String tension varying with distance

  • #1

Homework Statement



A system consists of a rope of length l and mass m with two blocks of mass M1 and M2 (with M1 < M2) attached to the end points. The block of mass M1 lies on a plane inclined at an angle alpha, while the second block hangs at the other end of the rope. The rope
can slide without friction, but the friction coeffcient between the first block and the plane is non-zero and equal to u.

The system is subject to gravity and initially half of the rope lies on
the plane.

Compute the minimum value, umin, of the friction coeffcient between
M1 and the plane for which the system remains in equilibrium when
released.
Assume now that u < umin. The system is released and the block of
mass M2 starts to descend. Compute its acceleration when the length
of the portion of rope lying on the plane is x.

Homework Equations



F=ma

The Attempt at a Solution



Right, the only problem I have with this is the fact that the tension varies along the rope.
My plan was to consider each halfs of the rope to be individual strings with individual tensions, and find one tension in terms of the other and the masses, and solve.

But, I'm thinking that if I take one section of the rope, and then a small section after that, and the component of it's weight at that section, then integrate to find the tension at the points pulling the masses, I'll be able to find the forces pulling the masses.

Any advice?
 

Answers and Replies

  • #2
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A system consists of a rope of length l and mass m with two blocks of mass M1 and M2 (with M1 < M2) attached to the end points. The block of mass M1 lies on a plane inclined at an angle alpha, while the second block hangs at the other end of the rope. The rope can slide without friction, but the friction coeffcient between the first block and the plane is non-zero and equal to u.

[...]

Right, the only problem I have with this is the fact that the tension varies along the rope.
My plan was to consider each halfs of the rope to be individual strings with individual tensions, and find one tension in terms of the other and the masses, and solve.

But, I'm thinking that if I take one section of the rope, and then a small section after that, and the component of it's weight at that section, then integrate to find the tension at the points pulling the masses, I'll be able to find the forces pulling the masses.

Any advice?
I think you may be misinterpreting the problem statement. The way that I read it, the only thing experiencing friction is the mass M1 with the inclined surface. The rope itself experiences no friction, therefore the tension on the rope is uniform.
 
  • #3
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Oh, wait. I misread the problem statement. The rope itself has mass (even if it is frictionless). So yes, the tension on the rope isn't uniform after all. Sorry for the confusion! :redface:

Back to your original question then. I don't think its necessary to find the tension at every possible point on the rope via integration. But you do need to account for how much of the rope's mass is on the incline (not for friction purposes, but for gravitational) and how much hangs over the other side.

But you can do this by dividing the rope into two sections, one on the incline and one hanging over the edge. Then, all you need to do is find the tensions in three places: at each end of the two sections, noting that the tension is the same at the top of the incline where the two sections meet.
 
  • #4
For u-min I'm getting
u = [M2 + (.5)(1 - sin@) - M1sin@]/M1cos@

@ = alpha

Is this correct?
 
  • #5
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For u-min I'm getting
u = [M2 + (.5)(1 - sin@) - M1sin@]/M1cos@

@ = alpha

Is this correct?
Oh, so close! :cry:

Everything looks fine, except you forgot to put the mass of the rope m in there.

It shouldn't be too hard for you to figure out what went wrong though. μ is a unit-less constant. And since there the denominator has units of mass, it means each term in the numerator must also have units of mass. Which one of the three terms in the numerator doesn't have units of mass? :wink:
 
  • #6
Oh, so close! :cry:

Everything looks fine, except you forgot to put the mass of the rope m in there.

It shouldn't be too hard for you to figure out what went wrong though. μ is a unit-less constant. And since there the denominator has units of mass, it means each term in the numerator must also have units of mass. Which term in the numerator doesn't have units of mass? :wink:

Sorry, that was a typo, the answer I have is

u = [M2 + m(.5)(1 - sin@) - M1sin@]/M1cos@
 
  • #7
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Sorry, that was a typo, the answer I have is

u = [M2 + m(.5)(1 - sin@) - M1sin@]/M1cos@
Yes, that is exactly what I got. :approve:
 
  • #8
Yes, that is exactly what I got. :approve:

For the acceleration I got

a = [lg(xM1sin@ - M2) + m(l + x)]/[m - M1x - lM2]

How's this?
 
