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Help with acceleration on incline

  1. Oct 20, 2004 #1
    An inclined plane that makes an angle of 28° to the horizontal is mounted on wheels. A small block of mass m = 1.3 kg rests on the plane, held there by a coefficient of static friction µ = 0.7.

    The plane is accelerating to the right. What is the minimum acceleration in order that the block slides down the plane?

    I'm confused on where to even start. help please
     
  2. jcsd
  3. Oct 20, 2004 #2
    Because coefficient of static friction=F x R(normal reaction),hence
    0.7=F x mg
    Force = 0.7/mg
    =0.055 N

    You say the plane is accelerating to the right, what is the plane acceleration ?
     
  4. Oct 20, 2004 #3

    Doc Al

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    Newton's 2nd law

    Draw a diagram of the mass showing all the forces on it. (There are three forces acting.) Now consider the conditions just before the mass starts to slide down the plane: The vertical forces are in equilibrium; the horizontal forces must produce an acceleration. Apply Newton's 2nd law and solve for the acceleration.
     
  5. Oct 20, 2004 #4
    i drew fbd, and set x-axis in the direction of the acceleration, and y-axis straight up.
    Then i applied the newtons second law, and this is what i got.

    ma = fs - mg(cosx)
    1.3a = 0.055 - (1.3*9.8*cos28)
    when i solve for a, i get a negative number.
    i cant figure out what went wrong.
     
  6. Oct 20, 2004 #5

    Doc Al

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    I don't know what that equation represents.

    Do it step by step:
    (1) Identify all the forces on the mass.
    (2) Write the equation for vertical equilibrium.
    (3) Apply Newton's 2nd law for the horizontal force components.

    Combine the equations from (2) and (3) and solve for "a".
     
  7. Oct 20, 2004 #6
    for vertical, i got this: F (force y) = N (normal) - W(weight)SinX
    for horizontal, i got : F (force x) = fs(static friction) - W(weight)CosX
    i replaced F with MA (mass x acc) and tried solving it. this is what i did for the previous post.
     
  8. Oct 20, 2004 #7
    you have you axes tilted...so with that tilted axis, you have accelertaion going in both the x and y directions....use a straight axes (were x is in the direction of acceleration)...that way you have acceleration in only one direction
     
  9. Oct 20, 2004 #8

    Doc Al

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    As Spectre5 notes, you need to analyze vertical and horizontal components, not components perpendicular and parallel to the surface.

    The three forces that I see are: normal force, static friction, and weight. All three have vertical components. (Note that since we are finding the point at which the mass just overcomes the static friction, that friction force must equal [itex]\mu N[/itex].) Similarly, the normal force and friction both have horizontal components.

    Redo your equations.
     
  10. Oct 20, 2004 #9
    is the frictional force for x component and y component equal in magnitude?
     
  11. Oct 20, 2004 #10
    i reworked the equations as stated above and this is what i got:
    Fy(force y) = fs(static fric y) + Ny (normal y) - W(weight)cosX
    Fx(force x) = fs (static fric x) + Nx (normal x)

    im kind of getting it but still alittle confused on how to calculate Ny and Nx and whether static friction force for y component and x components are equal.
     
  12. Oct 20, 2004 #11
    man i suck at this :(
     
  13. Oct 20, 2004 #12

    Doc Al

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    You are not taking full advantage of the given information. The only things you don't know are N and a; all else is given (mass, [itex]\theta[/itex], [itex]\mu[/itex]). Use this information to find the vertical and horizontal components of the three forces.

    For example. The normal force, N, has a vertical component of [itex]N cos\theta[/itex] and a horizontal component of [itex]-N sin\theta[/itex]. Due a similar component breakdown for the friction force ([itex]\mu N[/itex]) and the weight (mg).

    Now redo your equations once again. Don't give up.
     
  14. Oct 20, 2004 #13
    one more question. the angle for feta always going to be 28 degrees for all components?
     
  15. Oct 20, 2004 #14
    isnt the vertical component Nsine(feta) and horizontal Ncos(feta)?
     
  16. Oct 20, 2004 #15

    Doc Al

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    Draw a picture and figure it out. Remember that [itex]sin(28) = cos(62)[/itex], etc.
     
  17. Oct 20, 2004 #16

    Doc Al

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    I was using [itex]\theta = 28[/itex]; sorry if that was confusing.
     
  18. Oct 20, 2004 #17
    this is the final equation i was able to come up with.
    Fx = f*cos(feta) - N*cos(feta)
    Fy = f*sine(feta) + N*sine(feta) - W (W=mass*gravity)
    is this correct? you said above that for horizontal it should be sine, and vertical cos, but i dont understand this. i think it should be other way around.
     
  19. Oct 20, 2004 #18
    oh i see what you did. yeah that was alittle confusing. anyways does my equations look okay?
     
  20. Oct 20, 2004 #19
    dang it. its still wrong. i sux. :(
    Fx = Mass*A
    Fy = 0
    for friction i used 28degrees as feta
    for normal i used 62 degrees as feta
    i get negative number for an answer.
     
  21. Oct 20, 2004 #20

    Doc Al

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    Rather than use [itex]\theta[/itex], specify the particular angle you mean. It doesn't matter whether you use sin(28) or cos(62), but you need to specify the angle you are using. And instead of f, put what it equals [itex]f = \mu N[/itex]; same for W, use [itex]w = mg[/itex].
     
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