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Homework Help: Simple Newton's 2nd Law problem Why isn't this answer right, though?

  1. Dec 12, 2012 #1
    1. "A 40.0-kg crate is pulled along a horizontal floor by the ideal arrangement
    shown in figure below. The force F is 280 N. The coefficient of kinetic friction between the crate and the floor is 0.170. What is the acceleration of the crate?"


    Okay, this is a weird drawing of the situation. Just a box on a flat surface.
    It's attached by a string to a pulley and the other end of the string is anchored to a
    cone (^).

    So I thought the solution is just this:

    εF(y) = ma(y)

    -w + N = ma(y)
    -392 + N = (40)(0)
    N= 392N

    εF(x) = ma(x)
    -f + F = ma(x)
    [mu(k)](N) + 280 = (40)(a)
    (.17)(-392) + 280 = (40)(a)
    a= 5.33m/s^2

    BUT, this is not the right answer on my test!
    The right answer is 1.83m/s^2.
    How do you get that answer???

    Thank you so much! :D
  2. jcsd
  3. Dec 12, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Can you post the diagram? It's not clear what the arrangement is.
  4. Dec 12, 2012 #3

    Okay, this is it.
  5. Dec 12, 2012 #4
    Okay. If the pulley is massless, how is the tension in the cord related to the force F?

    (a) Tension = F
    (b) Tension = F/2
  6. Dec 12, 2012 #5

    Doc Al

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    Staff: Mentor

    Good. So now figure out the tension in the rope. That will tell you the force on the crate.
  7. Dec 12, 2012 #6
    Um, okay...

    So I looked up the formula for tension in a rope.

    They say it's: F(tens.) = ma - (mu)(-mg)

    To find the a, I did: F = ma
    280 = (40)(a)
    a = 7m/s^2

    So, that means it's: F(tens.) = (40)(7) - (.17)(-392)
    F(tension) = 346.64N

    So, then I thought I would use the tension force instead of the F...

    -f + T = ma(x)

    -392 + 346.64 = (40)(a)
    a = -1.134m/s^2 BUT THIS IS NOT RIGHT...

    It's negative. I don't really know what's wrong. I'm thinking the F(tension) is wrong... =_=
  8. Dec 12, 2012 #7

    Doc Al

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    Staff: Mentor

    To find the tension in the rope analyze the forces acting on that massless pulley.
  9. Dec 12, 2012 #8
    It seems to me that the tension would be equal to the force, F.

    ƩF(x) = ma(x)
    -T + 280 = (0)(a)
    T = 280

    So I still don't see why the Tension isn't the same thing as the Force F...

    ƩF(x) = ma(x)
    -f + T = ma(x)
    (.17)(-392) + 280 = 40(a)
    a = 5.334m/s^2.... The same wrong answer... Is there some force on the Pulley I'm leaving out?!
    It is massless, so there would be no N force or weight.
  10. Dec 12, 2012 #9
    If you look at the pulley, then you have two tension forces acting to the left.
    So, 2T = F
    → T = 140N
  11. Dec 12, 2012 #10
    WOOO! Hallelujah!

    Thanks, ap123!!! I've been working on this problem FOR-EVER!!!
    But what the heck?!
    2 tensions?
    Is this because of the tension where the string attaches to the box AND where the string from the pulley attaches to the cone?
    -Both are pointing in the left direction.
  12. Dec 12, 2012 #11
    Last edited: Dec 12, 2012
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