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!

Coefficients of friction

  1. Oct 16, 2003 #1
    *note* This question has been deleted.
     
    Last edited: Oct 19, 2003
  2. jcsd
  3. Oct 17, 2003 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You appear to have correctly calculated the components of gravitational force acting on the two masses normal to and along the incline, though you may have gotten your right triangle confused (it is gravity itself that is "directly down", not the pull on the box: For the 8 kg mass they are FS= 8(9.8)sin(35) (FS is the force in the direction of the incline) and
    FN= 8(9.8)cos(35). For the 3.5 kg mass they are
    FS= 3.5(9.8)sin(35) and FN= 3.5(9.8)cos(35).

    You almost said it! The normal force is mg cos(35) (not just mass times sin- mg times cosine-look at your diagram) and the friction force is the coefficient of friction times that. The problem is that, here, you are not given the coefficient of friction- that's the "unknown" you are asked to find. Let's call it "c". The friction force on the 8 kg mass is 8(9.5)cos(35) and for the 3.5 kg mass it is 3.5(8.9)cos(35).

    The "S" component of gravitational force on the 8 kg mass is pulling it down the incline, the "S" component of gravitational force on the 3.5 kg mass is holding it back (through the rope) and both frictional forces are holding it back. Taking the positive direction down the slope, we have for the total force
    F= 8(9.8)sin(35)- 3.5(9.8)sin(35)- 8(9.8)c cos(35)- 3.5(9.8)c sin(35)
    = (8-3.5)(9.8)sin(35)- (8-3.5)(9.8)c cos(35)
    = 4.5(9.8)sin(35)- 4.5(9.8)c cos(35).
    We are told that the 8 kg mass has acceleration 1.5 m/s2 so we must have (F= ma) 4.5(9.8)sin(35)- 4.5(9.8)c cos(35)= 8(1.5).
    Solve that equation for c.

    Of the four forces, two (the "S" component of gravity on the 8 kg mass and the friction force of the 8 kg mass) are acting directly on the 8 kg mass. The other two (the "S" component of gravity on the 3.5 kg mass and the friction force on the 3.5 kg mass) have to act through the string connecting the two mass. The total of those two forces is the tension in the string.
     
  4. Oct 17, 2003 #3
    I found the normal force on the 3.5 kg block to be 28.1 N and the normal force on the 8.00 kg block to be 64.2 N. Does this play at all into the equation? I was using F=coefficient of friction(normal force) before. When I went through the formula you suggested, I found the coefficient of friction to be .493.

    In the book, the coefficient of kinetic friction is .0871 and tension is equal to 27.4 N.

    When I try to get tension using the "S" component of gravity on the 3.5 kg mass and the friction force on the 3.5 kg mass, I figured the "S" component to be (3.5)(9.8)cos(35). I wasn't sure about the friction force, though. The friction force calculated above was for the whole system .. can still take (4.5)(9.8)sin(35)-(4.5)(9.8)cos(35) to get it?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Coefficients of friction
  1. Friction coefficient (Replies: 1)

Loading...