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Homework Help: Physics Dynamics Question homework help

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the acceleration in the system of a fletcher’s trolley given m1 = 1.0 kg and m2 = 9.0 kg and a 50 N force of friction exists.
    θ = 33°

    See the image below, to explain the diagram. This one has an angle of 33 degrees that is why it is hard.


    2. Relevant equations
    I am not sure.

    3. The attempt at a solution
    Okay so first I found out the horizontal component of m1 by the formula cos(33)*mass(which is 1.0kg) and got the value 0.838 , then I put the formula Fnet=[(M2*A)-Ff]/(m1+m2)
    which means 9.0kg*9.81m/s2 - 50 N divided by (0.838+9) .. the answer was 3.9 M/s2 but this is the wrong answer, i asked my teacher, he said you are somewhere close but this is not right. I cannot figure this out, I have been trying different things since hours.
  2. jcsd
  3. Mar 23, 2012 #2


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    welcome to pf!

    hi huzjm! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    sorry, but even if you mean a = Fnet/(m1+m2) = [(M2*A)-Ff]/(m1+m2), that's still completely wrong :redface:

    try calling the tension "T", and doing two F = ma equations (one for each block)

    alternatively, if you're treating the two blocks as a single system, your m in ma has to be the total (unadjusted) mass, and you have to use the component of m1g parallel to the string

    try it both ways … what do you get? :smile:
  4. Mar 27, 2012 #3
    Thank you very much :) In fact, thank you really very much :)
    This is what I did
    Fnet = ma = 9kg * a = mg - T = 9kg * 9.8m/s² - T = 88.2N - T
    → T = 88.2 - 9a

    For the lighter mass,
    Fnet = ma = 1kg * a = T - mgsinΘ - Ff = T - 1kg * 9.8m/s² * sin33º - 50N
    1kg * a = T - 5.34N - 50N = T - 55.34N
    → T = a + 55.34

    Since T = T,
    88.2 - 9a = a + 55.34
    32.86 = 10a
    a ≈ 3.3 m/s²

    Is the answer right now?
  5. Mar 27, 2012 #4


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    excellent! :smile:

    (btw, you'll notice you could get the same result by treating it as a one-dimensional motion, with a single body with a mass of 10 kg, friction of 50 N and gravitational forces of 9g N and -gsin33° N :wink:)
  6. Mar 27, 2012 #5
    Oh so the formula should have been acceleration= [(9.0kg*9.81m/s) - 50N - (9.81m/s * sin 33°)] / 10 KG

    Thank you really very much :)
  7. Mar 27, 2012 #6
    I mean m/s(square)
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