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Pulley, Force and TEnsion

  1. Nov 9, 2009 #1
    The figure looks like:

    There is a 8.2 kg (m) block hanging vertically from a pulley, which is connected to a block (M) 15.7kg. Initially the block M is moving to the left (because of the force of block m). A force F, with magnitude 75.3N, acts on it directed at an angle of 35.0 degrees above the horizontal, in the opposite direction of block m. There is no friction and the pulley and string are massless.

    b)What is the tension in the string?

    I used the equation (may)=Tcos(35.0)-mg (m0)=Tcos(35.0)-153.86 T=188 which was wrong. The correct answer was 70.4

    c)What is the acceleration of M?

    I used the equation (max)=Tsin(35.0) obviously I got this wrong because I didn't have the right value for T. the correct answer is -0.78

    I thought I was using the right equations but I guess not! Please help? Could you be specific because I am very confused by this problem. Thank you
  2. jcsd
  3. Nov 9, 2009 #2
    Could you plz draw a sketch?Maybe then i could help
  4. Nov 9, 2009 #3
    I am not sure how to load a picture on here. Any pointers?
  5. Nov 9, 2009 #4
    What i do is go to paint in my pc draw na dthen upload through the attachment icon on the reply
  6. Nov 9, 2009 #5
    I think this worked

    Attached Files:

    • 5.JPG
      File size:
      4 KB
  7. Nov 9, 2009 #6
    Is something missing?Because you said initially it went back the body with the mass M and then you said about the force F
  8. Nov 9, 2009 #7
    Could you plz tell me the whole excercise?
  9. Nov 9, 2009 #8
    The problem says that M is initially moving to the left. Then a force, F acts on it in the direction of the sketched F in the attachment. Sorry if it is not clear.
  10. Nov 9, 2009 #9
    If i got it right then you have two equations with two unknowns.
    for the body m: mg-T=ma
    for the body M: T-Mg=Ma

    If you solve these two you will find an accelaration close to 0.78
    i find 0.80 aproximatelly
    And then you can find the tension if you substitute the acc on one of the equations

    I hope i helped
  11. Nov 9, 2009 #10
  12. Nov 9, 2009 #11
    Ok, I really struggle solving these equations

    I solved one of the equations for T T=Ma+Mg and put this in the other equation for T

    mg-(Ma+Mg)=ma 80.36-(15.7a+153.86)=8.2a 80.36+153.86=8.2a+15.7a a=9.8
    What am I missing here?
  13. Nov 9, 2009 #12
    Where is the minus in the 153.86?
  14. Nov 9, 2009 #13
    Sorry my mistake in the equation for body M there is no Mg there is Fcos35
  15. Nov 9, 2009 #14
    I am sorry i misguide you
  16. Nov 9, 2009 #15
    The acceleration is supposed to be -.78. I am getting .78. That is what you got too?
  17. Nov 9, 2009 #16
    I got minus.Probably you forgot one minus.Chech your equations again.
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