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Car on a hill - reasoning for static friction

  1. Apr 14, 2017 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Problem :

    A car of mass ##m## is parked on a slope.

    What does static friction ##\mu_s## have to be for the car not to slide down if
    the incline of the hill is ##30## degrees?

    *******

    For this problem I have the following forces, where the hill is inclined upwards
    to the right:

    Upwards : normal force of ##mg \cos(\theta)##

    Downwards : Force due to gravity of ##mg \cos(\theta)##

    Right (so up hill): Resistive Force due to friction ##\mu_s mg \cos(\theta)##

    Left (down hill): Force due to gravity : ##mg \sin(\theta)##.

    For the car to remain in place I need to have the frictional force to be greater
    than the gravitational, if I set these up as

    $$
    \mu_s mg \cos(\theta) = mg \sin(\theta)
    $$

    Then reduce to
    $$
    \mu_s = \tan(\theta)
    $$

    Where ##\theta = 30## (degrees) , so ##\mu_s = \frac{\sqrt{3}}{3}##, which is around
    ##0.58##. I *think* that this is a reasonable coefficient for friction? It's less
    than 1, though I've not really got anything to go on there.

    I'm not sure what else I could do / say about this problem either. I think that
    I've considered all the forces, noted that the maximum that friction could be is
    that of the equality ##\mu_2 \cos(\theta) = \sin(\theta)## (I've dropped ##mg##
    there...)

    Static frictional force can change - whereas the kinetic friction is constant.
    However - both coefficients ##\mu_s## and ##\mu_k## are themselves constants, so my
    equation to find the coefficient ##\mu_s## is valid.

    I've attached sketches of this problem as they were requested in the equation.

    One other thing that I'm slightly unsure of is the expression for the force of
    friction here.

    I know that the force of friction is the normal multiplied by the frictional
    coefficient. So in this case I have the normal as ##mg \cos(\theta)##, giving
    ##\mu_s mg \cos(\theta)## as the force of friction. I was slightly unsure because
    it's actually ##\sin(\theta)## which is in the direction of the hill here.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2017 #2

    kuruman

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    Homework Helper
    Gold Member

    Your analysis is correct. Note that there is no reason for μs to be less than unity.

    On edit: Note that μs FN is the maximum force of static friction before the object starts sliding.
     
  4. Apr 14, 2017 #3
    Oh ok - i thought that the frictional coefficient was less than one for some reason.

    When might I encounter something greater than one?

    Thanks
     
  5. Apr 14, 2017 #4

    kuruman

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    Homework Helper
    Gold Member

  6. Apr 14, 2017 #5
  7. Apr 14, 2017 #6

    kuruman

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    Homework Helper
    Gold Member

    Yes, greater than 1 is not encountered often, but it is not an upper limit for some physical reason. Don't forget that if you place an object on an incline and you gradually increase the angle, sliding occurs at the critical angle θc such that μs = tan(θc). Having μs greater than 1 implies that the critical angle is greater than 45o. Everyday experience has us believe that this is not seen often.
     
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