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!

Force of friction help

  1. Jan 23, 2013 #1
    Force of friction help!!

    1. The problem statement, all variables and given/known data
    A force of 755N is exerted on a rope to pull a 45.0N wooden crate across a concrete floor. If the rope makes an angle of 35.0° with the floor and the crate is moving at a constant velocity what is the amount of the force of friction?


    2. Relevant equations
    F=ma


    3. The attempt at a solution
    So far I have x: F=Fcosθ-Ff=ma im not sure if thats the right equation given mg, Fa and angle
     
  2. jcsd
  3. Jan 23, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Re: Force of friction help!!

    Please don't use the same label (F in this case) for two different entities. Let's call the rope tension T. So, yes, you have Fx = Tcos(θ) - Ff = ma. What do you know about a?
     
  4. Jan 23, 2013 #3
    Re: Force of friction help!!

    It equals 0 because its at a constant velocity? I think haha
     
  5. Jan 23, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Re: Force of friction help!!

    Quite so.
     
  6. Jan 23, 2013 #5
    Re: Force of friction help!!

    So that means that Ff=Tcosθ?
     
  7. Jan 23, 2013 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Re: Force of friction help!!

    Yes. I assume there's more to this question. Do you have to find the coefficient?
     
  8. Jan 23, 2013 #7
    Re: Force of friction help!!

    no, just the force. im confused on how the weight of the box is not used whatsoever in the end equation?
     
  9. Jan 23, 2013 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Re: Force of friction help!!

    The only horizontal forces are friction and the horizontal component of the applied force. Therefore, the weight of the box is not involved.

    It is possible to find the coefficient of friction from the given information. To so this, the weight of the box would be involved.
     
  10. Jan 24, 2013 #9
    Re: Force of friction help!!

    Due to the weight of the box we get that the force of friction is a certain magnitude. From the fact that the box is not accelerating we then know that horizontal component of the pull must be equal to the force of friction and because we can calculate this from the information we have the force of friction. This is exactly how the force of friction is measured experimentally - pull the block along at a constant speed with a spring scale and take its reading. The force of friction is then eual to the reading on the spring scale if it was held horizontally during the pull.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force of friction help
  1. Force of friction help (Replies: 11)

  2. Friction Forces help (Replies: 8)

Loading...