1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding distance traveled up a ramp

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Untitled.png

    2. Relevant equations
    ΣFΔt=mΔv
    ΣF=ma
    v=d/t

    3. The attempt at a solution
    I already found the following for previous questions:
    mg=-661.9
    mgx=-3594.37
    mgy=-649.91
    Fx=2192.69
    Fy=-1250.69
    FF=-365.44
    Vf=257.18
    ΣFx=-1767.12

    ATTEMPT #1
    v=d/t
    d=vt
    d=(257.18)(10)
    d=2571.80 (incorrect)

    ATTEMPT #2
    ΣFΔt=mΔv
    ΣFΔt=m(d/t)
    [ΣFΔt/m](t)=d
    [(-1767.12)(10)/67.4]10=d
    -2621.84=d

    ATTEMPT #3
    ΔV=Δd/Δt
    ΔvΔt=Δd
    (257.18-5)(10)=Δd
    2521.80=Δd

    I am not sure where I am going wrong!
     
    Last edited: Feb 17, 2016
  2. jcsd
  3. Feb 17, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Your x and y components of the weight are greater than the weight.
     
  4. Feb 17, 2016 #3
    IMG_0105.JPG

    I keep getting the same things (Sorry about the bad quality photo, but I figured this was best so the triangles I made can be seen).
     
  5. Feb 17, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    In your pink triangle drawn on the inclined plane, note that mg is the hypotenuse. But in the work shown in the blue rectangle you have mgx as the hypotenuse.
     
  6. Feb 17, 2016 #5
    If I use the pink triangle again for mgx and use cos I get mgx=121.63.
    But I'm confused because I still got 3594.37 for mgy & thats much larger than mg (661.19) in the first place, so is that one incorrect?

    EDIT: mgy has to be correct because I had to calculate the force of friction in a previous question with it and obtained the correct answer!
     
  7. Feb 17, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    In the pink box you got the correct value for mgy.

    Your value for mgx now looks correct also.
     
  8. Feb 17, 2016 #7
    Okay, thanks!

    So am I supposed to use both when solving for distance or just the stuff in the x direction? I was only using the stuff in the x direction before.
     
  9. Feb 17, 2016 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You need to use the y direction information to help get the friction force. But once you have the friction, you just need to consider ΣFx = max.
     
  10. Feb 17, 2016 #9
    Okay so:

    ΣFx=max
    Fx-FF-mgx=max
    2192.69-365.44-121.63=67.4[(vf-vi)/Δt]
    (1705.62/67.4)10=vf-vi
    253.06=Δv

    And this is where I get lost... I am stuck solving for a distance, I don't know how to proceed.
     
  11. Feb 17, 2016 #10
    Never mind, I figured it out!

    d=[(vi+vf)/2]t
    d=[(5+257.18)/2]10
    d=1310.90m
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted