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Forces help (friction, tension, etc.)

  1. Mar 3, 2015 #1
    First off, I want to apologize if anything is made unclear. I'm ultra-stressed right now, I'm sure I can get over this physics problem. I just need a layman's explanation of this.

    1. A taut massless string connects two boxes as shown below. The boxes are placed on an incline plane at θ = 33.00. What is the acceleration of boxes as they move down the incline, given m2 = 1.2 kg, m1 = 6.8 kg, μ2 = 0.10 and μ1 = 0.20 QujSMww.gif .


    2. Constructed free body diagram for M1 (mass 1) with normal force (n) along the positive y-axis, friction (f) along the negative x-axis, tension (T) along the positive x-axis, and Force of gravity (Fg) in quadrant IV, 33 degrees from the negative y-axis. (a=acceleration, mu=coefficient of friction)

    a.) Summed forces of x as Fx = T - f + Fgx = M1*a ->
    -> M1 * a - mu*n + M1*g = M1 * a
    -> (This is where I think I'm screwing up. I want to isolate the first (a), but I have no idea how.)

    b.) Summed forces of y as Fy = n - Fgy = M1 * a ->
    -> n - Fgy = 0 (acceleration is 0)
    -> n = Fgy

    3. Constructed free body diagram for M2 (mass 2) with (n) along the positive y-axis, (f) along the negative x-axis, tension (T) along the positive x-axis, and Force of gravity (Fg) in quadrant IV, 33 degrees from the negative y-axis. (a=acceleration, mu=coefficient of friction)

    a.) Same as 2a
    b.) Same as 2b

    4. Basically I have no idea where to go from here. Did I screw up on the diagram? Is something wrong with my algebra? What can I do next?
     
  2. jcsd
  3. Mar 3, 2015 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    It sounds like you are setting up your x-y coordinates so that x is parallel to the inclined plane. It is more traditional to have x be horizontal, parallel to the ground, with y vertical. Then you write the forces on the inclined plane for the FBDs in terms of sin(θ) and cos(θ) components.

    Can you re-write your FBD equations with the x-y axes in the more traditional orientation? That will make it a lot easier to check your work. :smile:
     
  4. Mar 3, 2015 #3

    TSny

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    Homework Helper
    Gold Member

    Welcome to PF!

    Look's like you assumed T = M1a. Only the net force on M1 equals M1a. For Fgx it looks like you didn't get the x component correct.

    Good. You'll just need to get an explicit expression for Fgy.

    What direction is the tension on M2?

    Treating T and ##a## as unknowns you should be able to solve for ##a##.

    [EDIT: I see I wasn't quick enough. berkeman posted while I was still constructing my response. I think it's good to choose the orientation of the axes as you did. Then the acceleration has only an x component.]
     
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