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First sem. calc physics: Cable tension (vectors and acceleration)

  1. Sep 3, 2012 #1
    So, I'm trying to understand how tension works between a cable.
    No, not homework. A book example. I have the answers.

    a = 0.679 m/s2
    t = 12s

    So, I've seen the free-body diagrams. There is tension, sure, but the tension is in perpendicular directions. The book gives examples of tension in a linear fashion with

    1N <--- (hand)-----cable---------(hand) ---> 1N

    So, since each hand is pulling on the ends of the cable with 1N of force, there is a net force of zero. Ok, sure. But as a vector force, it seems like this is constrained to being within the realm of the x-axis.

    Now, the problem I listed has a cable.. sure.. and the difference is that the vector forces are pulling in different directions. The cable has tension caused by the "pull" of the climber, and it has tension caused by the pull of the rock. Thus, there becomes a netforce of zero.

    Well, I have a couple of questions about this:
    1) Does tension exist between the rock and the climber?
    Or does it exist for the climber and the rock separately? I think like this is an English issue of expressing how the tension exists with these two objects. Is tension shared between the climber and the rock? Because it's in two different directions... With the linear version, it seems like it is obviously shared. It is, right?

    Because if I were to draw free-body diagrams, it seems like the climber has his own tension in opposition to gravity. And the rock has tension in the positive x-axis direction... But the rock's tension is caused by the climber... so it makes me think that these tensions do not belong specifically to each individual object and instead is shared. With the way I see it, there are two free-body diagrams. And then again, because the free-body diagrams have their own tension, it makes me think the tension is not shared between the two objects.

    2) Why does the linking object (rope, in this example) not accelerate? There would be a change in the climber's position over time, and with that, the rope would definitely move throughout time. And movement would lead to velocity, which can be derived to find acceleration... So, I don't see why there couldn't be an acceleration of the rope..

    Maybe there is more to my questions, but I'll see what the responses are for now.

    Attached Files:

    Last edited: Sep 3, 2012
  2. jcsd
  3. Sep 3, 2012 #2

    Doc Al

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    Think of the tension as the magnitude of the force that the rope exerts throughout its length and thus at its two ends. (Assuming the rope is massless, the tension is the same throughout its length.) Since the rope changes direction, so does the force it exerts at each end.

    Yes, to analyze the forces involved, and to solve for the tension, you would create separate free body diagrams for each mass involved. The magnitude of the tension force will be the same in each diagram.

    Why do you think the rope does not accelerate?
  4. Sep 3, 2012 #3
    The rope would <edit> have negligible mass. Also, see attached image.

    The problem says to neglect the rope's mass. I'm assuming that means it has negligible mass.

    So, the magnitude of the tension is shared between the objects, not the tension itself? Because to say tension alone would imply that I'm also including direction? So, it would be best to say that the two objects share the magnitude of the tension?

    Attached Files:

    Last edited: Sep 3, 2012
  5. Sep 3, 2012 #4

    Doc Al

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    Yes, you would neglect the mass of the rope. But it still accelerates. (Not that you care, since it's massless.)
    I would call the 'tension' in the stressed rope the magnitude of the force that it exerts throughout its length. Of course, the vector force exerted at each end of the rope will be directed parallel to the rope at that point.

    It's a bit confusing to call the force exerted by the rope on some object the 'tension', but it's common practice.
  6. Sep 3, 2012 #5
    So, you're saying that a rope exerts tension?

    I'm trying to read more into this.
    Would it not be better to say that objects that pull on the rope cause tension, which is the magnitude of force shared throughout the length of the rope?

    Or to use your terminology, how about

    "tension is the force extracted from the rope by some object."

    or maybe better yet...

    "tension is the force exerted onto the rope by some object, which is the magnitude of force shared throughout the length of the rope?"

    But if there are two objects on the rope, such as the climber and the rock, then wouldn't it be fair to say that "tension is the magnitude of force extracted by some object."
  7. Sep 3, 2012 #6

    Doc Al

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    I'd say that the force exerted by the rope is equal in magnitude to the tension in the rope. But it's OK if you call that force 'tension'.

    Yes, that's fine. As long as you recall Newton's 3rd law: The object cannot pull on the rope without the rope pulling back with an equal and opposite force.

    That seems an awkward phrasing. (And not mine.)

    Still seems a bit awkward.

    It still seems awkward to me. I like to think of the tension as a property of a rope that is put under stress (by pulling its ends). The fibers of the rope are pulled apart (a bit) and the rope is exerting a force (between each segment) to hold itself together. That force is the tension in the rope.
  8. Sep 3, 2012 #7
    Ok, so another thing that is confusing me is how the free-body vector diagrams would look.

    In the first example, the Tx is moving to the right. Ty is moving upward.
    And since Tx = Ty, then I could label them Tx = T1; Ty=T2

    If I were to connect the free-body diagrams, it seems like things look like this:

    T1 -----> <----- T2

    And so the forces are moving toward each other. Which doesn't seem to make too much sense.
    This orientation is different from the other example.

    1N <------- -------> 1N

    At least, it appears different. Image attached.
    In the rock/climber example, the magnitudes are the same, and the values are positive.
    However, I would imagine in the hands on rope example, the values would be negative and positive: they cancel out.

    In the second example, the vector force arrows of 1N are in opposite directions.

    This makes things seem really weird. Because I can conceptualize that there is simultaneous pull from the rock putting stress into the rope and pull from the fall of the climber putting stress into the rope, thus causing tension in the rope.

    But it would seem to me that the vectors should always be in opposition of each other rather than leading toward a nexus, or what would appear to be the origin (the edge of the cliff).

    Why is the orientation of these examples like this?

    Attached Files:

    Last edited: Sep 3, 2012
  9. Sep 3, 2012 #8

    Doc Al

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    I'm not quite sure what you are referring to here. The first of your attached diagrams? Realize that ropes only pull, so I don't know what you are trying to represent with your diagrams.

    Sure. That diagram is showing the forces on some rope segment. They are equal and opposite.

    If you want to analyze the force acting on the segment of rope that is at the edge of the cliff, then you must also include the force from the cliff itself. The net force on that rope segment will be zero. You would have a vertical force from the rope below pulling downward on the segment and a horizontal force from the rope above pulling to the left. But also a force from the cliff edge itself. The vector sum would be zero.

    But you don't need to analyze the rope segments to solve the problem. (But it's good to think about.)
  10. Sep 3, 2012 #9
    Sorry about the editing. But there is an image attached if that helps.

    Perhaps the reason the orientation of the tension arrow in the way they are has to do with the condition of acceleration. If acceleration is not going to occur, then the arrows would be in opposition, thus there would be no movement. But if they are pointing to each other, then it implies there will be movement. That's the most sense I can make out of it.
    Last edited: Sep 3, 2012
  11. Sep 3, 2012 #10

    Doc Al

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    If you are referring to the attachment in your last post, I don't quite understand it. What are T1 and T2 and why are they pointing towards each other? When you draw a force diagram, the forces drawn must be acting on something. What something are you analyzing?
    Let's try again. In your diagram, what does T1 represent? A force on what by what?

    Does T1 represent the force of the rope on the rock? And does T2 represent the force of the rope on the climber?
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