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Crane pivot force calculation help

  1. Jun 21, 2015 #1
    Here's a diagram of what I'm talking about: http://i.imgur.com/irgHuqq.jpg

    Basically, the only variable altered is the value for x. I tried googling for hours now, have read everything in my book relating to this topic, I still have absolutely no clue how to answer this. I tried studying sites like these: http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html and these http://www.wikihow.com/Calculate-Tension-in-Physics, I still am no closer to knowing how to figure out this specific problem.

    Can you please give me the answers for those 3 values and show me how you did it?

    Thanks
     

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  3. Jun 21, 2015 #2

    haruspex

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    We don't give answers. Please post some attempt, or list any standard equations or principles you know which you think may be relevant and say what is blocking you from applying them.
     
  4. Jun 21, 2015 #3

    ehild

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    I think there was written some text together with the figure, you should have copied it, and also using the template you got when starting the thread.
    If you look at the picture you can see that the force read by the detector is the question, at three different positions of the load. And the detector reads the tension in the string, that is you have to figure out. And the ruler is pivoted about the point A. Was there no mass given for the ruler?
    Draw the forces acting on the ruler and find their torques about point A. The sum of the torques must be zero, as the set-up is steady. You can determine the tension from that condition.


    cranetorque.JPG
     
  5. Jun 21, 2015 #4
    Okay, so here's my attempt at answering it. I have no idea whether this method is correct or not.
    http://i.imgur.com/xvZaGWl.jpg
    http://i.imgur.com/RnKmNL7.jpg
    Btw, the ruler weighs 166g and the added weighted mass weighs 203g. I know how to work out the force for a simpler diagram. This diagram is just too complex for me.

    Note: this is the first year that I've ever done physics. I jumped into 12th grade without any background knowledge at all in physics. But, I'm a quick learner.
     
    Last edited: Jun 21, 2015
  6. Jun 21, 2015 #5

    haruspex

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    ##T=rF\sin(\theta)## is the right formula, but you need to understand what each of those variables means.
    Suppose a force F is applied at point P and there is an axis of interest at point Q. We want the torque T of the force about the point Q.
    r is the distance PQ, ##\theta## is the angle between the direction of the force and the line PQ.
    So, you have an axis at the lower left of the diagram. You need to consider one force at a time. Start with the weight of the ruler. Through what point does that act? How far is that from the axis? What is its magnitude? What is its direction? What is the angle between its direction and the line joining it to the axis?

    Oh, and please don't post working as images. That should be reserved for diagrams and textbook extracts.
     
    Last edited: Jun 21, 2015
  7. Jun 21, 2015 #6
    "Note: this is the first year that I've ever done physics. I jumped into 12th grade without any background knowledge at all in physics." As in, please assume that I do not understand the jargon and references that you are using.

    "Where does this act?" You mean the added weight? If so, it acts at xcm away from the axis. Let's say x=20cm. Do you mean the force of the string tugging on the ruler? That acts at 100cm away from the axis.
    "How far is that from the axis". Same answer as above^ why repeat the same question?
    "What is it's magnitude?" You mean, what is the magnitude of the weight of the ruler? That's 166g, which is 1.6N. Or the magnitude of the added weight? The magnitude of the added weight is 203g, which is 2N. Assuming 'magnitude' in this context means the force..
    "What is its direction?" The direction of what? The ruler? It's direction is 55 degrees from the retort stand. The direction of the force of the added weight? That's parallel to the retort stand.
    "What is the angle between its direction and the line joining it to the axis?" The 'direction' of what? What 'line'? The line is the ruler, right? Well, the ruler is at 55 degrees to the retort stand.
    "Please post some attempt" How do you attempt a question which you've never done in your life and have absolutely no clue how to approach it? It's purely a guessing game, for me, at this point, walking blind.
    "or list any standard equations or principles you know" The only equation I know is the T=rFsin(angle) equation, which I saw for the first time yesterday. And have no idea what to plug into the equation. (in this specific circumstance, only. In much simpler problems, I might be able to figure them out.)

    Please do not assume that I understand perfectly everything you are referring to. I need a diagram with arrows pointing to what you're talking about. Since this is the first time in my life I've ever dealt with these types of questions.
     
    Last edited: Jun 21, 2015
  8. Jun 21, 2015 #7

    haruspex

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    No, I said let's start with just the weight of the ruler. Where does that act? Strictly speaking, of course, it acts at every particle of the ruler, but assuming it to be uniform we can treat it all as acting at what point on the ruler?
    Because I'm trying to break it down into the individual steps. The first step is to say at what point of an object gravity can be taken as acting, the second step is to say where that is in relation to the axis.
    Its mass is 166g, its weight is 1.63N - don't confuse mass with weight, and don't reduce precision to less than that given.
    Anyway, it's better to keep everything symbolic for now, so let's just say its weight is Mrg.
    Well, I meant the ruler's weight, but same answer.
    Right. So we have a force F magnitude Mrg, a distance from axis which we've yet to pin down, and the angle between them. Once you have the distance you can apply ##T=rF\sin(\theta)## to find the torque due to the ruler's weight.
    After that, we can move on to the hanging mass.
     
