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Coefficient of Static Friction = tan (angle of incline)

  1. May 10, 2016 #1
    1. The problem statement, all variables and given/known data
    I am trying to prove that the coefficient of static friction is equal to the tan of the angle of incline. (You can find the proof of this from )

    I set the angle of incline as my independent variable and had an angle range from 10 to 37.5 degrees. After setting the slope to different angles, I measured the extra force required to cause the wooden block to begin to move on the slope. I did this by connecting a string to the wooden block and to a container that could be filled with sand (using a pulley to connect them).

    2. Relevant equations

    μ = (mg sin(θ) + Mg)/(mg cos (θ))

    where m is the mass of the wooden block and M is the mass of the handing container and sand.

    This simplifies down to μ = tanθ + M/(m cosθ)

    However, it is also known that μ = tanθ

    Equating the two equations we get tanθ + M/(m cosθ) = tanθ, which is impossible. Can anyone explain what I've down wrong here?

    3. The attempt at a solution

    I tried manipulating the equation,

    M/(m cosθ) = μ - tanθ

    => M = μmcosθ - msinθ

    => M = m(μcosθ - sinθ)

    Ultimately, I aim to draw a graph which shows μ = tanθ, however, with the values I obtained so far, no such graph can be drawn.

    I would really appreciate it if someone could help me!
     
    Last edited: May 10, 2016
  2. jcsd
  3. May 10, 2016 #2

    BvU

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    Hello LCC, :welcome:

    Nice experiment ! Well described in this your first post, kudos !
    Sounds like you filled until m started to move upwards along the slope. am I right ? If so, do ##\mu m g \cos\theta## and ##mg\sin\theta## point in opposite directions, as your equation suggests ?
     
  4. May 10, 2016 #3
    Hi BvU,

    Yes, you got the idea right, except I set it up so that m could start to move downwards along the slope.

    I am not sure to be honest. I do know however, mgcosθ and mgsinθ are perpendicular to each other because they represent the vertical and horizontal components of force due to the weight of m. Does this help?
     
  5. May 10, 2016 #4

    BvU

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    This is for the situation as in the video when there is no extra force involved !
    Ok, so your μ = tanθ + M/(m cosθ) has the right sign and you have a set of observations of M as a function of ##\theta##. You can investigate if ##\mu## depends on ##\theta## (*). But if you want to show that ##\mu = \tan\theta## directly, you'll have to find a way to vary ##\mu## and work with M = 0.

    (*)
    ##\mu = \tan\theta## doesn't mean that ##\mu## varies with ##\theta##; it means that the angle at which sliding is about to start has a tangent with a value that is equal to ##\mu##.
     
  6. May 12, 2016 #5
    If the pulley is at the top of the incline and you are adding mass M to try to pull the block up the incline, then the force balance on the block is
    $$Mg-mg\sin{\theta} = F \leq mg\mu cos{\theta}$$where F is the friction force. So, the coefficient of friction satisfies the inequality
    $$ \mu \geq \frac{M}{m}\sec{\theta}-\tan{\theta} $$and M satisfies the inequality:
    $$M\leq m(\sin{\theta} + \mu cos{\theta})$$
    The equal sign applies when the block is just on the verge of sliding.
     
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