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!

Find coefficient of kinetic friction of block and ramp.

  1. Dec 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A small block is on the point of slipping down a rough ramp inclined at 35 degrees to the horizontal.
    a) What is the coefficient of static friction between the block and the ramp?
    Answer μ_s = 0.70
    b) The ramp is attached to the edge of a 0.8 m high table. After the block is given a small nudge it slides down the ramp and lands on the floor a horizontal distance of 0.4 m away from the edge of the table. What is the coefficient of kinetic friction between the block and the ramp?
    Answer μ_k = 0.62

    2. Relevant equations
    For part a)
    ∑ F_parallel to ramp = 0 N
    ∑ F _perpendicular to ramp = 0 N
    Static friction force F_f = μ_s* R

    For part b)
    ∑ F_parallel to ramp = ma
    ∑ F _perpendicular to ramp = 0 N

    Work energy theorem
    W_exerted = ΔE_mech +ΔE_friction = 0 J
    ΔE_mech = ΔKE + ΔU
    ΔE_friction = μ_k* R *d

    3. The attempt at a solution
    For part a)
    ∑ F_parallel to ramp = 0 N
    Therefore if the direction pointing down the ramp is positive, and perpendicular to ramp is positive. Then...
    0 = mg sin (35) - μ_s* R

    ∑ F _perpendicular to ramp = 0 N

    0 = R - mg cos(35)
    Hence mg sin (35) = μ_s*mg cos (35)
    Finally μ_s = tan (35) = 0.70
    Which is the correct answer, the major issue occurs within part b).

    For part b)

    Note the following leads to a line such that μ_k will cancel out!!! Hence another approach to
    achieving the solution μ_k = 0.62 must be used.


    I started by assuming the following;
    Vertical height of the block initially is y = 0.8 m, and when the block reaches the floor y = 0 m.
    Horizontal distance between edge of the table and the other end of ramp is x = 0.4 m.
    Ramp length is d = √ ( 0.8^2 + 0.4^2 ) = (2√5)/5 m.
    Finally incline is not 35 degrees but in fact arctan (0.8/0.4) ≅ 63.4 degrees.

    Using W_exerted = ΔE_mech +ΔE_friction = 0 J
    ΔKE + ΔU + μ_k* R *d = 0 J
    0.5 m v_f^2 - 0 + 0 - mgy + μ_k* R *d = 0 J note v_f is velocity final
    0.5 m v_f^2 - mgy + μ_k* R *d = 0 J
    ∑ F _perpendicular to ramp = 0 N
    R - mg cos (63.4) = 0 N
    Therefore
    0.5 m v_f^2 - mgy + μ_k* mg cos (63.4) *d = 0 J , d = (2√5)/5 m
    0.5 m v_f^2 - mgy + μ_k* mg cos (63.4) *(2√5)/5 = 0 J
    Eliminating mass
    0.5 v_f^2 - gy + μ_k* g cos (63.4) *(2√5)/5 = 0 J

    0.5 v_f^2 = gy - μ_k* g cos (63.4) *(2√5)/5
    v_f^2 = 2gy - 2μ_k* g cos (63.4) *(2√5)/5
    Replacing v_f^2 with u^2 + 2ad, where u^2 = 0 as initial velocity is zero parallel to ramp, and d = (2√5)/5 m
    2a(2√5)/5 = 2gy - 2μ_k* g cos (63.4) *(2√5)/5
    a(2√5)/5 = gy - μ_k* g cos (63.4) *(2√5)/5

    Using ∑ F_parallel to ramp = ma to substitute for a

    ma = mg sin (63.4) - μ_k* mg cos (63.4)
    a = g sin (63.4) - μ_k* g cos (63.4)
    a = g (sin (63.4) - μ_k* cos (63.4) )

    g (sin (63.4) - μ_k* cos (63.4) ) * (2√5)/5 = gy - μ_k* g cos (63.4) *(2√5)/5
    (sin (63.4) - μ_k* cos (63.4) ) * (2√5)/5 = y - μ_k* cos (63.4) *(2√5)/5 The line above μ_k cancels.
     
  2. jcsd
  3. Dec 30, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Welcome to PF!

    Could the setup for part (b) be as shown:
    upload_2016-12-30_14-53-56.png
    Was any information given about the initial height of the block on the ramp?
     
  4. Dec 30, 2016 #3
    Perhaps but no there is no info regarding the height of the block on the ramp if it is that set up.
     
  5. Dec 30, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    In part (b) it says that the block is given a small nudge to get started. This means that the block is initially on the verge of slipping as in part (a). So, I'm "inclined" to think that the ramp is still sloping at 35 degrees in part (b).

    Anyway, I don't think enough information is given to work the problem if the ramp is on top of the table as shown in the figure I posted.
     
  6. Dec 30, 2016 #5
    I agree, there isnt enough info for that figure. Also the reason i assumed the angle changes is due to the fact that if theta is 35 degrees length x and height y do not correspond to the correct values.
     
  7. Dec 30, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Too bad they didn't provide a figure. If the setup is as I indicated, then working backward from the given answer, I get that the block must have been placed at a vertical height of 1.0 m from the base of the ramp (i.e., 1.0 m above the table top).
     
  8. Dec 30, 2016 #7
    And what was your approach?
     
  9. Dec 30, 2016 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    How would you approach it?
     
  10. Dec 30, 2016 #9
    As before using work energy theorem, however what of the 0.8 m height of the table itself?
     
  11. Dec 30, 2016 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    What can you learn about the motion of the block at the bottom of the ramp from the fact that the table is 0.8 m high and the block lands 0.4 m horizontally from the edge of the table?
     
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: Find coefficient of kinetic friction of block and ramp.
Loading...