Finding the Tension in a Hanging Rope: A Mathematical Approach

In summary, the tension in the rope is a function of the distance y measured vertically downward from the top of the rope. The force due to gravity at the halfway point is the same as the force at the top of the rope, but the fraction of the mass of the string below the halfway point is smaller.
  • #1
bmarvs04
12
0

Homework Statement



A block of mass M hangs from a uniform rope of length L and mass m. Find an expression for the tension in the rope as a function of the distance y measured vertically downward from the top of the rope.

Homework Equations



For the System:
W = (m + M)g = T

The Attempt at a Solution



I summed all the forces in the Y direction to arrive at the equation above, but I don't know how to make it in terms of 'L', or as they state 'y'.
 
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  • #2
bmarvs04 said:

Homework Statement



A block of mass M hangs from a uniform rope of length L and mass m. Find an expression for the tension in the rope as a function of the distance y measured vertically downward from the top of the rope.

Homework Equations



For the System:
W = (m + M)g = T

The Attempt at a Solution



I summed all the forces in the Y direction to arrive at the equation above, but I don't know how to make it in terms of 'L', or as they state 'y'.

Your expression for W is incorrect, the units of (M+m)g are units of Force (Newtons), not units of work/energy (Joules). Mathematically, what is the definition of work?

More importantly for this problem, what is the force due to gravity on the rope at the halfway point? Is it really the same as the force at the top of the rope? Why or why not? What fraction of the mass of the string is below the halway point? How about below a distance y below the top?
 
Last edited:
  • #3
I'm sorry, I forgot to define my variables. W is the weight of the system, not the work done. And T is the tension of the rope.

Now since the rope is uniform, I know the mass and length have a direct relationship and that the tension will increase when the length (L) is increased.

I still think T = (M + m)g is the right equation if I could just substitute an expression for L in for m. But I am having trouble finding a fraction or what not in order to do so..
 
  • #4
bmarvs04 said:
I'm sorry, I forgot to define my variables. W is the weight of the system, not the work done. And T is the tension of the rope.

Now since the rope is uniform, I know the mass and length have a direct relationship and that the tension will increase when the length (L) is increased.

I still think T = (M + m)g is the right equation if I could just substitute an expression for L in for m. But I am having trouble finding a fraction or what not in order to do so..

That's not really the correct way to do that. In the question, m is defined as the mass of the entire string (A.K.A. a constant)...you could however use a different variable to represent the mass of the string below the point y. For example you could use [tex]T=(M+{\mu}(y))g[/tex] where [tex]{\mu(y)}[/tex] is the mass of the string below the point y...As for finding an expression for [tex]{\mu}(y)[/tex], you know that there will be a linear relationship between [tex]{\mu}(y)[/tex] and [tex]y[/tex] so you can immediately write [tex]{\mu}(y)=Ay+B[/tex]. You then need to determine what the constants A and B are...how much mass is below the point y=0 (i.e. what is [tex]{\mu}(0)[/tex])? How about at the end of the string (i.e. what is [tex]{\mu}(L)[/tex])? Those answers should allow you to easily find A and B.
 
  • #5
Ok I think I understand. Would it make sense for T = (M + ((y/L)*m))*g then?

This makes sense to me because 'y/L' would give you the fraction of rope you are working with, then you could multiply it with 'm' to find the mass of the length you are working with.
 
  • #6
bmarvs04 said:
Ok I think I understand. Would it make sense for T = (M + ((y/L)*m))*g then?

This makes sense to me because 'y/L' would give you the fraction of rope you are working with, then you could multiply it with 'm' to find the mass of the length you are working with.

Not quite, y/L doesn't give you the fraction of the rope you are dealing with, because y is the distance below the top, not the distance above the bottom...so if y is the distance from the top to the point P and the total length is L, what is the distance from the bottom to the point P? This should give you your correct ratio.
 
  • #7
So it would be (L-y)/L? I forgot to re-read the problem to see what 'y' was actually measuring.

Thanks for everything.. I really appreciate it.
 
  • #8
bmarvs04 said:
So it would be (L-y)/L? I forgot to re-read the problem to see what 'y' was actually measuring.

Thanks for everything.. I really appreciate it.

Yup, now you've got it. And you're welcome.:smile:
 

FAQ: Finding the Tension in a Hanging Rope: A Mathematical Approach

What is Rope Tension expression?

Rope Tension expression is a mathematical formula that is used to calculate the force applied to a rope or cable when it is stretched between two points.

Why is Rope Tension expression important?

Rope Tension expression is important because it allows scientists and engineers to determine the amount of force that is being applied to a rope or cable, which is crucial in designing and building structures such as bridges and cranes.

How is Rope Tension expression calculated?

Rope Tension expression is calculated by multiplying the rope's Young's modulus (a measure of its stiffness) by its cross-sectional area and dividing by its length. This gives the amount of force (in newtons) applied to the rope.

What factors can affect Rope Tension expression?

Several factors can affect Rope Tension expression, including the material and thickness of the rope, the temperature, and the amount of stretching that occurs. Other factors such as external forces (such as wind or gravity) and friction can also affect the tension in a rope.

How is Rope Tension expression used in real-world applications?

Rope Tension expression is used in a variety of real-world applications, such as in the construction of bridges, cranes, and other structures that require the use of ropes or cables. It is also used in rock climbing and sailing to ensure the safety and stability of the equipment being used.

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