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Acceleration of a Block Down a Ramp

  1. Jun 11, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    A block with mass m1 = 9.2 kg is on an incline with an angle θ = 36° with respect to the horizontal. For the first question there is no friction, but for the rest of this problem the coefficients of friction are: μk = 0.33 and μs = 0.363.

    1.) When there is no friction, what is the magnitude of the acceleration of the block?

    2.) Now with friction, what is the magnitude of the acceleration of the block after it begins to slide down the plane?

    2. Relevant equations
    F=MA
    Force of Gravity: Fg=-Mg
    Normal Force: N = Mg

    3. The attempt at a solution

    I'm so confused. For question 1, I found the acceleration to be 5.77 m/s2 by finding the X-component of the normal force (53.05 N) and dividing by 9.2 kg. According my homework program, this is correct. However I don't really understand why.

    My work:
    Fx = sinθ(mg)
    Fx = sin36(9.2*9.81)
    Fx = 53.05 N

    A = F/M
    A = 53.05/9.2
    A = 5.766 m/s2

    For question 2, I've tried a couple of different things with no success. I don't even remember them all. I've just been floundering about for an hour. :confused: I'd appreciate any help.
     
  2. jcsd
  3. Jun 11, 2015 #2

    Simon Bridge

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    For each of these questions, start by drawing a free-body diagram for the block, then apply Newton's laws.
    What are the forces on the block for Q1?
    Which direction do you expect the net force to point in?

    In Q2 - it's just the same only there is an extra force.

    (Little puzzled as to why this thread is started by "Drakkith".)
     
  4. Jun 11, 2015 #3

    haruspex

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    Did you draw a FBD? What are the forces parallel to the plane? For (2), which coefficient will you use?
     
  5. Jun 11, 2015 #4

    Drakkith

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    For Q1, the forces acting on the block are the force of gravity and the normal force.
    For Q2 there's also the force from kinetic friction, which is parallel to the plane of the incline.
    I used 0.33 as my coefficient since the block is sliding.

    Because I'm taking a Physics 210 class.
     
  6. Jun 11, 2015 #5

    haruspex

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    Sure, but I asked specifically about parallel to the plane. What are the components of those forces parallel to the plane? What does ##\Sigma F=ma## give you for that direction?
     
  7. Jun 11, 2015 #6

    Simon Bridge

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    Considering this question:
    ... don't we need to motivate finding the component parallel to the plane?

    Weight and the normal force must add (head to tail) to the net force - you know the magnitude of the weight and the direction of the normal and the resultant forces. This lets you carefully draw out a right-angled triangle ... and you know how to deal with those.

    It's just easier to work in components and follow the rules.

    You are teaching a physics 210 class, or you are a student in one?
     
  8. Jun 11, 2015 #7

    Drakkith

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    I think its 55.78 N. The X-component is 53.05, the net force in the vertical direction is 17.232, so √(53.052 + 17.2322) = 55.78 N.
     
  9. Jun 11, 2015 #8

    Drakkith

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    Student.
     
  10. Jun 11, 2015 #9

    Drakkith

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    I think that's what I did in post 7. But that gives me an acceleration of 6.06 m/s2, which apparently is wrong according to the homework program.
     
  11. Jun 11, 2015 #10

    Drakkith

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    Wait... apparently I've severely misunderstood something here. When you take gravity and divide it into its two components, are they parallel and perpendicular to the surface of the plane?
     
  12. Jun 11, 2015 #11
    If by plane you mean the ramp, then yes.
     
  13. Jun 11, 2015 #12

    haruspex

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    It's not that a force, such as gravity, has components, in any absolute sense. You can choose some direction and find the component of the force in that direction. For this you use the cosine of the angle in between the force and the chosen direction. If you choose two directions at right angles, and find the component of the given force in each of those two directions, then the original force is equal to the vector sum of the two components.
    In ramp problems, it is often convenient to represent every force by its components parallel to and perpendicular to the slope. In particular, the component of gravity perpendicular to the slope will generally equal the normal force from the slope on the object (since there is no acceleration in that direction). It gets more complicated if there are other forces perpendicular to the slope, or if the ramp can move.
     
  14. Jun 11, 2015 #13

    Drakkith

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    Okay, that solved all of my issues. For Q2, the answer is 3.15 m/s2
    Thanks all.
     
  15. Jun 11, 2015 #14

    Simon Bridge

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    Well done.
    You'll know from reading PF that just learning the rules for manipulating a problem is not good enough around here :)
    Have you tried the problem geometrically - adding the vectors head-to-tail?
    That will show you why the components make sense mathematically.

    Physically - you want one component in the direction you'd expect to find the final acceleration because that it the direction of the net force.
    The other component has to be at right angles because that is what "component" means.
     
  16. Jun 11, 2015 #15

    Drakkith

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    I'm not sure what you mean by this.

    I haven't gotten out a ruler and protractor and made sure the angles and lengths of the arrows match up, but I can see how the vector components add together.
     
  17. Jun 12, 2015 #16

    Simon Bridge

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    No worries - enjoy.
     
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