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Can't figure out solution -- Space probe with engine thrust vectors

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data
    A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 25 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular?
    V0=0
    t1=25



    2. Relevant equations
    D=V0t+1/2at^2
    a=2T/m




    3. The attempt at a solution
    D= 0*t+1/2(2T/m)*25^2
    D=(T/m)*25^2







     
  2. jcsd
  3. Feb 11, 2016 #2

    berkeman

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    Welcome to the PF.

    You will need to use vectors to figure this out (or at least express the thrusts and distances in two dimensions). Can you draw two diagrams for the vector thrusts involved?
     
  4. Feb 11, 2016 #3
    How do I draw the force vectors on this site I have them on paper
     
  5. Feb 11, 2016 #4

    berkeman

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    If you can take a good picture of them (clear, bright, not fuzzy, etc.), you can use the UPLOAD button in the reply window to upload a JPG file from your PC or laptop. :smile:
     
  6. Feb 11, 2016 #5
    Here they are the upside down T is the perpendicular vector with the F in the middle and parallel are the ones are below.
     

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  7. Feb 11, 2016 #6

    berkeman

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    So what is the magnitude of the resultant force for the 2nd case? What is the ratio of the forces for the first and second case? What equation would you then use to figure out the ratio of the times it takes to travel a certain distance?
     
  8. Feb 11, 2016 #7
    I think it is pythagorean theorem for the magnitude of the resultant force of the second case. To solve for the distance I would use the equation D=Vo+1/2at^2 and since the distance is the same I can plug in the distance for the second case
     
  9. Feb 11, 2016 #8

    berkeman

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    You are on the right track. Use the fact that Vo=0 to help simplify that equation, so you can take ratios.

    Also, on your diagram, re-label the right-angle drawing with each of the two right-angle forces being "F", and the middle resultant will then be what by the Pythagorean theorem?
     
  10. Feb 11, 2016 #9
    The Magnitude of the vector in the middle labeled M in the picture
     

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  11. Feb 11, 2016 #10

    berkeman

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    So in the first case, M=2F. What is M in the second case? What does that give you for the ratio of the two times to go the same distance? What does that give you for the 2nd time? You're almost done! :smile:
     
  12. Feb 11, 2016 #11
    I have D=1/2(2T/m)*t^2 which reduces down to D=(F/m)*t^2 and for the second case is M equal to F?
     
  13. Feb 11, 2016 #12
    I have D=1/2(2T/m)*t^2 which reduces down to D=(F/m)*t^2 and for the second case is M equal to F?
     
  14. Feb 11, 2016 #13

    berkeman

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    No, M for the second case is not F. It is more than F and less than 2F. Look at your diagram and use the square you drew with F as each side and M on the diagonal (hypotenuse of the 2 triangles...).
     
  15. Feb 11, 2016 #14
    Would it be F/M meaning the cosine of 45 degrees
     
  16. Feb 11, 2016 #15

    berkeman

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    I'm not sure what you mean.

    If you have a right triangle with two sides of length 1, what is the length of the hypotenuse?

    I have to bail for a couple of hours. You are very close to solving this. I'll check back in later tonight to be sure you got the right answer. :smile:
     
  17. Feb 11, 2016 #16
    Would it be the square root of 2F
     
  18. Feb 11, 2016 #17

    berkeman

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    Written more precisely, yes it would be F√2.

    So finish it off now (if you haven't already). What is the ratio of the times? What is the time for the second scenario?
     
  19. Feb 11, 2016 #18
    I know that this answer is probably wrong but is it 84.089
     
  20. Feb 11, 2016 #19

    berkeman

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    Please don't make me do extra work :smile: Please show your detailed work with units included. Thanks.
     
  21. Feb 11, 2016 #20
    I have the ratio of the two and I have 2F/F
    I have the ratio of 2F/F(sqrt(2)) which reduces to the sqrt(2) sorry my computer doesn't show me the mathematical functions. so I put in what I got for distance earlier which is (F/m)*t^2=1/2*sqrt(2)*t^2
     
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