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B Pulley Force Vector

  1. Apr 22, 2016 #1

    inv

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    321.jpg


    I'd like to ask the question as above, I think it successfully acted as a lever and redirected the force vector perpendicular to the arm of the machine, but it does've > friction, reducing the force acting on the 10kg Weight, could anyone that knows pls help check for me?


    *Edit, ... I think it successfully acted as a *pulley ...
     
    Last edited: Apr 22, 2016
  2. jcsd
  3. Apr 22, 2016 #2

    BvU

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    Hello **-1,

    Is this the complete text of the exercise ? What is the aim of the depicted contraption ?
     
  4. Apr 22, 2016 #3

    inv

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    What's **-1?

    Yes, complete. There's no aim, it's just an exercise to test lever and force vector knowledge.
     
  5. Apr 22, 2016 #4

    BvU

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    So the exercise is to compare the torque on the machine arm when the block is present with the torque when the block is absent ?
    If so, the thing to do is calculate that torque in both cases !
     
  6. Apr 22, 2016 #5

    inv

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    No, my friction comment was just a trivia, the question is only as in the picture.
     
  7. Apr 22, 2016 #6

    BvU

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    Ah, there is another way to read this: the machine is meant to lift the weight and you want to know if the weight is lifted more easily (with more accleration) when the block is present than with the block absent ?

    In that case convert the torque from the machine (should be the same in both cases) to a tension in the cable for both situations.

    (Can we assume the machine arm orientation is given as drawn ?)
     
  8. Apr 22, 2016 #7

    inv

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    No, the question is simply "Did the block of wood successfully act as a lever directing the force perpendicular to the arm of the machine, as the red arrow indicates?".
     
  9. Apr 22, 2016 #8

    BvU

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    Yes, that's what it said. Is the weight a means to handle the arm or is the arm a means to lift he weight ? Not that it matters (symmetry).
    Why should it ? The wire also exerts an upward force on the block, counteracting the torque from the tangential component. Still need a calculation to show if one of the two changes wins.
     
  10. Apr 22, 2016 #9

    sophiecentaur

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    I cannot see any mention of a pivot for the lever. Is it supposed to be somewhere, out of shot, at the other end of the 'arm'? Its position is important if you want to calculate the torque.
    Presumably that circle on the left is a Pulley and not a "lever"?
    If all of the stuff that has been added to the arm is rigid then the force marked in red is not relevant because the string tension where it joins the rigid arrangement is being shared by internal forces in the block etc.. The torque without the block is the string tension times the perpendicular distance from the pivot that we cannot see. The torque with the block is the string tension times the new perpendicular distance (the angle has changed). The details of the arm and block are not relevant; the arm could be any shape you like - it's just the direction of the string that counts and that affects the length of a perpendicular line, dropped from the pivot to the string.
     
    Last edited: Apr 22, 2016
  11. Apr 22, 2016 #10

    inv

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    Sry typo, there're no levers in this question, all levers typed here were supposed to be pulleys, edited the picture w/ the question.
     
  12. Apr 22, 2016 #11

    A.T.

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    The force from the string at the red dot is 90° to the arm, if the string is 90° to the arm.
     
  13. Apr 22, 2016 #12

    BvU

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    What is the block of wood attached to :smile: ?
     
  14. Apr 22, 2016 #13

    sophiecentaur

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    This "arm" has to be a lever, doesn't it?
     
  15. Apr 22, 2016 #14

    CWatters

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    I think the short answer is no! The block does not achieve the desired effect if it is fixed to the arm.

    As proof, consider what happens if the round pulley is moved up and to the right so that it's at the pivot point of the arm.
     
  16. Apr 23, 2016 #15

    inv

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    Another thinner block of wood screwed to the machine arm, as shown.


    It has to be attached to something at its end, so yes.


    I think there's no difference to the perpendicular force vector to the arm, what do you mean, maybe a picture can explain better?
     
  17. Apr 23, 2016 #16

    Merlin3189

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    I have to agree with Cwatters. I can't see why people are making such a meal of this.
    IF the string and block are smooth (ie. frictionless) then the tension in the string at the red dot is the same as in the rest of the string. It will behave as if it were a pulley on a block attached to the arm.
    BUT there will still be the additional forces on the block being transferred to the arm. So it will not behave like the *pulley on the left.
    Practically it will not be frictionless, so it doesn't make sense to think of it as a pulley.
    IMO, all the unnecessary calculations would yield the same results whether it was a rough block attached to the arm or a smooth pulley (of very small diameter!) attached to the arm.

    Why don't you just get rid of the bits and pieces and tie the string to the arm, or to a nice solid extension if you need to offset the attachment?
     
  18. Apr 23, 2016 #17

    sophiecentaur

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    The question has not been defined fully enough for a proper answer. That is one of the problems here; we have to make assumptions where the information has been omitted.
    If you replace the corner of that block with a tiny pulley, it would make no difference to the force that the deflected string exerts on the 'arm'. The force on the arm is just the tension in the string times the sine of the angle between the string and the arm (the component of the tension at right angles to the arm). Whether you follow my original assumption of the arm being a lever or not, the final answer is the same.
    It is important [NOT!!! Edit] to rely on any 'gut' reaction that one has, due to the way the picture has been drawn and to get to the basics of it - that is, a string pulling at an arm at an angle. I was assuming that the arm is a lever because it's more convenient and because it is very unlikely that the arm would be attached rigidly to 'something' without being intended to turn it. In any case, whether intentional or not, there will be a torque on the arm. Unless the object it is attached to is shaped so that its cm is actually behind the contact point of the string (i.e. somewhere outside the picture, to the right) then there is a torque, by definition.
    It is rather disappointing that this question about such a concrete bit of Mechanics seems to be treated in such an arm waving way. Given the facts, there is only one solution.
     
    Last edited: Apr 23, 2016
  19. Apr 23, 2016 #18

    CWatters

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    OK try this diagram. The rope is in red.

    Modified.jpg

    If you think the force on the arm is still to the left (causing it to rotate clockwise) then try this one. Which way will this one rotate?
    Modified1.jpg
     
    Last edited: Apr 23, 2016
  20. Apr 23, 2016 #19

    sophiecentaur

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    The force on the string is at right angles but what about the force of the block on the arm, too? At the corner where the string bends around the block, there will be another force component that cannot be ignored. You cannot just pick a force that you fancy and use only that force in your conclusions. :smile:
     
  21. Apr 23, 2016 #20

    inv

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    So the force vector on the block's perpendicular to the red dot?


    If I'm correct, if a force is applied on an object's centre of gravity( generally center of mass), it will displace; Elsewhere, it'll turn.

    Are you using the right angle triangle Sin Y°= Opposite/ Hypothenus formula?

    But is the Force Vector at the point perpendicular to the arm?


    I think it'll pull up on the pivot and destroy it.

    To the right. What's the reason for these 2 results?




    Am I right to say you're saying the block's friction is causing the same force vector on it on the arm? If so, if the block's replaced with a pulley, it still causes the same "force vector" on the arm, unless it's separate from that arm, ie anywhere from its pivot, even while still on the machine?
     
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