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Dynamics help: Spring force and dynamic motion

  1. Apr 3, 2013 #1
    The 20-kg sphere, attached to a spring of stiffness 500 N/m
    is released from an un-stretched position of the spring, as
    shown in the figure. Determine the velocity of the
    block “A” of mass 10 kg, at the instant the ball has fallen
    25 cm. A roller mechanism keeps the spring in a vertical
    position. Ignore the size of the ball and the mass of the
    spring and roller mechanism. The coefficient of kinetic
    friction between the block “A” and the ground is 0.25

    Here is an image of the problem:
    http://s1306.photobucket.com/user/wellthisisdumb/media/1_zps80a2c31a.png.html

    All angles in the triangle are 60°.

    I started by making free-body-diagrams for the ball and the block, then I made separate diagrams for the Normal force between the two objects.
    Here are the diagrams:
    http://s1306.photobucket.com/user/wellthisisdumb/media/2_zpsf6514a00.png.html

    I know that as the ball falls, the spring force increases, so the effect of the ball on the block's movement will decrease with distance, so I know I need an equation for motion of the block in terms of x(the displacement of the ball).

    Wb = Weight of Ball(B)
    Wa = weight of block A
    Fs = spring force
    Ff = frictional force
    Na = Normal force of A
    Nb = Normal force of B

    Wb = mg = 20(9.8) = 196N
    Wa = mg = 10(9.8) = 98N
    Fs = 500(.25) = 125 N

    Nb = Wb - Fs
    = 196 - 500x

    Na = Wby + Wa
    = Nbsin60 + 98
    = (196 - 500x)sin60 +98
    = 267.7 - 433x

    Wbx = Nbcos30
    = (196 - 500x)cos30
    = 169.7 - 433x

    Ff = μNa
    = .25(267.7 - 433x)
    = 66.9 - 108x

    ƩFx = Wbx - Ff = MaAx
    (169.7 -433x) - (66.9 - 108x) = 10Ax
    102.8 - 325x = 10Ax
    10.28 - 32.5x = Ax

    I have my equation for the acceleration of the block in terms of the motion of the ball, so next I used ads = vdv to solve for velocity

    ads = vdv
    ∫ads = ∫vdv (ads from 0 to .25, vdv from 0 to V)
    ∫(10.28 - 32.5x)dx = 1/2(v^2)
    10.28x - 16.25x^2 = 1/2(v^2)
    10.28(.25) - 16.25(.25)^2 = 1/2(v^2)
    1.55 = 1/2(v^2)
    3.1 = v^2
    1.76 m/s = v

    So I got v = 1.76 m/s as my answer. My professor said to use ads = vdv to solve the problem, and this is the only way that I could get to even using ads = vdv. I feel like I've done it correctly, but we've never done a problem even similar to this in class, so I'm just kind of lost.

    If someone could verify that I got the right answer that would be great, and if it's wrong I'd really appreciate a step-by-step solution. Thank you!
     
  2. jcsd
  3. Apr 3, 2013 #2
    I apologize in advance for how long this post is. I'm a mediocre formatter.
     
  4. Apr 3, 2013 #3

    haruspex

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    In your diagrams (top right) you have a Wb which I assume should be Nb. Anyway, Na = Nb, right?
    In your equations you have Nb = Wb - Fs. Do you mean Nby = Wb - Fs? Even then, you're ignoring that there has to be a resultant vertical force on the ball to account for its acceleration. I haven't checked the rest of your work.
     
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