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

In summary, the problem involves a 5m rope attached to two hooks 4m apart on a horizontal ceiling. A 10kg mass is hung from the rope, creating segments of 3m and 2m. The goal is to find the tension in the rope. The sine and cosine rules can be used, and a triangle of forces can be drawn to find the angles and tensions. Another method involves finding the vertical and horizontal components of the tensions from a diagram. Both methods result in tensions of 88.5N and 69.5N.
  • #1
verb tense
19
1

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.
 
Physics news on Phys.org
  • #2


Let us imagine we are hanging the mass to the rope.Right?
 
  • #3


Yes.
 
  • #4


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


We are trying to do it together!
 
  • #6


A triangle is formed.
 
  • #7


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


Ok let me try it out.
 
Last edited:
  • #9


I got the angles to be 104,47 and 29 am i right and what next?
 
  • Like
Likes Nerve
  • #10


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


Here is a diagram.
 

Attachments

  • ddiag.doc
    24.5 KB · Views: 386
  • #12


i got the tension to be 73.5N and 49.1N
 
Last edited:
  • #13


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


Here it is.
 

Attachments

  • ji,.jpg
    ji,.jpg
    15.9 KB · Views: 705
  • #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


Need help don't know what to do.:confused:
 
  • #17


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: 331
  • #18


Ok is this it?
 

Attachments

  • ji,.jpg
    ji,.jpg
    3.4 KB · Views: 553
  • #19


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


For example tension T1 will make angle 43.4deg with the vertical.
 
  • #21


I think i got it
 

Attachments

  • ji,.jpg
    ji,.jpg
    3.4 KB · Views: 579
  • #22


That is correct!

Remember. There is a diagram for the distances and a diagram for the forces.
 
  • #23


Okay , so this is what i got for tension 88.5N and 69.5N
 

Attachments

  • ans.jpg
    ans.jpg
    17.8 KB · Views: 604
  • #24


Correct.

There is another way how to do it.
 
  • #25


Ok, Thanks.

Please can you show me the other way to do it?
 
  • #26


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: 263
  • #27


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

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

1. How much tension is there on the rope?

The tension on the rope can be calculated using the formula T=mg, where T is the tension, m is the mass of the rope, and g is the gravitational acceleration. In this scenario, the tension on the rope is equal to the weight of the rope, which can be found by multiplying the mass of the rope by the gravitational acceleration (9.8 m/s^2).

2. What is the maximum weight the rope can hold before breaking?

The maximum weight the rope can hold before breaking depends on the strength and material of the rope. To determine the exact weight, the breaking strength of the rope would need to be known. This can be found by consulting the manufacturer or conducting a strength test.

3. How much force is being exerted on each hook?

The force being exerted on each hook can be calculated using the formula F=T/d, where F is the force, T is the tension on the rope, and d is the distance between the two hooks. In this scenario, the force on each hook would be equal to half of the tension on the rope.

4. Can the distance between the hooks be increased without affecting the tension on the rope?

Yes, the distance between the hooks can be increased without affecting the tension on the rope. As long as the weight of the rope and the gravitational acceleration remain constant, the tension on the rope will not change.

5. What is the angle between the rope and the ceiling?

The angle between the rope and the ceiling can be calculated using the formula sinθ = h/L, where θ is the angle, h is the height of the ceiling, and L is the length of the rope. In this scenario, the angle between the rope and the ceiling would be equal to the inverse sine of 4/5, which is approximately 53.1 degrees.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
858
  • Introductory Physics Homework Help
Replies
22
Views
3K
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
9K
  • Introductory Physics Homework Help
Replies
7
Views
8K
Back
Top