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Physics Lab Question - Where To Begin

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    In class, we conducted a lab experiment in which we placed an object on the end of a piece of wood and lifted the same end of the wood until the object slid down to the bottom. We measured the height the wood was lifted to. (to assist in calculating Coefficient of Friction for kinetic force) We then repeated this step, and quickly brought the piece of wood down so as to stop the object from moving, and measured this height. (to assist in calculating Coefficient of Friction for static force) The length of the wood was measured (which would act as "r" or "radius" or "hypotenuse" in this case).

    We repeated the whole routine above for four additional different objects, for a total of five objects' values.

    The trouble I am having is a question asked of us as part of the lab questions, based on this experiment and the collected data.

    Q1: From the value of [tex]\theta[/tex] (which I calculated already), determine the coefficient of friction [tex]\mu[/tex] (static) and [tex]\mu[/tex] (dynamic), as indicated below:

    a) Draw the force diagram of a weight in equilibrium that is on the inclined plane. Set up the x and y axis parallel to the surface respect to the inclined plane. Then set up two equations from these forces from the equilibrium equations

    [tex]\Sigma[/tex]Fx = 0
    [tex]\Sigma[/tex]Fy = 0

    then solve for [tex]\mu[/tex] (coefficient of friction)

    b) Using the formulae derived in (a) above, solve for [tex]\mu[/tex].

    2. Relevant equations

    F = [tex]\mu[/tex]*N
    [tex]\mu[/tex] = F/N = m*g*sin[tex]\theta[/tex] / m*g*cos[tex]\theta[/tex]
    [tex]\mu[/tex] = sin[tex]\theta[/tex] / cos[tex]\theta[/tex]
    [tex]\mu[/tex] = tan [tex]\theta[/tex]

    3. The attempt at a solution

    No attempt has yet been made, as I do not know where to begin, nor understand what I am being asked to do. I've read the textbook, which does not help me determine what to do...

    Do I take the sum of all 5 values for static and dynamic friction coefficients and determine an average [tex]\mu[/tex] ?
    Do I draw 5 force diagrams of a weight in equilibrium for each object, and then solve for [tex]\mu[/tex] by using specifically only the values pertaining to that one object?
    Am I going to subtract static friction force from dynamic friction force? Average them?
    And by "formulae derived" are they referring to something I have to derive myself from the given formulae, or the very same formulae they gave me (i.e.: did they derive the formulae for me already)?
     
  2. jcsd
  3. Oct 14, 2009 #2

    Delphi51

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    Homework Helper

    The tan formula you have will do the whole job. Just use it twice for each object.

    Looks like you are asked to show in detail how you got that formula, which will include the force diagram.

    It seems to me that the first run (lift until it goes) is a measure of the static friction rather than the kinetic.
     
  4. Oct 14, 2009 #3
    Thank you for the clarification.

    If the first run is a measure of static friction, how is the second run (lift up til it goes then lower until it stops) kinetic friction? It seems they would be the opposite, but it's possible I have opposing ideas compared to reality. (confusing/mixing up each concept)
     
  5. Oct 14, 2009 #4
    When the thing is moving, and you lower the board, you're dealing with kinetic friction because it's sliding. The angle will be less than it was before, because the kinetic frictional force is weaker and can't balance as big a gravity component as the static could.
     
  6. Oct 14, 2009 #5

    Delphi51

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    Homework Helper

    What merryjman said - for the second part you use the angle that just stops it. Kinetic.
     
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