A rope of 5m is fastened to two hooks 4m apart on a horizontal ceiling

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  • #1
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Homework Statement


A rope of 5m is fastened to two hooks 4m apart on a horizontal ceiling. To the rope is attached 10kg mass so that the segment of the rope are 3m and 2m.Compute the tension in the string


Homework Equations



Sine rule

The Attempt at a Solution


I get stuck at the part that says '10kg mass so that the segment of the rope are 3m and 2m'. Please i just need the right diagram i know what to do.
 

Answers and Replies

  • #2
998
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Let us imagine we are hanging the mass to the rope.Right?
 
  • #3
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Yes.
 
  • #4
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So if one fastened a 5m rope to two hooks at a horizontal distance 4m apart, the rope will hang in a curve. But when the mass is hung at the specified point, what will happen to the rope? (assume the rope does not extend)
 
  • #5
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We are trying to do it together!
 
  • #6
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A triangle is formed.
 
  • #7
998
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So one knows the all the sides of this triangle. But I do not think that he sine rule can be used because there are to many unknowns for this rule to be used. Hence one can use the cosine rule to find the angles.
 
  • #8
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Ok lemme try it out.
 
Last edited:
  • #9
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I got the angles to be 104,47 and 29 am i right and what next?
 
  • #10
998
15


Correct but it is better to work with 104.5, 46.6 and 29.0 deg.

Now you can either use the sine rule or use components of the tensions.
 
  • #11
998
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Here is a diagram.
 

Attachments

  • ddiag.doc
    24.5 KB · Views: 297
  • #12
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i got the tension to be 73.5N and 49.1N
 
Last edited:
  • #13
998
15


if you show the working I can tell you where there is something not correct.
 
  • #14
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Here it is.
 

Attachments

  • ji,.jpg
    ji,.jpg
    15.9 KB · Views: 587
  • #15
998
15


Using the sine rule is ok but it is being used in the wrong triangle. Note that the tiangle of the set up of the apparatus is not necessarily a traingle of forces. In fact the triangle you are using is a traingle of distances not forces.

Try to draw a triagle of the forces showing the directions of the weight, tension1 and tension2. Then put the angles required for this traingle of forces. If you need further help just ask.
 
  • #16
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Need help don't know what to do.:confused:
 
  • #17
998
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Look at the diagram of the forces and fit in its angles using information form the diagram of the apparatus.
 

Attachments

  • DIAGR.doc
    24.5 KB · Views: 257
  • #18
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Ok is this it?
 

Attachments

  • ji,.jpg
    ji,.jpg
    3.4 KB · Views: 468
  • #19
998
15


How can the angle at the bottom of your diagram be 104deg if it must be LESS than 90deg.
Try again to get the angles of the FORCES diagram. The information required for this can be obained from the diagram of the strings attached to the mass.
 
  • #20
998
15


For example tension T1 will make angle 43.4deg with the vertical.
 
  • #21
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I think i got it
 

Attachments

  • ji,.jpg
    ji,.jpg
    3.4 KB · Views: 493
  • #22
998
15


That is correct!

Remember. There is a diagram for the distances and a diagram for the forces.
 
  • #23
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Okay , so this is what i got for tension 88.5N and 69.5N
 

Attachments

  • ans.jpg
    ans.jpg
    17.8 KB · Views: 512
  • #24
998
15


Correct.

There is another way how to do it.
 
  • #25
19
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Ok, Thanks.

Please can you show me the other way to do it?
 
  • #26
998
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One starts from the diagram in the attachment. Then the vertical and horizontal components of the tensions are drawn.Try to do that.
 

Attachments

  • ddiag[1].doc
    24.5 KB · Views: 198
  • #27
998
15


... and then one can compare the two methods.
 

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