How to Calculate Sag in Terms of Weight, Pulleys, and Distance?

  • Thread starter leonkhasi
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In summary: Theta is the angle of the hypotenuse. That means that the angle with the horizontal is 90- theta. So, on the left side of the pulley, the wire makes an angle 90-theta with the horizontal and on the right side, it makes an angle of 90+theta. In summary, the conversation is about a pulley with two weights attached to it: W1 on the left and W on the right. The goal is to express the sag d at the center in terms of W, W1, and the horizontal distance L between the left support and the pulley. The equation for d is given as d = 1/2L/√(2w/w1)^2
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leonkhasi
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Homework Statement


A pulley to which is attached w1 rides on a wire that is attached to a support at the left and that passes over a pulley on the right to weight w.the horizontal distance between the left support and the pulley is L.Express the sag d at the center in terms of w,w1 and L.
The answer is d = 1/2L/√(2w/w1)^2-1.Please help. how do you arrive at the equation for the sag d. sag is the sink or downward bulge.
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Homework Equations

The Attempt at a Solution

 
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  • #2
Hello Leon, :welcome:

The answer to your question is: make a drawing and identify all the forces that play a role. Then proceed with working out the free body diagrams of the relevant parts -- that are in a steady state, so they must be in equilibrium. In yout list of relevant equations :rolleyes: you should have relationships that you can apply to that equilibrium state. Post an attempt at solution, even if you get stuck (it is one of the conditions to get help in PF, see the guidelines)
 
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  • #3
i understand that w=1/2w1 is it not?
 
  • #4
No they are not necessarily the same, that's why the two weights are differentiated in the answer. BvU is requesting you to upload your attempt at solving the question, please do that so that we may help you :approve:
 
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  • #5
ouch I'm stuck.:frown:
 
  • #6
leonkhasi said:
i understand that w=1/2w1 is it not?
You can check for yourself by considering the limit case: For d >> L it's somewhat true, but the smaller d is, the less it's true: Why should it be true for d ##\rightarrow## 0 ?
 
  • #7
Good picture. What forces work on W1 ? What forces work on W ? what force is in common ?
 
  • #8
[for d →0 i must say w is maximum.or i must say w1 is zero.
 
  • #9
Snapshot1.jpg
I'm just guessing how to arrive at the solution but i think its nowhere at all close to the solution :wideeyed:
 
  • #10
How about draw some forces? In particular, tension will be useful here.
 
  • #11
leonkhasi said:
i understand that w=1/2w1 is it not?
No, but the moving pulley might be at the middle.
 
  • #12
is d=tanθ⋅L/2? how do i determine how θ changes when i do not know basedwhich of the weights are heavier?
 
  • #13
You are asked to say something about d at equilibrium.
W is heavier than W1
##1\over 2##W is not right for the tension in the wire.
 
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  • #14
leonkhasi said:
is d=tanθ⋅L/2?
Yes,

Both weights are in equilibrium. The forces the string exerts on them is the tension in the string. Draw the force vectors both at w and w1. They need to add up to zero.
 
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  • #15
do we have to show the equilibrium of the forces at the point at the center ?
 
  • #16
Hello there
is w1.g=2t1cosθ? and wat about should t2=w=t1? where t1 and t2 are tensions and should t1=t2 for tension tobe same in the system since force is transmitted on the same rope.
 
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  • #17
leonkhasi said:
Hello there
is w1.g=2t1cosθ? and wat about should t2=w=t1? where t1 and t2 are tensions and should t1=t2 for tension tobe same in the system since force is transmitted on the same rope.
Yes, t1=t2=w. If theta is the angle the string makes with the horizontal then w1.g=2t1cosθ is not true.
 
  • #18
i assume θ is the angle between the sag d and the hypotenuse..
 
  • #19
leonkhasi said:
i assume θ is the angle between the sag d and the hypotenuse..
Then your equation d=tanθ⋅L/2 is not true . (Post #12)
 
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  • #20
if i assume the angle θ like i mention then d=L/2/tan θ yes..but i guess i won't arrive at the solution so i must take θ as angle between hypotenuse and horizontal and d=tanθ.L/2. am i right?
 
  • #21
leonkhasi said:
if i assume the angle θ like i mention then d=L/2/tan θ yes..but i guess i won't arrive at the solution so i must take θ as angle between hypotenuse and horizontal and d=tanθ.L/2. am i right?
You can use any angle, but do it consequently.
 

What is the concept of two weights and two pulleys?

The concept of two weights and two pulleys involves using two weights connected by a string or rope that passes over two pulleys. The weights are used to create a mechanical advantage, allowing for easier movement of heavy objects.

How does the mechanical advantage work in a two weights and two pulleys system?

The mechanical advantage in a two weights and two pulleys system is achieved through the use of a simple machine called a pulley. By distributing the weight of an object between two pulleys, the force required to lift the object is reduced, making it easier to lift. The mechanical advantage of this system is equal to the number of ropes supporting the weight.

What are the different types of two weights and two pulleys systems?

There are two main types of two weights and two pulleys systems: the fixed pulley system and the movable pulley system. In a fixed pulley system, one end of the rope is attached to a fixed point, while the other end is attached to the weight. In a movable pulley system, both ends of the rope are attached to the weights, and the pulley is able to move along the rope.

What are the advantages of using a two weights and two pulleys system?

One of the main advantages of using a two weights and two pulleys system is that it allows for the distribution of weight, making it easier to lift heavy objects. This can also reduce the amount of force needed to lift an object, making the task less physically demanding. Additionally, these systems are relatively simple and inexpensive to set up and use.

Are there any limitations to using a two weights and two pulleys system?

The main limitation of using a two weights and two pulleys system is that it is only effective for lifting objects in a vertical direction. It also requires a certain amount of space for the pulleys to move freely, making it less practical for use in tight or confined spaces. Additionally, the weight of the pulleys themselves can add to the overall weight that needs to be lifted.

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