Tension in Strings HELP thanks-very simple.

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    Strings Tension
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SUMMARY

The discussion focuses on calculating the tension in a string supporting a 16 kg mass in equilibrium and determining the frequency required for a standing wave pattern. The tension in the string is derived from the weight of the mass, calculated as 156.96 N, and the equilibrium condition leads to the equation 1.491F = mg, resulting in a tension of approximately 105.30 N. For part B, the user initially miscalculated the frequency, which is linked to the wave properties of the string.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly equilibrium conditions.
  • Knowledge of trigonometry for calculating angles and components of forces.
  • Familiarity with wave mechanics, specifically standing waves and their frequencies.
  • Ability to apply the wave speed formula: v = √(T/μ), where T is tension and μ is linear mass density.
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  • Learn how to derive tension in strings using equilibrium conditions in physics.
  • Study the relationship between wave speed, tension, and linear mass density in strings.
  • Explore the concept of standing waves and their formation in strings.
  • Investigate the calculation of frequencies for different harmonic modes in vibrating strings.
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  • #31
the trick is in the x and y components
 
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  • #32
yes but when i tried to use the length for the x-component it was wrong
 
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  • #33
physicsbhelp said:
yes but when i tried to use the length for the x-component it was wrong

Well, what did you actually try? If you don't give us any working then it's incredibly difficult to give you any advice on where you may have gone wrong!
 
  • #34
well i did 1 times 9.81
and 1.5 * 9.81
then i tried to add up all the lenghts: 1+1.18+1.5 and multiply that by 9.81
 
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  • #35
SOMEONE please help me with this problem! thanks i appreaciate all the effort/help!
 
  • #36
You need to learn how to find the components of a vector. If you have some vector with magnitude F that makes an angle \theta with the x-axis (the horizontal), the components are:
  • vertical (y) component = F \sin\theta
  • horizontal (x) component = F \cos\theta
I recommend looking at the examples on this page, especially the one with the dog on a leash: http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/vectors/u3l1e.html" . In fact I strongly recommend that you spend a good bit of time exploring the rest of that site to better understand forces, components, and equilibrium.
 
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  • #37
oh okay i think i get what you are saying. thank you!

so what i basically have to do is this: 41.81*2*T= 156.96?
is that what it means. i think?
 
  • #38
because the Tension in the x equals mass times gravity= 156.96
and the Tension in the y equals sin (theda) = 1.118
and the Tension in the x of the angle equals cos (theda) = 1/1.5
 
  • #39
No. What angle does each string make with the horizontal?
 
  • #40
48.187
 
  • #41
physicsbhelp said:
48.187
Good! Now use what I told you in post #36.
 
  • #42
so sin(48.187) + cos (48.187) = 0
 
  • #43
wait no sorry it should be: sin(48.187) + cos (48.187) = mg
 
  • #44
right?
 
  • #45
No. In post #36, I told you that:
Doc Al said:
vertical (y) component = F \sin\theta
In your problem, the F would represent the tension in the string. So write down the vertical component. (Just plug in the angle and compute the sin.)

Since you have two strings, you'll have to double it.

Remember: F, the tension, is an unknown. Once you get the equilibrium equation written, you will solve for F.
 
  • #46
oooo now i get what you mean so its 2F(sin(theda))
so basically i got 2*(sin(48.187))F
which = 1.491F
 
  • #47
but what i don't get is how you have 2 STRINGS because i am jus ttlaking about diagram a) so i hope u are not trying to figure out the tension in both the pictures.
or are you saying that since the big traingle can be divided into two triangles, there are two strings?
 
  • #48
physicsbhelp said:
oooo now i get what you mean so its 2F(sin(theda))
so basically i got 2*(sin(48.187))F
which = 1.491F
Exactly! Now set that equal to what?

physicsbhelp said:
but what i don't get is how you have 2 STRINGS because i am jus ttlaking about diagram a) so i hope u are not trying to figure out the tension in both the pictures.
or are you saying that since the big traingle can be divided into two triangles, there are two strings?
As far as the forces go, you can view the mass as being pulled by two strings--one left and one right. (Of course, it's just two parts of the same string, but that doesn't matter.)
 
  • #49
okay so now i totally get you.

thanks
so now the next step i set 1.491F=mg?
or 1.491F=.6667--cos(48.187)
 
  • #50
physicsbhelp said:
so now the next step i set 1.491F=mg?
Yes! Then solve for the tension, F.
or 1.491F=.6667--cos(48.187)
No!
 
  • #51
PhysicsB, though this may seem unrelated to this thread, what book are you working from? I know what course you are taking (The same one as myself) but what book are you working from?
 
  • #52
THANK YOU

ooo okay so let's see. (btw nice smiley face haha)

okay so mg=156.96
so 1.491F=156.96
so F= 105.2963827Newtons?

is that the correct answer?! THANK YOU
 
  • #53
_Mayday_ said:
PhysicsB, though this may seem unrelated to this thread, what book are you working from? I know what course you are taking (The same one as myself) but what book are you working from?

i am not using a book. my physics teacher believes that we should not use books because it hinders your learning of physics. so he gives us random questions.
 
  • #54
Yes but do you use a book in class to learn from or to use as a reference?
 
  • #55
No, that is what i just said. our teacher did not tell us to buy a book. we do not use any book. he just gives us notes on equations and then we get problems that he sends us through webassign or something
 
  • #56
so i hope you got my answer Doc Al, is it right?
 
  • #57
physicsbhelp said:
okay so mg=156.96
so 1.491F=156.96
so F= 105.2963827Newtons?
Looks good! (Round off your final answer to some reasonable number of significant figures.)

physicsbhelp said:
i am not using a book. my physics teacher believes that we should not use books because it hinders your learning of physics. so he gives us random questions.
You've got to be kidding! :eek:
 
  • #58
I would say buying a book does the opposite, it does not hinder your progress it guides you in the right direction. These questions on vectors are not easy a first but with the help of a book you can work through some excercises and make sure you are doing the correct thing.
 
  • #59
OMG OMGOMGOMGOMGOMGOMGOMGOMG thank you soooo much DOC AL YOU
ARE THE BEST OF THE BEST!
!
!
!
!
 
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  • #60
Doc Al said:
You've got to be kidding! :eek:

Exactly what I thought, it's insane! You need something to reference all your work or to use as a base for all your knowledge...
 

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