Tension in Strings HELP thanks-very simple.

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The discussion revolves around calculating the tension in a string supporting a 16 kg mass in equilibrium and determining the frequency for a standing wave pattern. Users analyze the forces acting on the mass, emphasizing that the tension must equal the gravitational force due to equilibrium conditions. The importance of understanding angles and vector components is highlighted, as they are crucial for resolving forces correctly. After several calculations and clarifications, the tension is found to be approximately 105.3 Newtons. The conversation also touches on the need for reference materials in learning physics effectively.
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[SOLVED] Tension in Strings HELP thanks!--very simple.

Homework Statement



A 16 kg object hangs in equilibrium from a string with a total length of L=5m and a linear mass density of u= 0.0010 kg/m. The string is wrapped around two light, frictionless pulleys that are separated by a distance of d=2.0m.

A) Determine the tension in the string.
B) At what frequency must the string between the pulleys vibrate in order to form the standing-wave pattern shown in Figure b)?



Picture is attached!




Homework Equations





The Attempt at a Solution



If there was an axis at the bottom of the inverted triangle and the angles were measured from the axis to the strings;

I have also tried using;
v = Square root of (tension/u), but I don't have v,
So i tried to find v using f = v/ wavelength, but i don't have frequency.

I have tried so many things and I can't seem to get a start. Any hints?
 

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Since the 12kg mass is stationary, what can you say about the sum of the forces acting on it?
 
since its in equilibrium the net force must be 0.

acknowledge all the forces in the y direction.
tension upwards
gravities pull downward..

since their total combined force is 0 due to mr.Newton

then tension must be equal to gravities pull, besides direction

(assuming the weight is connected to one string)
 
rsala said:
since its in equilibrium the net force must be 0.

acknowledge all the forces in the y direction.
tension upwards
gravities pull downward..

since their total combined force is 0 due to mr.Newton

then tension must be equal to gravities pull, besides direction
Wow physicsbhelp! You said that without even moving your lips...:rolleyes:
 
wait so is the tension equal to gravity? i typed in 9.81 and it said it was wrong so i typed in -9.81 and it said that was wrong too! i am still in need of help!
 
9.81 m/s^2 is the acceleration due to gravity (for an object in freefall). What's the weight of the 16 kg mass?
 
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ooo the weight would be 16*9.81? = 156.96. i typed that in and it was also wrong.
 
so what else could it be?
 
If I understand the setup correctly, then that would be the tension in the string (in N). How fussy is the system about significant figures? Round it off. (You certainly don't have 5 significant figure accuracy!)
 
  • #10
no that is not the answer. i didnt even round up on the real answer.
 
  • #11
okay i attached the diagram. i think that will help. part a) has to do with picture a).
 
  • #12
Providing a diagram makes a huge difference! Now it's clear. :wink:

Analyze the forces acting on the hanging mass. You know it must be in equilibrium--that will allow you to figure out the tension. (First figure out the angle the strings make.)
 
  • #13
what angle are you talking about? like the angle that the string makes with the object? i thought it was a 45 45 90 triangle. but i still don't see how angles will help find tension. is there an equation for that.
 
  • #14
How did you figure out those angles? Don't guess--they gave you the length of the string for a reason. Once you have the angles, then you can draw yourself a free-body diagram of the forces acting on the mass. (The tension acts along the strings--that's why you need the angles.) The equation you need is the equilibrium condition, specifically: The sum of the vertical force components must equal zero.
 
  • #15
i am still not quite understanding this equilibrium angle equation.
so the base of the inverted trianlge is 2 meters and the two sides are 1.5 meters long each. then i still don't get the angles. or basically how to find them.
 
  • #16
Draw your own diagram, marking the lengths. Then use a little trig to find any angle you might need. Hint: Draw a vertical line passing through the mass. Use that as a reference.
 
  • #17
i drew my triangle the way u said. but i still don't seem to get it. like the vertical line passing throught the mass is 1.118 meters. i figured that out. but i can't tell how to do trig? like do i do cos (1/1.5)?
 
  • #18
physicsbhelp said:
i drew my triangle the way u said. but i still don't seem to get it. like the vertical line passing throught the mass is 1.118 meters. i figured that out.
OK.
but i can't tell how to do trig? like do i do cos (1/1.5)?
What's the cosine of the angle that the string tension makes with the vertical? You'll need that when you write your force equation. (Write that equation!)
 
  • #19
ooo wait i think i just got it! the whole time my calculator was in radian mode, that is why i kept getting weird answers. haha.
okay so for the ablge between the side that equals 1meter and 1.5 meter i got 48.187 degrees and for the other angle i got 41.81 degrees.
now i really don't know how i can find out the tension.
 
  • #20
Call the unknown tension in each string T. Draw a diagram showing the forces. Add up the vertical components. Solve for T.
 
  • #21
i did that but i have no numerical values. i got that 2Ft=Fg? is that right?
 
  • #22
or is it not?
 
  • #23
physicsbhelp said:
i got that 2Ft=Fg? is that right?
That would be true if the strings were vertical, but they are not. If the tension in each string is T, and the angle that the string makes with the vertical is \theta, what is the vertical component of the force that each string exerts on the mass?
 
  • #24
Ft(theda)
or
0
 
  • #25
Read this: http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/vectors/u3l3c.html"
 
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  • #26
well that sort of helped. but I am asking for the tension in the string, not for the Net Force of the string on the OBject
 
  • #27
Can you find the net force of the string on the object? What's the only other force on the object? Add them up to get zero and you can solve for the tension.

Look at the problems described on that page. Almost the exact problem that you have is discussed several times.
 
  • #28
so gravity plus the tension of the string on the object equals 0
 
  • #29
physicsbhelp said:
so gravity plus the tension of the string on the object equals 0
Yes, the net force on the mass is zero. As I've stated several times over.

Find the vertical components. Learn how to do that or you're just wasting time. Read the examples on that website.
 
  • #30
but i don't know how to do that. i read the website you sent
i don't get it. like i konw what vecotrs are ut can you say they are distnace?
 
  • #31
the trick is in the x and y components
 
  • #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!
 

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