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Problem with tension on two strings at angles

by lilmissbossy
Tags: angles, strings, tension
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lilmissbossy
#1
Aug23-08, 11:59 PM
P: 14
mass m suspended by a pair of vertical ropes attached to the ceiling
what are the rope tensions if they comprise a V shape each at an angle theta with the ceiling?

i have played with this problem for a while and i cant seem to find what the relevant equations are really only first time dealing with physics i am a biology student...
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rock.freak667
#2
Aug24-08, 12:28 AM
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Split the two tensions into vertical and horizontal components. At equilbrium, the sum of the vertical forces=0 and similarly, the sum of the horizontal forces=0
dynamicsolo
#3
Aug24-08, 12:29 AM
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You should start with a picture of the mass and the ropes (which are not vertical, but canted), showing the forces acting on the mass. There will be the mass' weight force and the tensions from each rope. Is the angle theta measured from the vertical or from the horizontal?

lilmissbossy
#4
Aug24-08, 12:46 AM
P: 14
Problem with tension on two strings at angles

I have drawn a picture which kinda looks like this

_________________
\o o/
\ /
\ /
\___/
[_m_]
didnt know how to inuput a paint drawning. o represents theta

if the ropes were hanging vertically i have the equation T1= 1/2mg and T2=1/2mg as they both equally support the mass.
I am thinking of using sine theta to work out the answer but i am stuck at that point, im not sure what to use.
lilmissbossy
#5
Aug24-08, 12:46 AM
P: 14
ok that picture didnt work at all
sorry
dynamicsolo
#6
Aug24-08, 01:09 AM
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Quote Quote by lilmissbossy View Post
if the ropes were hanging vertically i have the equation T1= 1/2mg and T2=1/2mg as they both equally support the mass.
That would be correct; this also provides a clue to how to deal with the tilt of the ropes.

I'm assuming from the description (and the way you marked your diagram) that the ropes make an angle theta to the horizontal. So the tensions are still equal, but now the weight W = mg acting downward is counterbalanced by the vertical components of the two tensions, rather than the full tensions T = T1 = T2. If the tension in one of the tilted ropes is now T' , what is the component of its tension that acts vertically?
lilmissbossy
#7
Aug24-08, 01:56 AM
P: 14
Ok so i guess instead of using 1/2 i factor in the angle it hangs on.
maybe something like T1= sin(theta)mg...

again i am not very good at physics at all
tiny-tim
#8
Aug24-08, 03:28 AM
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Quote Quote by lilmissbossy View Post
I have drawn a picture which kinda looks like this

_________________
\o o/
\ /
\ /
\___/
[_m_]
didnt know how to inuput a paint drawning. o represents theta
Hi lilmissbossy!

(Have a theta: θ )

Use code tags, 'cos they preserve spacing :

_________________
       \θ        θ/
        \         / 
         \       /
          \___/
          [_m_]
hmm … still needs adjusting … I wonder why?

    _________________
       \θ       θ/
        \       / 
         \     /
          \___/
          [_m_]
dynamicsolo
#9
Aug24-08, 10:32 AM
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Quote Quote by lilmissbossy View Post
Ok so i guess instead of using 1/2 i factor in the angle it hangs on.
maybe something like T1= sin(theta)mg...
You're getting close. If [tex]\theta[/tex] represents the angle each rope makes to the horizontal ceiling, then the vertical component of the tension T' in each rope will indeed be [tex]T' sin(\theta)[/tex].

Now there are two such ropes connected to the suspended mass pulling upward, and the mass' weight mg is pulling downward. Since the mass is not accelerating (indeed, not moving), the sum of the forces acting on it in the vertical direction are

[tex]T' sin(\theta) + T' sin(\theta) - mg = ma_y = 0[/tex]

T' is the only unknown, so you can now solve for that, which is the magnitude of the tension in each rope. (How does it compare to the tension in the case of vertical ropes -- is it larger or smaller? Why would that be?)
tiny-tim
#10
Aug24-08, 12:39 PM
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Quote Quote by lilmissbossy View Post
Ok so i guess instead of using 1/2 i factor in the angle it hangs on.
maybe something like T1= sin(theta)mg...

again i am not very good at physics at all
Hi lilmissbossy!

i take it you're not happy with components? You don't know when to use them, or whether to use sin or cos?

    _________________
       \θ    | θ/
        \    Ξ / 
         \   |/
          \__/___
          [_m_]
Look at the diagram … I've drawn the two components for the tension force in the right-hand string.

The vertical line has a sort-of arrow pointing up, and the short horizontal line should have an arrow pointing right (but doesn't ).

Together, they make up the tension in the string , and they are the right lengths!

