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Free body diagram of an incline

  1. Nov 22, 2014 #1
    1. The problem statement, all variables and given/known data
    For the maximum angle for which you have data draw a free body diagram and explain how the forces add to give the resultant (net) force and show the calculations required to determine the acceleration.

    The lab was about using Galileo's inclined plane to measure acceleration due to gravity. We rolled the cart down the incline and measured its acceleration with a sensor. We did 10 trials and and raised the height of the incline 4 cm each trial.
    jNb96NW.png
    2. Relevant equations
    a= gsin(Θ)
    a×=AcosΘ (?)

    3. The attempt at a solution
    Oq7F5r0.png

    I drew the inclined plane at 21 degrees and drew the cart with arrows. But I do not know how to continue from there. I know that the force of gravity and the normal force will cancel each other out. I know that there is a force causing the cart to go down the plane. But I don't know how to find any of these forces and draw a scaled diagram without the mass of the cart, which was not given.
    I tried doing 3.04= 9.81 x .358 to make sure the acceleration was correct but it wasn't. I thought maybe I could use a×=AcosΘ but I can't even plug anything in correctly.
    I've been crying for an hour because I don't know how to do this problem for my report...
     
  2. jcsd
  3. Nov 22, 2014 #2

    Simon Bridge

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    In the absence of friction, there are only two forces on the cart.These forces must add to a result that points in the direction of acceleration. Use trigonometry.
     
  4. Nov 22, 2014 #3
    The thing is, I don't know how to do that... Thank you though
     
  5. Nov 22, 2014 #4

    Simon Bridge

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    Don't know how to do what?
    Please be specific.
     
  6. Nov 22, 2014 #5

    gneill

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    Your drawing has an error. The force due to gravity is always directed straight downwards (towards the center of the Earth). So the force that you've labelled ##F_g## is actually just the component of the force due to gravity that happens to be perpendicular to the plane of the wedge. It's the wedge's reaction force to this (Newton's 3rd law) that is the normal force ##F_n##.

    You need to find a way to split up the (known) force due to gravity into components that are perpendicular to and parallel to that plane. That will give you ##F_n## and ##F_a##.
     
  7. Nov 22, 2014 #6
    I don't know how to use trig to find the forces and draw them? If that's what you meant... I'm sorry for being so dumb that I don't even know what I'm talking about.
     
  8. Nov 22, 2014 #7

    Simon Bridge

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    Start by drawing a good diagram.
    Draw a gravity arrow pointing the right way.
    You know which way gravity points right?
    You also need to go through your class notes looking for your work on adding and finding components to vectors.
     
  9. Nov 22, 2014 #8
    Well I have the the diagram on a sheet of paper and it is much neater. I corrected the direction of the gravity arrow but I don't know how long it should be since it says that it must be scaled diagram in which 1N = 2cm.
     
  10. Nov 22, 2014 #9

    gneill

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    If you did not measure or were not given the mass or weight of the cart, then assume some value that will suit your diagram's scale. The actual value of the mass is not that important as it will cancel out during calculations of the acceleration (as you'll find out!). A mass of 0.41kg has a weight of about 4N (what formula applies to finding the weight of a given mass?).
     
  11. Nov 22, 2014 #10
    Isn't it fg=mg? So I drew an 8 cm line. And now I'm not sure how to continue. Do I draw the normal force? Is it (0.41kg)(9.81m/s^2)(cos21)?
     
  12. Nov 22, 2014 #11

    gneill

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    Yes.
    Yes. The normal force and the down-slope force due to gravity can both be obtained from the weight of the object. You've written the formula for the normal force (or rather it's 3rd Law "reflection"). So what's the corresponding formula for the down-slope force? Remember that the combination of all three forces form a right angle triangle:
    Fig1.gif
     
  13. Nov 22, 2014 #12
    erm.. I think then its (.41kg)(9.81m/s^2)(sin21) which is 1.44N? So that's the applied force?
     
  14. Nov 22, 2014 #13
    I think I've got it now.. Thank you soooooooooo much. I am failing physics so it was hard for me. Now l understand it better.
     
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