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Lecturer made a mistake ?

  1. Jul 31, 2010 #1
    Hi guys,

    I am currently studying a basic college physics course at a university in New Zealand.

    The course is only going over what we were taught in college so far. I came across a practice problem and was unsure how to go about solving it. So I looked up my lecturers online notes and I believe he has made a mistake, I would like someone to confirm whether I am correct.

    [PLAIN]http://img192.imageshack.us/img192/84/17681455.png [Broken]

    I am only interested in the reaction force equation he has given, his trig manipulation seems to be the problem, someone please check the picture and confirm whether it is incorrect or correct.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 31, 2010 #2

    cristo

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    Looks correct to me. What do you think is wrong?
     
  4. Jul 31, 2010 #3
    i think that the reaction force equation should be N = W/(cos θ)
    the trig dosn't make sense to me this way
     
  5. Jul 31, 2010 #4

    Doc Al

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    You need to review how to find components of a vector.

    Some tips: The component of a vector (in this case, W) can never be greater than the vector itself, so you would never divide by cosθ to find a component.
     
  6. Jul 31, 2010 #5

    arildno

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    Remember that the normal force opposes that component of the weight so that there is [no acceleration of the object along the normal direction.

    A component of the weight can never be greater in magnitude than the weight itself.
     
  7. Jul 31, 2010 #6
    i understand what u guys are telling me but when i try to derive that equation that is what i get maybe this should be in maths not in physics. ill review my working, and MAYBE talk to the lecturer

    at any rate ty for your time guys i really appreciate it

    its hard to find good help at university without dishing out serious cash which i cant afford
     
  8. Jul 31, 2010 #7

    Doc Al

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    http://www.physicsclassroom.com/Class/vectors/U3L1e.cfm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Jul 31, 2010 #8

    arildno

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    Hi, chris!

    I will try to illuminate this further to you:

    When you take the components of normal force N (and of course, there is nothing really "wrong" with that), what do these components actually MEAN??

    What they mean, is quite simply, the amount of normal force along the VERTICAL and HORIZONTAL directions. Agreed?

    But, if you agree to this, you must ask yourself:
    Is there any non-zero ACCELERATION going on in either of these directions?

    If there is such non-zero acceleration, for example in the vertical direction, then obviously, you cannot say that the net sum of forces in that direction equals 0!!!


    Thus, it is more PRUDENT to decompose along the tangential and normal directions, because then you KNOW that the acceleration along the normal must be zero.
    You thereby get a simpler system to work with!

    I'll do the decomposition for you properly now:

    DECOMPOSITION ALONG HORIZONTAL AND VERTICAL DIRECTIONS:
    a) Horizontal direction (with positive direction to the left):
    [tex]N\sin(\theta)-F\cos\theta=ma_{x}(1)[/tex]
    b) Vertical direction:
    [tex]N\cos\theta+F\sin\theta-mg=ma_{y}(2)[/tex]
    c) Relation between N and F:
    [tex]F=\mu_{k}N(3)[/tex]
    d) Motion restricted to tangential direction:
    This means that the vector (a_x,a_y) is orthogonal on the normal vector (sin(theta), cos(theta)), so that we get the equation:
    [tex]a_{x}\sin\theta+a_{y}\cos\theta=0(4)[/tex]
    Now, the simplest way to solve this system of equations is to multiply (1) with sin(theta), (2) with cos(theta), and adding them together. Using (4) on the resultant right-side gives you the eqation:
    [tex]N-mg\cos\theta=0(5)[/tex]
    (Remember: [itex]\sin^{2}\theta+\cos^{2}\theta=1[/tex]

    (5) is nothing else than Newton's second law, as decomposed in the normal direction!
     
    Last edited: Jul 31, 2010
  10. Jul 31, 2010 #9

    Born2bwire

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    Probably the simplest thing to do first is look at the limiting cases. Obviously, when \theta=0 the normal force should be mg and when \theta = \pi/2 the normal force should be zero. The lecturer's equation satisfies this while your own derivation does not.
     
  11. Jul 31, 2010 #10
    ty arildno and born2bwire i get it now, and thanks to born2bwire i get my equation is incorrect. Also ty born2bwire for the simple check.

    after going through the trig a hundred times i figured i was looking at the wrong triangle a simple rookie mistake. I ty legends for your time the maths works now.
     
    Last edited: Jul 31, 2010
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