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Homework Help: Finding tension of a rope pulling a crate

  1. Oct 15, 2011 #1
    Did this problem out, but answer doesn't look right.
    1. The problem statement, all variables and given/known data
    Find the tension of the rope.

    Coefficient of kinetic friction=.244
    Rope is at an angle of 22.2° above the horizontal
    Pulled at constant speed, meaning a=0

    2. Relevant equations
    I figured that...

    So, N=mg+Tsin∅

    Then I did a simple substitution with the N equation...


    When I got T by itself, I got....


    3. The attempt at a solution

    Then I just plugged the given numbers in, and I got T=69N (Approximately)

    Am I correct? Because I figured the tension of the rope has to be greater than the mg of the crate.
  2. jcsd
  3. Oct 15, 2011 #2

    Replace -Tsinangle by +Tsinangle
  4. Oct 15, 2011 #3
    Okay, so that means that my final equation for T changes to...


    I get an answer of about 164.5 N. It's higher, but is that right. I'm still wondering if T is supposed to be greater than the mg of the crate. If this is correct, then why did I have to change the -Tsin∅, to a positive Tsin∅?
  5. Oct 15, 2011 #4
  6. Oct 15, 2011 #5


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    When you sum forces in the y direction, N acts up (+), mg acts down (-), and T sin theta acts up (+). That is why grzz corected the signage for Tsin theta.

    Beyond that, your algebra is incorrect. There should be a uk in front of the Tsin theta term (note that a(b + c) = ab + ac).

    Why do you feel T has to be greater than mg? Consider pulling a heavy crate on a near frictionless surface. The pulling force is much less than mg.
  7. Oct 15, 2011 #6
    Ah, I see now.

    Didn't catch the algebra error. Thanks for that.

    Meh. I don't know what I was thinking earlier. Was doing this problem early in the morning.

    Okay, so now that I corrected my errors, I get a tension of approximately 108 N.
  8. Oct 15, 2011 #7


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    looks good!
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