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Using newton laws

  1. Jul 19, 2007 #1
    1. The problem statement, all variables and given/known data
    The crates are on a horizontal frictionless surface. The woman (wearing golf shoes so that she can get traction) applies a horizontal force F = 50.0N to the 6.00kg crate.
    a.) what is the magnitude of the acceleration of the 6.00kg crate?
    b.) what is the tension T in the rope (of negligible mass) connecting the two crates?

    2. Relevant equations
    [tex]F = ma[/tex]

    3. The attempt at a solution

    a.) since the horizontal component of force has only 50 newtons with zero friction:
    [tex]F_x = ma[/tex]
    [tex] a = \frac{m}{F_{x}}[/tex]
    [tex] a = \frac{6.00kg}{50.0N}[/tex]
    [tex] a = 8.33 \frac{m}{s^2} [/tex]

    b.) Tension?
    No other x component of force
    so i believe the tension is 50.0N
    Last edited: Jul 19, 2007
  2. jcsd
  3. Jul 19, 2007 #2


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    Staff Emeritus
    Science Advisor

    But she's not moving only the 6 kg crate! Yes, the 50 N force is on the 6 kg crate but there is also a force (that you don't know yet) caused by the drag of the 4 kg chart.

    In essence she is moving both crates: a total of 10 kg. There common acceleration is
    [tex]\frac{50}{10}= 5 m/s^2[/itex]
    (you have your fraction upside down! Since F= ma, a= F/m, not m/F.)

    Now, what what NET force would produce that acceleration in the 6 kg crate alone? (F= ma= (6)(5)) The difference between that and the 50 N is the "drag" of the 4 kg crate and is the tension in the rope between.
  4. Jul 19, 2007 #3
    you have the acceleration of both the blocks. draw the free body diagram of the two blocks saperately and find the tension
  5. Jul 19, 2007 #4
    hey pretty much beautiful drawing of the lady there:)
  6. Jul 19, 2007 #5
    hey thanks a lot :smile:
    so 20N is the tension between the crate

    i think when you already have an acceleration you can instantly compute for the tension between 4kg and 6kg mass... using the 4kg mass,

    so 4kg(5m/s^2) = 20N
    Last edited: Jul 19, 2007
  7. Jul 20, 2007 #6
    right on. the tension is the only force moving the 4kg mass.
  8. Jul 22, 2007 #7
    w8 how would i draw a free body diagram on the two blocks including the woman? and the forces acting on each body?
  9. Jul 22, 2007 #8
    F is 50N from the problem, you have the acceleration of the blocks(that is 5m/s/s).
    only force on 4kg block is tension in the rope.
    on the 6kg block, net force in the direction of acceleration or motion is applied force - tension in the rope.
    on the lady, friction balances the 50 N
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