Two ropes hold a 200N weight. Find the tension in each rope.

In summary, the conversation is about finding the tension in two ropes holding a 200N weight using a vector diagram. The equations given by the teacher are used, but the student realizes that there may be something wrong with them. They try to solve the problem using the equations, but their answer does not match with their understanding of physics. They are advised to use their understanding of forces and vectors to construct the correct equations.
  • #1
Chrishtofu
1
0

Homework Statement


The question is;
Two ropes hold a 200N weight as shown here; http://i.imgur.com/BgeoIVq.png

Find the tension in each rope using a vector diagram.
The forces are in equilibrium in the diagram btw.

Homework Equations



Well they gave us 200N so no need for F = MA
Rx = R * cos(theta)
Ry = R * sin(theta)
The above are resolution to component equations.

Other equations the teacher gave us were:
tan(theta) = Ry/Rx
R = SQRT[(Rx)^2 + (Ry)^2]
The above are component to resolution equations.

The Attempt at a Solution



So I fished around on this site and there were some solutions to this question, but I never quite got it, partially because (theta) for R2 is 90.

R1y = R1 * sin(53)
R2y = R2 * sin(90)
Downwards force = 200N

Therefore; because of equilibrium;
R1sin(53) + R2sin(90) = 200N [First Equation ( Y Axis )]

---
R1x = R1 * cos(53)
R2x = R2 * cos(90)

Therefore; because of equilibrium;
R1cos(53) = R2(cos90) [Second Equation (X Axis)]

---
Okay, the problem here is cos(90) = 0,
So does that mean that R1 = 0? because 0* R2 = 0...

So the answer I got is
R1 = 0
R2 = 200

Im pretty sure that I am wrong, but then again i only just started physics in year 11 (australia)
 
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  • #2
You have applied the equations first, and then checked using phsics, and observed that these do no match. It follows there is something wrong with the equations or how you have applied them.

R1y = R1 * sin(53)
R2y = R2 * sin(90)
Downwards force = 200N
... so you have the wrong angles.

You are better just to use the physics first, with your understanding of forces and vectors, to construct the correct equations first go.
Less memorizing to do too. So throw away the equations you were given:

So look at the picture - can you see that the horizontal components of ##\vec R_1## is just going to be ##R_2##?
What is the role of rope 2 anyway? If you cut it, what would happen?

For R1, draw the triangle that shows the horizontal and vertical components adding up to make the vector ##\vec R_2##.
Draw in the 53deg angle you are given.
Now use trigonometry to work out the relationships.

(It may help you to draw in the x and y axes too.)

The other way is to draw the vectors for R1+R2 ... you know what they have to add up to, the weight 200N straight up. This gives you a triangle. You know how to deal with triangles.
 
Last edited:

1. How do you calculate the tension in each rope?

The tension in each rope can be calculated using the formula T = F/2, where T is the tension and F is the weight being held.

2. What is the unit of measurement for tension?

The unit of measurement for tension is Newtons (N), which is the same as the unit of measurement for force.

3. Does the length of the ropes affect the tension?

Yes, the length of the ropes does affect the tension. The longer the ropes, the more they will stretch and the lower the tension will be. Similarly, shorter ropes will have less stretch and a higher tension.

4. Can the tension in one rope be greater than the weight being held?

No, the tension in each rope can never be greater than the weight being held. The tension in each rope will always be equal to half of the weight being held.

5. How does the angle of the ropes affect the tension?

The angle of the ropes can affect the tension in each rope. The greater the angle, the more the weight is distributed between the two ropes, resulting in a lower tension in each rope. Conversely, a smaller angle will result in a higher tension in each rope.

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