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Component acceleration in free-body diagram

  1. Feb 15, 2012 #1
    Hi all, first post here!

    1. The problem statement, all variables and given/known data

    Find the x component of acceleration.
    Find the y component of acceleration.

    The mass of the particle is 2kg.


    2. Relevant equations

    3. The attempt at a solution

    I've only tried to find a_x. I thought that for finding component values you need to add up the relevant components. There are four forces in the diagram, but one has no x component so I ignored it when finding a_x.

    [tex] F=ma [/tex]
    [tex] F_x=ma_x [/tex]
    [tex] F_x=2a_x [/tex]
    [tex] a_x=\frac{F_x}{2} [/tex]

    So do I just add up the three forces with x components like this?

    [tex] a_x = \left(\frac{F_{x,1}}{2}+\frac{F_{x,2}}{2}+\frac{F_{x,3}}{2} \right) = \frac{1}{2}(F_{x,1}+F_{x,2}+F_{x,3})[/tex]

    Starting clockwise at 1 N,

    [tex] F_{x,1} = (-1)cos(20) ≈ -0.94[/tex]
    [tex] F_{x,2}=(-2.82)cos(110) ≈ 0.96[/tex]
    [tex] F_{x,3}=(5)cos(20) ≈ 4.7[/tex]
    [tex] a_x = \frac{1}{2}(-0.94+0.96+4.7) = \frac{1}{2}(4.72)=2.36 [/tex]

    I must have a fundamental misunderstanding about something(s), because this is way off from the accepted value of 1.49 for a_x. thanks for commenting! :)
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 16, 2012 #2
    Hello octowilli,
    Thank's for providing nice diagrams and explanations of your attempt, in particular well done for solving algebraically before finding a numerical answer.

    However you have become slightly unstuck when calculating the forces along x, I fear it may be a case that you need to reconsider the diagram. You have solved for an axis x' which lies horizontal in the page, but not along the x-axis that the question has defined. So you answer is correct in one set of axis but they have asked for a different view.

    Apply the same logic that you have already done so along the defined x axis (hint the 5N lays along it)

    In general when these types of questions come up it is wise to pick your axis so that you have the smallest amount of work, in the system that you solved the 3N force was perpendicular but all the others where at some angle, in the system they have requested 5N lies along the axis so that you only need to calculate 2 angles!
  4. Feb 16, 2012 #3
    Thanks for responding gash789!

    I guess I was confused by the way the axes are rotated. Looking at it again, the 2.82 N vector certainly doesn't have an x-component. Or maybe you'd like to say the x-component is 0. Anyway, let me take another shot at a_x.

    From the given diagram, the x-component of acceleration is given by
    [tex] a_x=\frac{1}{2}(F_{x,1}+F_{x,2}+F_{x,3}+F_{x,4}) [/tex] where

    [tex] (F_{x,1}), (F_{x,2}), (F_{x,3}), (F_{x,4}) [/tex] are the x-component magnitudes (or do you say x-component forces, or just x-components?) of the vectors with magnitude 5 N, 3 N, 1 N, and 2.82 N respectively,

    [tex] F_{x,1} = 5\cdot cos(0) = 5 [/tex]
    [tex] F_{x,2} = (-3)\cdot cos(110) ≈ 1.03[/tex]
    [tex] F_{x,3} = (-1)\cdot cos(0) = -1 [/tex]
    [tex] F_{x,4} = 2.82\cdot cos(-90) = 0 [/tex]

    [tex] a_x=\frac{1}{2}(5+1.03+(-1)+0) = \frac{1}{2}(5.03) = 2.515 [/tex]

    Alright, the next thing I think I'm confused about is what signs to use when finding (F_x,n). If you redo what I've typed out with a positive 3 instead of a negative 3 for (F_x,2), it comes out to the right answer of about 1.49. I'm using -1 and -3 because in the picture they're both pointing left of the y-axis, so it seems to me they're negative. Another thing, should the angles have signs when computing this? If (F_x,1) and (F_x,3) are at a positive 20 degrees, does that mean (F_x,2) is at a positive 110 degrees and (F_x,4) is at a negative 90 degrees? The signs are confusing me.
  5. Feb 16, 2012 #4
    Indeed you have spotted your own mistake, when writing out the component of the 3N force along x, try to imagine two vectors parallel to the y and x axis. The magnitude of these is what you are physically trying to find.

    You have written that
    [itex] F_{x,2}=(−3)⋅\cos(110)≈1.03 [/itex]
    I try not to work with positive and negative angles as it will always confuse you, if instead you simple consider a right angled triangle with 3N along the adjacent angle to the hypotenuse (which is the x axis). Then you can quickly see that the component along x is [itex] F_{x,2}=(−3)⋅\cos(70)≈-1.03 [/itex]

    I think geometrically by using the angle to the positive x direction you already accounted for the negative sign, so by adding it in again just confused matters. It is up to you but I always prefer to work with right angled triangles as once you have all the vectors parallel you can easily account for the relative signs.
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