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!

Coefficient of Kinetic Friction

  1. Sep 30, 2012 #1

    MG5

    User Avatar

    Heres the problem,

    A 1,440-N crate is being pushed across a level floor at a constant speed by a force of 270 N at an angle of 20.0° below the horizontal

    (a) What is the coefficient of kinetic friction between the crate and the floor?

    I know how to do this. The answer is .165

    Heres how I solved it..

    270sin20 = 92 N
    270cos20 = 254 N

    1440N + 92N = 1532N

    Ff=uk(n)

    254N=uk(1532)

    uk= .165

    The only thing I don't quite get is why do I use 254N, the product of 270cos20, as the Ff and why is 1532N the normal force?

    Thanks.
     
  2. jcsd
  3. Sep 30, 2012 #2
    The only thing I don't quite get is why do I use 254N, the product of 270cos20, as the Ff and why is 1532N the normal force?
    ...................
    You have chose the values. Surely you have reasons for choosing them.
    I think the questions should be are your reasonings correct?
     
  4. Sep 30, 2012 #3

    MG5

    User Avatar

    I knew someone was going to say this. The reason I don't know why I'm using those numbers that I solved for is because I saw the solution to the problem worked out. I understand everything but why those numbers are used for Ff and n. Other than that everything else seems easy and makes sense.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Coefficient of Kinetic Friction
Loading...