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2 quick questions (Force vector questions

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Two players kick a soccer ball at the same time. If one player applies a force of 100N[N25*W] and the 250g ball experiences an acceleration of 200m/s^2 [W15*S], determine the magnitude and direction of the force applied by the second player.





    2. Relevant equations
    Fnety=F1y+F2y
    Fg= Mg
    a=(Fx-Ff)/m
    Fnet=0=Fn+Fsin45-Fg


    3. The attempt at a solution
    I have no idea where to start
     
    Last edited: Sep 28, 2012
  2. jcsd
  3. Sep 27, 2012 #2

    Simon Bridge

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    You know that ##\vec{F}_{tot}=\vec{F}_1 + \vec{F}_2##

    You are given ##\vec{F}_{tot}## and ##\vec{F}_1##, your task is to find ##\vec{F}_2## ... what's the problem?
    Hint: how do you subtract two vectors?
     
  4. Sep 27, 2012 #3
    I have Ftotal by using a= (fx-Ff)/m but the answer in my textbook says its 104N [S3.3*W] I have no idea how they got there. Because Ftotal is 50N based on the equation a= (fx-Ff)/m

    I'm aware that the Mass times the Acceleration gives you to total force but i'm getting no where near 104
     
    Last edited: Sep 27, 2012
  5. Sep 27, 2012 #4

    Simon Bridge

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    ##\vec{F}_{tot}=m_b\vec{a} = (0.250\text{kg})(200\text{m}\cdot\text{s}^{-2} \text{ [W15*S]}) = 50\text{N [W15*S]}## ... is what I get. Notice that I used a different reasoning to you - and how I am being pedantic about the vectors having direction.

    But ##\vec{F}_{tot}## is not what you are asked to find is it?
     
    Last edited: Sep 28, 2012
  6. Sep 28, 2012 #5
    No, Thats the resultant force of the two. I'm asked to find the force exerted on the ball by the second player so its going to look something like 100N [N25*W]= Force of first soccer player

    104N[S3.3*W] is the second players exerted force on the ball somehow (I checked the back of my textbook)

    I have no idea how they got there

    I even broke down the X & Y components of it to see if in the end it would give me 50N but to no avail

    i did 100cos25+ 104cos3.3 = X
    100sin25-104sin3.3=Y
    (Because its in the opposite Y direction)

    But i don't end up with 50 when i do Pythagorean theorem
     
  7. Sep 28, 2012 #6
    I got √62+1032=103.17N
    θ=ArcTan(6/103)=3.3°(S3.3W)
     
    Last edited: Sep 28, 2012
  8. Sep 28, 2012 #7
    What was your method? what did you do?
     
    Last edited: Sep 28, 2012
  9. Sep 28, 2012 #8
    Resultant of x is 50Cos15°
    Force by first man is 100Sin25°

    50Cos15°-100Sin25°
     
  10. Sep 28, 2012 #9
    why are you using 100"Sin"25 if its the x coordinate you are looking for? because you are adding them
     
  11. Sep 28, 2012 #10
    If one player applies a force of 100N[N25*W]
    -------------------

    Start from pointing North and turn 25°(CCW) to the West(assume to be at pointing left).

    The resultant is [W15*S], start at pointing West and rotate 15° to South.
    F=ma=.250 x 200=50N
     
    Last edited: Sep 28, 2012
  12. Sep 28, 2012 #11
    I meant why are you using the SIN function, i was taught that you always use cos if you are doing X component and always use Sin for Y component, so i know why you used that force,

    But why Sin function?
     
  13. Sep 28, 2012 #12
    It depends on which angle you choose.
    Sinθ=Cos(90°-θ)

    100Sin25°=100Cos65°
     
  14. Sep 28, 2012 #13
    I'm having trouble understanding the addition part

    Firstly you came up with 50 by doing the Ftotal=mass x acceleration im assuming right?

    Resultant x would then yes be 50cos15 but the force of the first dude is 100N[N25*W] how are you to know to assume to use SIN ?

    2958rqq.jpg
    terrible but rough drawing
     
    Last edited: Sep 28, 2012
  15. Sep 28, 2012 #14

    Simon Bridge

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    To understand the method ... draw the vectors head-to-tail - only for a vector subtraction rather than a vector addition. The trick with subtraction is to realise that it is the same thing as addition by a negative number - and a negative vector points the opposite way.

    So you are doing:
    ##\vec{F}_2 = \vec{F}_{tot} - \vec{F}_1 = \vec{F}_{tot} + (^-\vec{F}_1)##

    You just add ##\vec{F}_{tot}## to ##^-\vec{F}_1##, both of which you know, in the normal way that you add vectors.
    Actually draw the vectors to see the geometry.

    If you don't know how to do vector addition then that's another problem.
    (Resolve vectors into N-S and E-W components.)
     
  16. Sep 28, 2012 #15
    Yes i know how to add vectors.... i'm pretty sure i wouldnt get a question like this for homework if i didnt
     
  17. Sep 29, 2012 #16

    Simon Bridge

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    :)

    In which case you also know how to subtract vectors.

    So the main issue is actually geometry?

    This is what I was trying to get you to talk about before. I would have liked to see your attempt to subtract vectors before continuing.

    Some people find that they work better if they change compass directions into x-y directions - and put all angles anticlockwise from the x axis. That way you just remember that the x component is the cosine and the y component is the sine.

    However - it is more powerful to be able to just look at the diagram and choose the angles you want to use. Here's an example:

    If a force X is magnitude a, angle A south of due west, then there is a component pointing to the west which has magnitude acos(A) and a component pointing south which has magnitude asin(A).

    If another force Y is magnitude b, angle B west of due south, then force (-Y) has the same magnitude pointing in the opposite direction - east of due north. That gives (-Y) a component bcos(B) to the north and bsin(B) to the east.

    X-Y= X+(-Y) = [acos(A)-bsin(B)]W + [bcos(B)-asin(A)]N

    It is confusing unless you clearly sketch the diagram.

    Once you have the triangle of the vectors you are adding - the rest should be a matter of choosing the method. There are many ways: you don't have to resolve into components. You could use the cosine rule to get the magnitude, for eg, and the sine rule to get the angle.
     
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