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Incline solving for xf

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15°. The sand enters a pipe 3.0 m below the end of the conveyer belt, as shown in the figure below. What is the horizontal distance d between the conveyer belt and the pipe?

    Xi = 0 m
    Ti = 0 s
    Viy = 6.0 sin θ
    Yi = 3 m
    Vix = 6.0 cos theta
    Xf = ???
    Tf = t
    Vf =
    Yf = 0

    2. Relevant equations

    Yf = Yi + Viy(t) - 1/2g(t)2
    Xf = Xi + Vix(t)

    3. The attempt at a solution

    3 + 6.0sin15t -4.9t2

    t= .956797

    Xf = 0 + 6*cos15*.956797 = 5.54 meters

    not it
     
  2. jcsd
  3. Feb 2, 2013 #2

    TSny

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    Did you mean to write an equation here? What is the sign of the initial y-component of velocity?
     
  4. Feb 2, 2013 #3
    Hi.

    Yf = Yi + Viy(t) - 1/2g(t)2

    0 = 3 + 6.0sin15t -4.9t2

    the sign on initial velocity is +
     
  5. Feb 2, 2013 #4

    TSny

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    The problem says that the sand moves down the conveyor. So, the y-component of velocity should be down.
     
  6. Feb 2, 2013 #5
    I thought that the direction was accounted for in setting Yi = 3 and Yf = 0.

    so my proper equation 0 = 3 - 6.0sin15t -4.9t2
     
  7. Feb 2, 2013 #6

    TSny

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    That looks good. Note that you took yf =0 and yi = 3 which is correct for taking upward as positive. You also used -9.8 m/s^2 for the acceleration which agrees with taking upward as positive since the acceleration is downward. Likewise, the y component of the initial velocity is downward, so it should be negative when taking upward as positive.
     
  8. Feb 3, 2013 #7
    Is this correct
    Yi = Yo + Voy (Tf - Ti) - (1/2)(g)(Tf-Ti)^2

    Because using -g turns my squared term positive
     
  9. Feb 3, 2013 #8
    Nevermind. Thx u were a great help
     
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