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Vector help please

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    #2
    Two people pull on a stubborn mule, as seen from a helicopter in Figure 1. Find (a) the single force that is equivalent to the two forces shown, and (b) the force that a third person ould have to exert on the mule to make the net force equal to zero.

    Figure 1:
    [​IMG]
    the blue on the left and right, represent for a person.


    #3
    A car travels due east with a speed of 50km/h. Rain is falling vertically with respect to the Earth. The traces of the rain on the side windows of the car make a angle of 60* with the vertical. Find the velocity of the rain with respect (a) the car and (b) the Earth.
    2. The attempt at a solution
    I am totally got no idea with the problem #1. Can you guys give me some hints or help me please?

    #2
    Is this right right picture for the problem ?
    [​IMG]
     
    Last edited: Sep 20, 2007
  2. jcsd
  3. Sep 20, 2007 #2
    Problem number 1: hint/method = find the components

    Problem number 2: it is right geometrically, but why do you have it oriented all strange like?
     
  4. Sep 20, 2007 #3

    learningphysics

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    For the first problem, the angles don't look right. The blue angle is the smaller angle in the picture... but it is 60, while the bigger angle is 45...

    For the second problem... Yes, your picture is correct. If you are unsure though, I recommend writing the vectors in i/j form... [tex]\vec{i}[/tex] for east west... [tex]\vec{j}[/tex] for the vertical direction... take r as the speed of the rain... so what is the velocity of the car with respect to the earth... what is the velocity of the rain with respect to the earth (in terms of r)... i,j are just unit vectors...
     
  5. Sep 20, 2007 #4
    Problem 1 fixed
     
  6. Sep 20, 2007 #5

    learningphysics

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    cool. now you should be able to get the components, as Mindscrape mentioned.
     
  7. Sep 20, 2007 #6
    What they mean by Find (a) the single force that is equivalent to the two forces shown, and (b) the force that a third person ould have to exert on the mule to make the net force equal to zero.
     
  8. Sep 20, 2007 #7
    They mean, if you could express the two forces at those two angles, as just one force at one angle, what would that force be? Then they want to know what force cancels it.
     
  9. Sep 20, 2007 #8

    learningphysics

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    the resultant of the two forces. The sum...
     
  10. Sep 20, 2007 #9
    They mean to find the resultant of the forces.In problem 1. While in problem 2 they are asking you to find a force which is equivalent to the resultant of the other two forces in magnitude but opp in direction. So for all the forces the following should hold.

    [tex]\sum{F_x}[/tex]=0

    [tex]\sum{F_y}[/tex]=0
     
  11. Sep 20, 2007 #10
    Got it !
    your explaination is a little complicated !

    Got the first one, but totally out of the second one ! It too complicated to me!


    Thanks you all ! You guys have a good day !
     
  12. Sep 20, 2007 #11
    Sorry if it is complicated, but you need to realize it's all relative. If the mule were in a black box how would he know whether it was two people pushing on him, 8 people pushing on him, or just one person pushing on him? The particle, mule in your problem, doesn't care how many forces are acting on it, it just cares about the sum of the forces, and how they are superposed onto it.
     
  13. Sep 21, 2007 #12

    learningphysics

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    You've got the resultant of the two forces... now you need a force out the original two forces (which is exactly the same as the force to cancel out the resultant of the first two forces)...

    If you have a vector... and you want another vector to add to it, to make the sum zero... then you add a vector of the same magnitude but opposite direction...

    so part b), is just the same as part a), except the opposite direction.
     
  14. Sep 26, 2007 #13
    I thought I got it, but you confused me !
    #1
    a) after i found the component of first force and second force, then just add the y component to get the resultant ! Is that right
    b) still confuse
     
  15. Sep 26, 2007 #14
    No, so for the first one you need to add all the x components and y components:

    [tex]F_{net_x} = F_1_x + F_2_x[/tex]

    and

    [tex]F_{net_y} = F_1_y + F_2_y[/tex]

    Then the net resultant is

    [tex]||F_{net}|| = \sqrt{F_{net_x}^2 + F_{net_y}^2}[/tex]

    What would the direction be?

    How would you find the force that cancels F_net?

    What are you confused about on the second one?
     
  16. Sep 26, 2007 #15
    F_net = 198 N
    I only get that
     
  17. Sep 26, 2007 #16
    What do you mean you only get that? I don't know if that is the only part you understand, or the only part you have gotten because it's the only one I explicitly told you how to do. If you want the direction, think about trig and how to find angles of a triangles. What do your x and y vectors look like with your resultant vector, a bit like a triangle?
     
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