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Visualizing Free body diagrams?

  1. Mar 11, 2008 #1
    Visualizing Free body diagrams? solved, thanks tiny-tim and doc Al!

    I ran into a problem in my HW, I found the answer but I just have a hard time seeing as to WHY my answer is correct.

    http://i163.photobucket.com/albums/t288/Kyin1/1-6.jpg

    Thats the problem and my FBD is for block B. My FBD are the ones I did in red/blue w/ paint. The red means I am pretty sure that is what is given / correct. The lines in blue are what I believe I can maybe do in order to make more right triangles to help me find magnitudes of forces.
    If my FBD was not clear enough please let me know so I can fix.
    I didn't put what forces are acting on what because I think my problem is seeing that the lines in black (according to my answer) is equal to the magnitude of my Weight force?

    Is there a better way of seeing why my answer is the way it is?


    Thank you for your time.
     
    Last edited: Mar 12, 2008
  2. jcsd
  3. Mar 11, 2008 #2

    Doc Al

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    I'm unclear as to what your question is exactly. Looking at your FBD for B, you have correctly shown the three forces acting on it. Since B is in equilibrium, those forces must sum to zero. That means that the sum of the force components in any direction must be zero. Often it's useful to take components parallel to the incline.
     
  4. Mar 11, 2008 #3
    Sorry, I'll try to make it more clear.
    The way I am thinking about it right now is like this

    For example, looking at part A alone (finding the tension).
    I got my answer as wSin([tex]\alpha[/tex])

    http://i163.photobucket.com/albums/t288/Kyin1/2-3.jpg

    I did a zoom on a certain section of my FBD, I guess what I am trying to say is, why was I allowed to assume that the imaginary blue line I made is equal magnitude with that of the Weight force?
     
  5. Mar 11, 2008 #4

    tiny-tim

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    Hi kyin01!

    You're asking why does the book show the lines in black, when you get the lines in red, but it's still the right answer?

    Can I check … are the black lines exactly opposite the red lines?

    Because, if so, the black vectors (forces) are all minus-one times the red ones.

    So it's the same diagram (rotated 180º), so the answer will be the same! :smile:

    However,
    I'm still mystified why the book does it that way round. :confused:

    Can you show us the original question?
     
  6. Mar 11, 2008 #5
    Oh no, sorry. The blue / black / red / green FBDs are all mine I drew on MS Paint. The original problem are the question in the picture and the only given diagram are the actual diagrams where they show the ramp and all that stuff. Everything else I drew on paint.

    The red lines I drew are the ones I'm pretty sure are correct.
    The blue lines are what I assume I can do? They are equal in magnitude and opposite direction to the already given forces (The ones in red)
    The black line is what I am having trouble seeing. In that right triangle the black line (opposite of the angle) is equal magnitude to the Weight force?
     
  7. Mar 11, 2008 #6

    tiny-tim

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    Oh I see!

    No no no … don't ever try to draw triangles like that little one (the one marked "is = to W??") on your FBDs! :frown:

    Only draw the forces (and maybe horizontal and vertical axes, to make the angles clear).

    Then choose a direction, and add up the components, in that direction, of all the forces … which you know must equal zero.

    In this case, the direction to choose is the slope itself.

    (The reason? … always choose a direction which makes the calculation as short as possible … which usually means a direction perpendicular to one of the forces, in this case perpendicular to the normal reaction, N.)​

    Then the componenents of the three forces are 0.N, 1.T, and -sin(alpha).W, which must add up to 0.

    In other words: T - sin(alpha).W = 0. :smile:
     
  8. Mar 11, 2008 #7

    Doc Al

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    Well, you can't! Note that doing so gives you the wrong answer: [itex]\sin\alpha = W/T[/itex] is equivalent to [itex]W = T\sin\alpha[/itex], which is incorrect.

    What you should be doing is finding the component of the weight parallel to the incline, which is [itex]W\sin\alpha[/itex]. Since the tension and that component of weight must cancel out, that gives you [itex]T = W\sin\alpha[/itex].

    One tip that might help when trying to identify a right triangle for taking components: The full vector is always the hypotenuse of the right triangle, while the other sides are the components. In this example, W must be the hypotenuse of some right triangle, while [itex]W\sin\alpha[/itex] and [itex]W\cos\alpha[/itex] are the components of W parallel and perpendicular to the incline.
     
  9. Mar 11, 2008 #8
  10. Mar 11, 2008 #9

    Doc Al

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    I'm not crazy about that diagram either (although it might technically be correct). What I want to see is a triangle with the weight as the hypotenuse. The weight is the vector pointing vertically down.

    You need to learn how to take components of the weight parallel and perpendicular to the incline. Read this: Inclined Planes

    Once you understand that the component of the weight down the incline is [itex]W\sin\alpha[/itex], then you can conclude that [itex]T - W\sin\alpha = 0[/itex].
    The incline makes an angle alpha with the horizontal. Since the tension is parallel to the incline, it also makes an angle alpha with the horizontal. The normal and the incline (and thus tension) are perpendicular, so they form a good coordinate axis. A little triangle geometry will show that the weight (vertically down) makes an angle alpha with the normal axis and 90 - alpha with the tension/incline axis.

    I hope that helps a little.
     
  11. Mar 11, 2008 #10

    tiny-tim

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    Hi kyin01!

    I don't like the new triangle any more than the old one … :frown:

    Your problem seems to be geometry, rather than physics - you can't see how to work out the angles.

    So you draw in extra right-angled triangles to help you.

    I agree that can work, but
    (a) you're likely to make a mistake, and
    (b) you already have eight right-angles (yeah … count 'em!) in the centre of the diagram, and you can use them.

    If all you want is a right-angle, you don't need it to be in a triangle.

    You're wanting to calculate the angle between W and T. If we call B the horizontal blue line to the left in your diagram, then angle BT = alpha (as your diagram says), and angle BW is obviously 90º, so angle TW = 90º minus alpha.

    Then all you have to remember is that, to get the component of a force in a direction, you always multiply by the cos.

    In this case, it's cos(90º minus alpha), which is sin(alpha). :smile:

    And no triangles!
     
  12. Mar 12, 2008 #11
    Ah!!!!!! So excited!!!!!

    Thank you much tiny-tim and Doc Al

    Ever since you guys explained in details on how to correct away of looking / thinking about it, the rest of the HW went pretty easy and I understood as to why it is the way it is.
    So now instead of making a lot of triangles I just 'delete' my Weight vector and only use the X and Y components of it. So much easier way of looking at it.

    Thanks a lot you two!!
     
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