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Find the acceleration of an object climbing an incline...

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    The only given data is that the incline is 60 degrees steep. I also presumed that there is friction. I made an attempt to solve the problem and would like to know if it is correct.

    2. Relevant equations
    Fnet=m*a

    Ffk-kinetic friction force
    k-friction coefficient

    3. The attempt at a solution
    judging by the Ox Axis:

    Ffk+Gx=m*(-a)
    m*g*k+m*g*sin(60degrees)=m*(-a)
    m*g(k+sin(60degrees)=m*(-a)
    (m gets cancelled out)
    a=-(gk+gsin(60degrees))
    a=-9.8k-8.5

    (The acceleration in my case has a negative value since I chose to make the Gx and Ffk positive)
    Any criticism or correction would be much appreciated.
     
  2. jcsd
  3. Dec 17, 2015 #2

    PeroK

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    The answer is not correct. Gravity is only partly pushing the object into the slope. You have calculated too much friction.

    Why not leave it as a general angle ##\theta## for the time being and see what comes out?
     
  4. Dec 17, 2015 #3
    What makes the problem a bit disturbing for me is that so far for the Ox axis I have only put Gx and Ffk as the only forces acting on the body. Perhaps I need to add another force vector on the opposite side showing that something is pushing the body up?
     
  5. Dec 17, 2015 #4

    PeroK

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    You have gravity down the slope as ##mgsin(\theta)##, which is correct. But, you then have gravity into the slope as the full ##mg##, which can't be correct.
     
  6. Dec 17, 2015 #5
    From closer inspection I think I got it! Instead of equating Ffk with Gy*k I equaled it with G*k, so:

    m*g*cos(60)*k+m*g*sin(60)=m*(-a)
     
  7. Dec 17, 2015 #6

    PeroK

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    Yes. Now, let me show you something. I want to keep your answer as:

    ##a = -gkcos(\theta) - gsin(\theta)##

    You could now check that answer as follows:

    1) What happens if ##\theta = 0##? I.e. there is no slope. You get ##a = -gk## which is all friction.

    2) What happens if ##\theta = 90°##? You get ##a = -g## which is all gravity.

    If you plug all the numbers in and get an answer like 12.135 or whatever, then there is no way to check this is correct in any way. Leaving the symbols in at least gives you the chance to see whether things have gone wrong and where they've gone wrong.
     
  8. Dec 17, 2015 #7
    Hmm, I see. So dealing with symbols until the very last moment is more efficient as it helps us check the steps one by one. We lose that benefit when we replace them with numbers right away. Something I'll keep in mind. Thank you for the help!
     
  9. Dec 17, 2015 #8

    haruspex

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