[\huge\sf 6.\]Now we have both \(F_{T_1}\) and \(F_{T_2}\). When we round to two significant figures, they are just what your professor said.

Answer 26

Let me know if you have any questions!

Like, why did I make it \(+m\ g\)? I did that because \(\left|m\ g\right|\) is the magnitude, and \(270^\circ\) is the angle. But, it works that the direction (negative \(y\)) is given by \(g\), which is really a vector itself. And we're adding \(m\ g\) because it is \(\sum F_y\) - we are adding them up to get 0. It's a different way of looking at it. For your way, you would use \(\left|m\ g\right|\) as the magnitude and the direction would be \(270^\circ\). For my way, I just used the direction given by \(g\), which is negative because \(g\approx -9.81\ [m/s^2]\)

Answer 27

Tension is an internal force trying to restore the original length of a rope or a rod pulled or pushed along its length:)

Answer 28

Haha, wonderful definition, @Deba_001 . And I notice that it applies to the original prompt :)

Answer 29

10x theEric