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Resolve force into components along different paths

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img5/629/meprob25.jpg [Broken]

    A. Determine magnitude of resultant force (of F1 and F2 acting on the point)

    B. Resolve F1 in to components along the u and V axes and determine the magnitudes of these components.

    C. Resolve F1 in to components along the u and V axes and determine the magnitudes of these components.

    2. Relevant equations
    law of sines,

    Pythagorean theorem




    3. The attempt at a solution

    I just need help with parts B and C.

    A goes :

    Fx = 150(cos60) + 200(cos45)= 216 lb

    Fy = 150(sin60) + 200(sin45) = -11.5 lb

    (2162 + 11.52)1/2 means

    R = 216 lb at 3 deg under the U axis because tan-1(-11.5/216) = -3


    For part B the only guide to find how to solve these before is use law of sines, but

    trying to solve for resolving the components of F1 along a and B:

    200 / (sin45) = F(u) / 1

    F(u) = 282 lb

    but it is supposedly wrong.

    Any hints?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 30, 2013 #2

    Simon Bridge

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    It is useful to draw the vector triangles to help you know what to do.
    Trying to remember equations won't help. But you do know how to do trigonometry - and you know how to handle triangles that are not right-angle triangles? And you know about dot and cross products?

    B. resolve along u and V axis
    This means either:
    1. ##\vec{F}_1=a\hat{u}+b\hat{V}## (where the hats indicate a unit vector)
    2. the amount of ##\vec{F}_1## along direction ##\hat{u}## and ##\hat{V}##.

    You'll have to figure out which depending on how your course is taught.

    The second one comes from the dot product.

    The first one is trickier - you you need to draw the vector head-to-tails summation diagram where one vector is along the V direction and the other vector is along the u direction (lynchpin: the magnitude can be negative - you can use the fact that the u-direction is horizontal to help your sketch). You can use the geometry of the resulting triangle to help you.
     
    Last edited: Jan 30, 2013
  4. Jan 30, 2013 #3
    I drew you this image to help you with part B. If you can get this you'll know how to do part C

    rvldfa.jpg

    Now you use law of sines to find F_u and F_v. let's see if that gives you an idea on how to go along with this. I hope you can see how i got the angles.
     
  5. Jan 30, 2013 #4

    Simon Bridge

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    That's the one! That's really close to doing the homework for the OP though.
    Hopefully OP can see how that was drawn.
    If not, it may be easier to do F2 first.
     
  6. Jan 31, 2013 #5
    Thanks, I had the triangle wrong. I did not really know that it was supposed to be parallel to the other axis. Yes, Simon, it was asking for the first one.

    Part B: 200 lb/(sin 30) = Fu/(sin 105) ---> Fu = 386 lb

    Part C: 200 / (sin 30) = Fv / (sin 45) ---> Fv = 283 lb


    The second one might be helpful to know later on too. Can I ask how the second case comes from the dot product? I'm guessing you would just be looking for the magnitude and it would probably go something like: (magnitude in direction) = (orginal force) / cos (angle between original force and direction) ?
     
  7. Jan 31, 2013 #6

    Simon Bridge

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    From the geometric interpretation of the dot product...

    i.e. for when you consider forces that do not act through the center of mass...

    The amount of ##\vec{u}## in direction ##\vec{v}## is ##\vec{u}\cdot\vec{v}/|\vec{v}| = |\vec{u}|\cos\theta## where ##\theta## is the angle between them.

    Note: the angles you were given, 30, 60, 45, are "nice" angles:

    ##\sin(30)=\cos(60)=\frac{1}{2}## ; ##\cos(30)=\sin(60)=\frac{\sqrt{3}}{2}## ... 2-1-√3 triangle

    ##\sin(45)=\cos(45)=\frac{1}{\sqrt{2}}## ... 1-1-√2 triangle

    ... it is worth memorizing these, and the angles in a 3-4-5 triangle.
     
    Last edited: Jan 31, 2013
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