1. Limited time only! Sign up for a free 30min personal 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!

Calculating force needed to push sled down incline

  1. Oct 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the force required to push a 22kg sled down a 6 degree hill and have an ending speed of 60km/hr after traveling 75m when the coefficient of friction is 0.20.

    2. Relevant equations
    Fnet= ma
    (V2)^2 = (V1)^2 + 2a(d2-d1)
    Ff= uN
    W= mg

    3. The attempt at a solution
    I'll use @ for theta, and u for mu. Fa= force applied to the sled, Ff=force of friction

    With N lying on the positive Y axis and Weight in the x direction lying on the positive x axis, I set Wy=wcos@ and Wx= wsin@, so N=wcos@.

    I tried summation of forces:
    xFnet = Wx + Fa - Ff = ma
    Fa= Ff - Wx + ma
    Fa = uN - wsin@ + ma
    Fa = u(wcos@) - wsin@ + ma
    Fa = umgcos@ - mgsin@ + ma
    Fa = m(ugcos@ - gsin@ + a)

    I then tried to solve for a using (V2)^2 = (V1)^2 + 2a(d2-d1)
    Since the starting velocity was 0 and the starting distance is 0,
    A = (V2)^2 / (2d2)

    I converted 60 km/hr to m/s, to get 16.66 m/s.
    A= (16.66^2)/ 2(75) = 1.85 m/s^2

    Plugging that in...

    Fa= 22 ( (0.2)(-9.8)(cos6) - (-9.8sin6) + 1.85 ))
    I got Fa = 20.36 N, though that isn't correct. I'm not sure what I did wrong.
     
  2. jcsd
  3. Oct 23, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Careful with signs. g is just a positive constant.
     
  4. Oct 23, 2016 #3

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Check the signs of the forces. Looks like you have the required net force (ma) the same sign as the friction force (uN)?

    May not be the only problem but that looks wrong to me.
     
  5. Oct 23, 2016 #4
    Thank you. I've used -9.8 m/s^2 before in questions and have gotten the correct answer. How do I know when to use the positive/negative version?
     
  6. Oct 23, 2016 #5

    Doc Al

    User Avatar

    Staff: Mentor

    I would always use g as a positive constant and add signs as needed. For example, using down as negative, the acceleration of a falling object is -g = -9.8m/s^2.

    When you set up your force equations, always assign your signs consistent with whatever sign convention you are using. (For example, taking down the incline as positive would make the weight component, the applied force, and the acceleration positive and friction negative.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculating force needed to push sled down incline
Loading...