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Forces on Particle, Find Third Force?

  1. Sep 26, 2004 #1
    A lemon has three horizontal forces that act on it as it is on a frictionless table. Force F1 has a magnitude of 4.50 N and is at Angle1 = 26°. Force F2 has a magnitude of 7.00 N and is at Angle2= 26°. The lemon half has mass 0.0250 kg.
    Picture is Attached.
    What is the third force if the lemon half has the following velocities?


    I did f1=4.5cos26i + 4.5sin26j
    f2=7cos(-64)-7sin(-64)

    THen I did f=(4.5cos26+7cos(-64))i+(-7sin(-64)+4.5sin36)j=7.11i+(-4.32j)

    (a)At Zero velocity i got the I component to be -7.11 and the j component to be 4.32.
    The j component is right but the i component is wrong and I don't know why.

    (b)constant velocity v = (13.0 i - 14.0 j) m/s
    (c)
    varying velocity v = (13.0t i - 14.0t j) m/s, where t is time in seconds

    What would I do to solve for B and C?
     

    Attached Files:

    Last edited: Sep 26, 2004
  2. jcsd
  3. Sep 26, 2004 #2
    Your reference angles may be wrong. I would make all angles with respect to the polar axis (positive x axis) so that 26 degree angle would become 116. I'm not sure if I understand what part a is asking. What is the force if the velocity is 0i+0j? Is that it?

    Just a little hint for part c (part b is essentially part a over again): take the derivative of their varying velocity equation to get the constant vector acceleration
     
    Last edited by a moderator: Sep 26, 2004
  4. Sep 26, 2004 #3
    yes that is what they are asking. I don't understand why you would use 116 as the reference angle, either way I still get the wronganswer. ANd I don't understand what you mean by part b being part a again.
     
    Last edited: Sep 26, 2004
  5. Sep 26, 2004 #4
    Sorry. I meant angle 1 is (180-26) degrees with respect to the polar axis and angle 2 is (-90+26) degrees with respect to the polar axis. use those and you should get a correct answer.

    Part b is exactly part A because from what I can see the velocity is still constant meaning no net force is being applied to it.
     
  6. Sep 26, 2004 #5
    I don't see how using 154 as angle 1 would make the answers correct, since using angle1 as 26 degrees gave me the right j component.
     
  7. Sep 26, 2004 #6
    Using it as an angle was correct because: sin(180-theta) = sin(theta) which you used in your j calculations correct? However, cos(180-theta) = -cos(theta) which messes you up on the i calculation. You basically found the force as if it was 26 degrees with respect to the polar axis. It is very important to choose the right angle.
     
  8. Sep 26, 2004 #7
    so the i component would be (4.5cos154)+7cos(-64)?
     
  9. Sep 26, 2004 #8
    I believe so.
     
  10. Sep 26, 2004 #9
    It's wrong:( I don't understand. HOw can I get the J component and not the I!
     
  11. Sep 26, 2004 #10
    hang on I'll work out the question. Give me a moment. Edit: is the homework online? "so the i component would be (4.5cos154)+7cos(-64)?"

    Ok I worked it out. the i component IS what you mentioned (approximately -0.975i) but remember this value is -F3*cos(theta) because F1*cos(theta)+F2*cos(theta)+F3*cos(theta) MUST equal 0.

    Now work out the j component. That would be -F3*sin(theta). See if you can't figure out the two unknowns given the two equations.
     
    Last edited by a moderator: Sep 26, 2004
  12. Sep 26, 2004 #11
    online through webassign
     
  13. Sep 26, 2004 #12
    (see above edit)
     
  14. Sep 26, 2004 #13
    i dont understand
     
  15. Sep 26, 2004 #14
    What don't you understand specifically? You know since the object is stationary that it is in static equilibrium meaning that the sum of all the i components of the forces must equal zero as well as the sum of all the j components of the forces must equal zero (by Newton's second law). You found what the sum of forces F1 and F2 were but now you have to find F3 with components such that F1+F2+F3 = 0 in both the i and j directions. Manipulating the equation we find F1+F2 = -F3 or -(F1+F2) = F3.
     
    Last edited by a moderator: Sep 26, 2004
  16. Sep 26, 2004 #15
    so if that's the case then wouldn't the i component of the third vector be just .976?
     
  17. Sep 26, 2004 #16
    Yep it sure would.
     
  18. Sep 26, 2004 #17
    It was right, thank you so much. I'm sorry, but I just really don't understand this. Now for part b you said it was essentially the same as part a, so I thought you would just take f1+f2+f3=(13.0 i - 14.0 j) m/s but it's not working.
     
  19. Sep 26, 2004 #18
    Well consider what you're doing. 13.0i - 14.0j is the velocity right? You can't just set forces equal to velocity. What does a constant 13.0i-14.0j velocity tell you about the net force?
     
  20. Sep 26, 2004 #19
    the net force is constant?
     
  21. Sep 26, 2004 #20
    Well technically it is constant BUT consider what the velocity-time graph would look like. A straight line across would be it right? If acceleration is given by dv/dt, what is the acceleration?
     
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