So the upward component is Tsinθ (opp/hyp), and the horizontal one is Tcosθ (adj/hyp).
You can always do it this way, just to check, if you're not sure you have the correct θ.
lilmissbossy
#11
Aug24-08, 03:37 PM
P: 14
Ok i understand it now, thank a heap guys. i wish i new about this place a lot earlier...
Cheers
lilmissbossy
#12
Aug24-08, 07:13 PM
P: 14
i know this again

i have done two different types of equations both giving the same answer except one is a -ve number. here are the equations

Tsin(θ)+Tsin(θ)-mg=ma=0
rearranged
T= sin(θ)+sin(θ)-mg
T=sin(53)+sin(53)-(55kg)(9.8ms-1)
T= -537.402

then i used the cos equation to check if there was any difference and this is what i got

Tsin(θ)+Tcos(θ)-mg
T=sin(θ)+Tcos(θ)-mg
T=sin(53)+cos(53)-(55kg)(9.8ms-1)
T= -537.6

i am not sure if i am rearranging the equations right or if i have the right method>>>
lilmissbossy
#13
Aug24-08, 07:21 PM
P: 14
i also get for the tension on the vertical ropes (no angles) at

T1=1/2mg
T1=1/2(55kg)(9.8ms-1)
T1=269.5

which is basically what i get when i divide the previous answers by 2

should they be the same???
dynamicsolo
#14
Aug24-08, 08:38 PM
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Quote Quote by lilmissbossy View Post
i have done two different types of equations both giving the same answer except one is a -ve number. here are the equations

Tsin(?)+Tsin(?)-mg=ma=0
rearranged
T= sin(?)+sin(?)-mg
This is not rearranged correctly: you cannot simply separate the sine factors from the T's. This should be

2 · T · sin(?) = mg

T = mg / [ 2·sin(?) ]

Now you won't get a negative force magnitude; magnitudes of forces can't be negative.

then i used the cos equation to check if there was any difference and this is what i got

Tsin(?)+Tcos(?)-mg
T=sin(?)+Tcos(?)-mg
T=sin(53)+cos(53)-(55kg)(9.8ms-1)
T= -537.6
Where would a cosine factor come from? From the angle theta being measured from the horizontal, Tcos(?) would be the horizontal component of the rope tension; that component is not involved in supporting the weight.


In the case of the ropes being vertical, the angle theta is 90Ί (not "no angle"). The tension in each rope is then T1 = T2 = T = mg / [2 sin 90Ί] = mg / (2 · 1) = mg/2 , as you said earlier.
lilmissbossy
#15
Aug24-08, 08:51 PM
P: 14
i got the cosine from tinytim, but i realise now that its not a factor.
ok i see where i rearranged wrong and i now have answers that show a difference in the tension at 90* and at 53*. with the 53* having more tension on the strings. i guess when you think about it, it makes sense that the angled string has a greater tension as it is strected further from its initial point.... thank you again this is all very confusing
dynamicsolo
#16
Aug24-08, 09:49 PM
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Quote Quote by lilmissbossy View Post
i guess when you think about it, it makes sense that the angled string has a greater tension as it is strected further from its initial point....
I'm not sure what you mean by the string being "stretched further": the tension in the rope has nothing to do with the distance between the points where it is fastened. The answer to your problem would be the same whether the ropes were half or twice the lengths that are used in the problem (as long as the mass was still suspended and not resting on the floor).

The reason the tensions for the angled ropes must be greater is that the forces along the ropes are not directed fully in the vertical direction, so some portion of those forces have no influence in countering the weight force acting on the mass. Only the component of the tensions that is directed exactly opposite to the weight force will act to balance it; the component of the tensions that acts perpendicularly (horizontally here) will have no effect against the weight.

Consider this situation. You have some object with a string tied to it that you want to lift. If you pull straight up on it, some amount of force will succeed in raising the object. How hard would you have to pull sideways on the string in order to make the object rise?
tiny-tim
#17
Aug25-08, 03:11 AM
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Quote Quote by lilmissbossy View Post
then i used the cos equation to check if there was any difference and this is what i got

Tsin(θ)+Tcos(θ)-mg
T=sin(θ)+Tcos(θ)-mg
T=sin(53)+cos(53)-(55kg)(9.8ms-1)
T= -537.6

i am not sure if i am rearranging the equations right or if i have the right method>>>
Hi lilmissbossy!

You have the wrong outlook, even before you get to the rearranging …

There's no such thing as a "cos equation" or a "sine equation".

There is a vertical equation and a horizontal equation (and of course, an equation for any other direction … but you only need two ).

The horizontal equation in this case is not Tsin(θ)+Tcos(θ) = mg …

it's Tcos(θ) - Tcos(θ) = 0 …

in other words, the horizontal components of the two tensions are equal and opposite, as expected.
Hint: whenever you do one of these problems, start each equation with the words "Taking components in the … direction".
oh … and when the string is at angle, it's pulling sideways as well as up, so there's more pull … that's all!
lilmissbossy
#18
Aug25-08, 03:19 AM
P: 14
thank you so much i get really confused with all this physics, biology is my degree so i find this all very stressful


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