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Simple Pulley Ratios

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data

    The 14.0 kg block in Figure P8.34 is held in place by the massless rope passing over two massless, frictionless pulleys. Find the following values.

    http://img26.imageshack.us/img26/4025/p834.gif [Broken]

    Find tension T1, tension T2, tension T3, tension T4, tension T5 and the magnitude of force vector F.

    2. Relevant equations

    T=ma


    3. The attempt at a solution


    I have figured out every single tension except for T_4 and it has me stumped.

    Here are all the other answers
    t_1=ma
    t_2=((ma)/2) = t_3 = t_5 = "Magnitude of Force F"

    t_4 has me stumped however. I thought it may be ((ma)/3) since it was being held up by t_2, t_4 and t_5 but that was wrong. Could anyone either explain it to me or link me to a decent explanation?

    edit: i realize i included no numbers for my correct answers.

    "m" times "a" comes out to 137.2N
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 17, 2009 #2

    tiny-tim

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    Hi myxomatosii! :smile:

    (try usng the X2 tag just above the Reply box :wink:)

    Your reasoning for T4 is correct (i assume you meant "t_2, t_3 and t_5") …

    but however did you get ma/3 out of it? :confused:
     
  4. Mar 17, 2009 #3
    Well I thought that since there were three ropes (I counted t4 and t3 as the same rope) it made sense.

    Why wouldn't they be? I mean, well yea, why wouldn't they be? Doesn't rope 4 just run through the metal pipe to rope 3?
     
  5. Mar 17, 2009 #4

    tiny-tim

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    Nooooooo :cry:

    it's not a pipe, it's a clampy thing …

    ropes T3 and T4 are completely separate!
     
  6. Mar 17, 2009 #5
    ohhhh, a clampy thingy.

    give me a minute and i may have it, we'll see.
     
  7. Mar 17, 2009 #6
    ok well in that case.. its supporting the whole system so it should be 137.2N but i've already tried that..

    those other 2 ropes are draped over it tho...ummm that might change things. let me think.
     
  8. Mar 17, 2009 #7
    http://img26.imageshack.us/img26/4025/p834.gif [Broken]

    ok here is my thought.

    let ma=f

    so rope 3 gets f/2 because it shares the force with rope 2

    rope 5 gets f/2 as well for the same reason

    so if rope 4 shares with rope 3 and rope 5 then maybe it would only be supporting f/4 because ropes 3 and 5 only had f/2 and rope 4 is only supporting half of that? or a third of that?
     
    Last edited by a moderator: May 4, 2017
  9. Mar 17, 2009 #8

    tiny-tim

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    do you mean mg = F?

    why should it?

    T2 = T3 = T5 = T, say, because they're the same rope and there's no friction …

    so F = how many T?

    and mg = how many T? :smile:
     
  10. Mar 18, 2009 #9
    I just got off work and am going to bed, I'll look at this more tomorrow before class.

    Maybe someone will ask to do this one, I haven't been to class since the last test in Febuary.. since I've had to borrow cars to get there but I'm still managing a mid 90s grade, anyway, off subject. I'll respond to this when my brain is functioning properly.
     
  11. Mar 18, 2009 #10
    http://img26.imageshack.us/img26/4025/p834.gif [Broken]

    Are you saying that the force of t4 is actually greater than force caused by the weight of the object since it has to support both forces?

    If so what would that be, something like apparent weight.. ?

    Well I looked at it and it doesn't look like its moving so I think my wapp idea is wrong. If they are massless pullies and ropes then the tension on t4 cannot be more than 137.2N since that is derived from the total weight of the object.. is that right?

    I think you are saying there are two systems at work.

    t1 holding up the mass being just 137.2N and

    t2, t3 and t5, but those all have 68.6N of tension in that one rope.. so.. I see that all three of these ropes have points of contact with rope 4 indirectly. Umm.. so rope 4 supports all of this.. I already knew that but I just am not sure if the tensions in other ropes would make the tension in rope 4 higher than the tensions in rope 1, which doesn't make sense to me..

    From what you wrote, I think you want me to say/realize F=3T and mg=1T?

    But what does that matter? t4 isn't in contact with any of them, all it does is supports everything.

    Okay so if I'm standing in a warehouse and there is a pully supported by a sewing thread and I say hey thats weird that is even holding and I go stand under it and pull even slightly, its going to break/fall... well I guess not if there was no friction in the system it might not do anything... and we're not talking about acceleration here anyway...

    I am trying to figure out verbally whether t4 can be greater than t1, maybe t4 is just both tensions added together? I like things to make sense in a realistic environment though.
     
    Last edited by a moderator: May 4, 2017
  12. Mar 18, 2009 #11

    tiny-tim

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    uhh? the force is the same as the tension in the rope: F = T.

    I think you'd better start again …

    this time, draw another mass on the "F" end of the rope, with mass M = F/g …

    that will be exactly the same system …

    so you have two masses, m and M …

    now can you see what all the forces will be? :smile:
     
  13. Mar 18, 2009 #12
    So I thought from all of this that maybe since the force is providing an equal push to make the object sit still maybe the newtons are 2(mg) but it wasn't..

    So I've used up 4 out of 5 of my tries and the next answer I come up with is my last.

    So I am thinking maybe the t4 is..

    1: the tension needed to hold up the mass. 137.2N

    plus

    2: the tension needed to hold up t2 (half of tension needed to hold mass) which would make it 205.8N

    Thats my current train of thought but I'm not going to submit it quite yet, I'm going to go check in on my other post about Tension on an Incline because that has me completely stumped and I'm not even halfway through this assignment... =(
     
  14. Mar 19, 2009 #13

    tiny-tim

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    in that case, before you do anything else, tell us what you think the forces would be with masses of m and M as in my previous post :smile:
     
  15. Mar 19, 2009 #14
    My idea now is that mass m (original mass) and mass M (mass representing the balancing force) add up to equal t4.

    The entire system is supported by t4 and that is m, the original mass plus M, which is strung in such a way that the force it has to support is not half.. umm how do I want to say this, its force is less but seems greater because its using 2 tensions..

    Yea, so I had enough confidence in this to submit it earlier and it did work. I'm still not 100% on it but considering I was super lost I consider this a pretty good gain..
     
  16. Mar 19, 2009 #15

    tiny-tim

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    Let's just go over this again while you're still thinking about it …

    there are three principles involved here:

    i] the tensions in T2 T3 and T5 are all the same, = T say (because there's no friction)

    ii] m is supported by 2T, and M is supported by T

    iii] T4 supports m+M

    so mg = 2T, Mg = T, T4 = (m+M)g = 3T


    If you use F instead of Mg, it's the same result (with F = T instead of MG = T) … the ony advantage of using M is that it's easier to visualise a hanging mass than a "magic force from nowhere" :wink:

    Another principle you could use is the work done by the tensions, resulting in "gearing" …

    if mass M were to go down one metre (I know it doesn't, but suppose), mass m would rise only half a metre (think about it!) … since the work done must balance, double the distance means half the force, and so the force on M must be half that on m. :smile:
     
  17. Mar 20, 2009 #16
    This helps a lot.. seriously :smile:

    I'll have to work a few by myself to practice but I'm not as worried.
     
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