  9. Jun 22, 2015 #8
    Ok, so:
    F=1.63N
    r=0.2m
    sin(θ)=sin(55)

    Therefore:
    T=(r)(F)sin(θ)
    T=(0.20)(1.63)sin(55)
    T=0.267N

    That is, the force that the ruler, alone, exerts on the string, right? Now for the main part that confuses me, how to take into account an added mass somewhere along the ruler? this is the crux of the problem, for me.

    Edit: Oh wait, no, I made a mistake. I wrote r=0.20m, when we were only dealing with the ruler. Should r=1.00m? So: T=(1.00)(1.63)sin(55)=1.34N
     
  10. Jun 22, 2015 #9

    haruspex

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    No, it's a torque, not a force. It's important to keep these concepts distinct.
    No, not 1m. Are you familiar with the concept of mass centre?
     
  11. Jun 22, 2015 #10
    As in, assuming the entire weight of a symmetrical object to be focused in the centre of said object? Not sure, though.

    Can't you please just give me the formula for getting the answer? As in, a formula where I can just insert the x variable and derive the force read by the detector. I've been at this for 3 days now, it's due tomorrow. I don't want to spend many more hours on this 1 problem. Writing out the formula for me will take you literally under a minute. I will memorize it and learn that way, I don't need a discussion on every single aspect of the question. Reading 'worked solutions' is how I learn best. I'm an independent learner who doesn't rely on any teachers or tutors, I just need a little nudge sometimes (rarely) when I hit brick walls, like right now. I will study the formula on my own once I get it.
     
    Last edited: Jun 22, 2015
  12. Jun 22, 2015 #11

    haruspex

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    yes, treat the weight of the ruler as acting through its centre, so r is only .5m.
    You now have all you need. You can calculate the torque from the ruler, and apply the same methods to get the torque from the hanging mass and from the cable. Combine them into one equation, and out will come your answers.
     
  13. Jun 22, 2015 #12
    huh? How do I 'combine' two equations? As in, F1+F2? So, the torque of the weight+the torque of the ruler?
     
  14. Jun 22, 2015 #13
    Are the following calculations correct?:
    Ruler alone:
    T=(0.500)(1.63)sin(55)=0.668N

    Added 203g at 20cm from axes (without ruler):
    T=(0.200)(1.99)sin(55)=0.326N

    Added 203g at 20cm from axes (with ruler):
    0.668+0.326=0.994N


    Added 203g at 40cm from axes (without ruler):
    T=(0.400)(1.99)sin(55)=0.652N

    Added 203g at 40cm from axes (with ruler):
    0.668+0.652=1.32N


    Added 203g at 60cm from axes (without ruler):
    T=(0.600)(1.99)sin(55)=0.978N

    Added 203g at 60cm from axes (with ruler):
    0.668+0.978=1.65N

    Correct?
     
  15. Jun 22, 2015 #14

    haruspex

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    Right, except the units will be Nm, Newton-metres. These are torques, not forces.
    Finally, if the tension is T, what torque does that provide? This must equal the combined torque of ruler plus mass.
     
  16. Jun 22, 2015 #15
    I don't understand the question. I know that tension is a force, F=m(9.8). But didn't I already find the torque in each one of those measurements? Wait, but I thought torque is T... So, "what torque does torque provide?"? How can both tension and torque both be T?

    Soo, equal to 0.994, 1.32, etc. for each measurement? Didn't I already 'combine' the torque for the ruler and the mass?
     
    Last edited: Jun 22, 2015
  17. Jun 22, 2015 #16

    haruspex

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    Sorry, I should have used a different symbol for the tension.
    (Commonly, T is used for tension and ##\tau## for torque, but you used T for torque so let's stick with that.)
    So let the tension be X, whatever. Write out the expression for the torque that produces at the axis.
    Then you can equate that to one of the total torques you calculated.
     
  18. Jun 22, 2015 #17

    billy_joule

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    (I hope this isn't giving too much away..)
    You've found the torques due to the weight & ruler. We know the sum of the torques must add to zero so we can say:

    ∑T = 0 = Tweight + Truler +T*string
    (* T for torque, not tension)

    If we choose clockwise to be positive, Tstring is negative as it is anticlockwise (We could also choose anticlockwise to be positive and reach the same answer)
    so we can rearrange to give:
    Tweight + Truler =Tstring


    You've calculated Tweight and Truler so you already know the left hand side. And we know T = Frsinθ so:

    Tweight + Truler =Fstringrstringsinθstring


    The only unknown is Fstring. So we can rearrange and solve for Fstring, then plug in the numbers three times for the three different Tweight values.
     
  19. Jun 22, 2015 #18
    Would it be Tweight + Truler =Fstring(1.00)sin(43)?
    If so, wouldn't that just give the torque of the string? How then, will I convert that to force in newtons?

    Edit: ignore that. I hallucinated and thought the F was a T (lack of sleep). So, that's it? That'll give me the tension in the string? Nothing more left to do? And the values are 1.00 and 43...right?
     
  20. Jun 22, 2015 #19

    billy_joule

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    What are the units of Fstring?
     
  21. Jun 22, 2015 #20
    Torque/meter=Tm-1 I think..

    Edit: I mean, newtons/meter=Nm-1 I think.. But I know that tension in general in a string is simply newtons=N
     
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