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Calculating the tangent angle

  1. Aug 15, 2016 #1
    1. The problem statement, all variables and given/known data
    SURbVVS.png

    Calculate the force on the cilinder. You need the angle of the tangent and r.

    2. Relevant equations
    Down under here. The solution for the angle is 30°. But why does the formula I used did not work out?
    My solution is 40,9°, why isn't this correct for this excercise?

    3. The attempt at a solution
    52DlRUH.png
     
  2. jcsd
  3. Aug 15, 2016 #2
    I do not understand the problem statement. It does not look like English to me.
     
  4. Aug 15, 2016 #3

    Krylov

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    It is the native language of your 8th president :wink:
    To the OP: It is polite (and, I believe, according to the rules here) to translate your question entirely.
     
  5. Aug 15, 2016 #4

    TSny

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    Why do you have a factor of ##\dot{\theta}## in the denominator?
    upload_2016-8-15_15-36-10.png
     
  6. Aug 15, 2016 #5
    Sorry guys, I didn't want to be inpolite. This is the translation:

    The cylinder C can only move in the slot. The movement is described by r = 0,6*cos(theta) m. The lever OA turns left (counter clockwise) with an angular speed of 2 rad/s and has a angular acceleration of 0,8 rad/s^2 at the moment when theta = 30°. What is the force on the cylinder C at that moment. The cylinder touches only one side of the slot (without friction). The movement is horizontal.

    I was solving this question. And I needed the angle between the lever and the tangentline of the cylinder. The formula is on my paper. I needed this to find the angles for my forces. But the fomula seems not to be correct. What did or do I wrong? Thanks guys! And sorry again, i did not want to be inpolite!
     
  7. Aug 15, 2016 #6
    Just derivate of the term above. Chain rule for derivates...?
     
  8. Aug 15, 2016 #7

    TSny

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    Note that tanΨ is dimensionless. Is the right hand side of your expression for tanΨ dimensionless?
     
  9. Aug 15, 2016 #8
    Just derivate of the term above. Chain rule for derivates...?

    EDIT: Okay, this is my fault. The second term doesn't need to be there.

    l5BMvP2.png

    I was confused because of the image above. I think that the derivate of above is dr/dt and thats why Theta has to be also be derived. (chain rule?). In this excercise it must be the derivate to theta and dsin(theta) /d(theta) = cos(theta). Because we have to derive to theta, when we apply the chainrule to theta, it is just a 1?

    I think I got it, thank you guys very much!!!
     
  10. Aug 15, 2016 #9

    Krylov

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    Geen zorgen (= don't worry), you fixed it just fine, nobody got hurt :wink:
     
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