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Torque: grindstone and axe

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A 50.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 160 N (Fig. P10.57). The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 N*m between the axle of the stone and its bearings. How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to in 9.00 s?

    2. Relevant equations
    Net torque = Iα

    3. The attempt at a solution
    α = 1.396 rad/s^2
    Torque from axle + Torque from friction + Torque from crank handle = Iα
    6.50 + (0.6)(.26)(160) - thandle = (1/2)(50)(.26)2(1.396)
    thandle = 29.1 N*m = rF
    F = 29.1/.26 = 111.926 N

    The answer is 67.6 N though. What am I doing wrong?
     
    Last edited: Feb 28, 2015
  2. jcsd
  3. Feb 28, 2015 #2

    gneill

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    Staff: Mentor

    You want to rethink how the various torques add up to yield the necessary total torque. Which ones add and which ones subtract from the net?
     
  4. Feb 28, 2015 #3
    Aren't the signs on the individual torques correct? The two signs of torques from friction will be the opposite from the torque from the handle because friction opposes motion, which is what I have (handle is negative because it rotates it CW, while the torque from friction wants to rotate it CCW)
     
  5. Feb 28, 2015 #4

    gneill

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    The torque from the handle must overcome both the axle friction and the axe friction and leave enough left over to yield the required total.
     
  6. Feb 28, 2015 #5
    I don't understand how this helps with the problem. I know that the net torque has to be greater than zero because the torque from the handle must over the two frictional torques. Isn't that what the equation I used means?
     
  7. Feb 28, 2015 #6
    I can't find net torque (the torque left over to create the acceleration), unless I find torque from the handle first, which is what I'm solving for.
     
  8. Feb 28, 2015 #7
    Sorry, I can find net torque from Iα, which I did. And I'm solving for the torque from the handle. My equation is saying that there is more torque going CW because otherwise it would be in equilibrium.
     
  9. Feb 28, 2015 #8

    gneill

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    Staff: Mentor

    You wrote:
    which translates to:
    $$\tau_{axle} + \tau_{friction} - \tau_{handle} = \tau_{required}$$
    Surely the handle torque must exceed the frictional torques, otherwise the grindstone would rotate backwards. Further, the frictional torques should subtract from the handle torque, with what remains giving the net required torque.
     
  10. Feb 28, 2015 #9
    The frictional force causes it to rotate in the counter-clockwise direction though, which means it should be positive right? Why would you subtract it then?
     
  11. Feb 28, 2015 #10
    Also, isn't τrequired the same as net torque?
     
  12. Feb 28, 2015 #11

    gneill

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    I wrote "required" in order to highlight the fact that it is the amount of torque needed to achieve the desired acceleration. Perhaps I should have left it as "net". Sorry for any confusion that may have caused.
     
  13. Feb 28, 2015 #12
    τnet = Iα
    τnet = (1/2)(50)(.26^2)(1.396) = 2.35924 N*m
    τnet = τaxle, friction + τaxle, friction - τhandle
    τhandle = τaxle, friction + τaxle, friction - τnet
    τhandle = 6.50 (given) + μnR (torque due to friction from the axe) - 2.35924 (found above)
    τhandle = 6.50 + (0.6)(160)(0.26) - 2.35924
    τhandle = 29.10 Nm
    τhandle = perpendicular force * distance from the axis = 29.10Nm
    FR = 29.10
    F = 29.10/0.26 = 111.80 N

    What am I doing wrong? I still haven't gotten the answer.... My work shows that I am solving for the torque of the handle. I've done this step by step many times and always end up with the same answer.
     
  14. Feb 28, 2015 #13

    gneill

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    Staff: Mentor

    You can choose your CW/CCW positive/negative convention either way, but in the end you have one torque that is providing impetus in the desired direction and two that are opposed. Thus:

    $$\tau_{handle} - \tau_{axle} - \tau_{axe} = \tau_{net}$$

    Note that the net torque has the same sign as the handle torque.

    Adjust the signs of all terms by multiplying through by -1 if you wish.
     
  15. Feb 28, 2015 #14
    This doesn't change my results though. It just changes the sign. I'll still end up with a magnitude of 110.80, which is incorrect.
     
  16. Feb 28, 2015 #15
    I'm trying to figure out why the magnitude of the the force is different, not the direction.
     
  17. Feb 28, 2015 #16

    TSny

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    Is there some information left out here?

    Does the stone have a CW or CCW angular acceleration α? Did you consider the sign of α?

    Also, did you use the correct radius for the crank in your calculation?
     
  18. Feb 28, 2015 #17
    Oh. For some reason it didn't show up when I copied and pasted it. The final angular velocity is 120 rev/min = 4pi rad/s. I still included the acceleration above though.
     
  19. Feb 28, 2015 #18

    gneill

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    Let's take a look using the magnitudes of the known torques. You have the following known numbers:

    Taxle = 6.50 Nm
    Taxe = 24.96 Nm
    Tnet = 2.359 Nm

    and one unknown, Thand

    You need Thand to exceed Taxle + Taxe by the amount Tnet. That is,

    Tnet = Thand - (Taxle + Taxe)

    Thand = Tnet + Taxle + Taxe
     
  20. Feb 28, 2015 #19

    TSny

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    Is the angular acceleration positive or negative?
     
  21. Feb 28, 2015 #20
    Negative. That would only increase the torque of the handle above 111 though.
     
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