  • #9
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For the acceleration I got
a = [lg(xM1sin@ - M2) + m(l + x)]/[m - M1x - lM2]
I don't think that's right. I came up with something quite different.

You can tell something is wrong with the answer too because it's not dimensionally correct. There are places where you are adding mass with length x mass, etc.

For starters, Newton's second law states that mass x acceleration = sum of all forces. But for this problem, the mass in question is the total mass of the system, M1+M2+m. Since we're looking for the acceleration, we need to divide both sides by the total mass. So at the very least we should expect to find M1+M2+m in the denominator. (And there are still some issues that need fixing in the numerator too.)
 
  • #11
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Any hints? :)
One of the trickier parts is expressing the mass of rope on the incline, and separately the mass of rope hanging over the other side.

The rope has a mass m, and length l, so its linear density is m/l.

The length of rope on the incline is x. So it's total mass is its length times linear density, x * m/l = mx/l.

The length hanging over the other side is l-x. I'll leave it to you to find its mass (same method I used above).

Beyond that, use Newton's second law. Mass x acceleration = sum of all forces.

For this problem, the forces in question include:

  • Gravitational force of block M2.
  • Gravitational force of section of rope hanging off the side.
  • Component of gravitational force parallel to the incline of the section of rope on the incline.
  • Component of gravitational force parallel to the incline of block M1.
  • Frictional force between block M1 and the incline.
 
  • #12
How about

a = [g(M2 + m - M1sin@ - uM1cos@) - (mgx)(1 + sin@)/l] / [M1 + M2 + m]
 
  • #13
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How about

a = [g(M2 + m - M1sin@ - uM1cos@) - (mgx)(1 + sin@)/l] / [M1 + M2 + m]
You got it! :approve:

Good job!

(Btw, you could simplify a little more by some minor factoring [such as a g to the outside of the square brackets in the numerator]. Either way though, your answer looks correct to me.)
 
  • #14
You got it! :approve:

Good job!

(Btw, you could simplify a little more by some minor factoring [such as a g to the outside of the square brackets in the numerator]. Either way though, your answer looks correct to me.)

Thank you very much for your help. :smile:
 
  • #15
Just for the record, my professor says
u-min = [M2-M1sin@]/[M1cos@]
 
  • #16
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Just for the record, my professor says
u-min = [M2-M1sin@]/[M1cos@]
Your professor is incorrect.

Your professor made the incorrect assumption that the gravitational force of the half of the rope on the plane is equal and opposite to the gravitational force of the other half of the rope hanging off the edge. Thus your professor incorrectly assumed that the gravitational force from both sides of the rope cancel, and the mass of the rope does not need to be considered.

But this isn't true. There is also a normal force on the part of the rope which is on the plane, from the plane. This normal force does not exist on the half of the rope hanging off the side.

Ask your professor to consider this scenario. Put a massive rope on the frictionless plane incline with angle α, such that half of the rope is on the plane, and half is hanging off the other end. In other words, the same thing as the original problem except M1 = M2 = 0, and m > 0, and there is no friction involved at all.

Does the rope stay still or does it slide off the edge? Of course it slides off the edge. To further the point, imagine if α is small or equals zero.

The mass of the rope cannot be neglected. The equation for μmin must contain a term which depends on the mass of the rope, m.

However, your instructor's equation would be correct, if the problem statement were written differently, stating that mass M1 and M2 are at the same height (i.e. same elevation). In that case (and only that case) can the mass of the rope can be ignored when calculating μmin. But problems statement, as written here says, "initially half of the rope lies on the plane," which is a different situation altogether.

[Edit: So in summary, if the actual problem statement said, "initially M1 and M2 are at the same height," then your instructor is correct. if the actual problem statement said, "initially half of the rope lies on the plane," your instructor is incorrect. But don't blame your instructor too much. We all make mistakes from time to time.]
 
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