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Sled pulled up incline at an angle with friction

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A child pulls a 3.6 kg sled at 25 degrees to a slope that is 15 degrees to the horizontal, as in the figure. The sled moves at a constant velocity when the tension is 16 N. What is the acceleration of the sled if the rope is released?


    2. Relevant equations

    F=ma, f=[itex]\mu[/itex]N


    3. The attempt at a solution

    Constant velocity means that friction + the component from Fg shall equal FTx (tension component parallell to plane) that is FTx = f + Fx

    Where FTx=cos(25)*FT=14.5 N and Fx=3.6*9.8=9.13 N

    Now I want [itex]\mu[/itex] so that I can calculate f. And we know that N=cos(15)*mg+T*sin(25)=40.84 N

    If we go back to the start, FTx=[itex]\mu[/itex]N+Fx so [itex]\mu[/itex]= [itex]\frac{ (FTx-Fx) }{N}[/itex]= [itex]\frac{14.5-9.13}{40.84}[/itex] = 0.131

    When we release the sled N will reduce to N=mg*cos(15)=34.08 and so f=0.131*34.08=4.46 N

    Fres=Fx-f=9.13-4.46=4.66 N

    And the acceleration is a=[itex]\frac{F}{m}[/itex]=[itex]\frac{4.66}{3.6}[/itex]=1.3 m/s^2

    But apparently this is not correct. Could someone please tell me what I've missed, this question has been very frustrating for me :(

    Also, negative signs tend to mess upp my Latex, I apologize.
     
  2. jcsd
  3. Oct 12, 2011 #2
    Here's the figure btw.
     

    Attached Files:

  4. Oct 18, 2011 #3
    Anyone?
    According to the answer sheet the correct answer should be 0.67 m/s^2 but i'm starting to wonder whether that's true or not?
     
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