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Coefficient of friction along an inclined plane

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data
    A horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin 7/25 (=16.3 degrees) to the horizontal. Find the coefficient of friction between the block and the plane and then acceleration with which the block will move when the force is removed.

    2. Relevant equations
    Friction = Coefficient of friction x Resistive force (parallel at a right angle to the slope on which the objects acts on)
    Force = Mass x Acceleration

    3. The attempt at a solution
    Well what I thought was that the force that the object has should be 2+gsin16.3 which equals 4.75N. This means that the acceleration will be 2.75N when the force of 2N is taken away. However the answer is meant to be 1.97ms^-2... Did I work out the acceleration wrong in the first place?

    Now attempting to work out the coefficient of friction, I simply did friction over resistive force: gsin16.3/gcos16.3 and that gives me 0.292 but the answer is supposed to be 0.0827... I do not understand what I am doing wrong.

    If someone can say what I'm doing wrong and lead me into the right direction then thanks!

    EDIT: Worked out coefficient of friction now but can't work out acceleration.
    Last edited: Nov 5, 2008
  2. jcsd
  3. Nov 5, 2008 #2
    check your units in all your calculations. Sometimes you're missing critical variables and units will help you.

    for instance your statement:
    "2 + g sin16.3 which equals 4.75N."
    is mathematically off in terms of units.

    You have the applied force (2N) plus 9.8 m/s^2 times a quantity that doesn't have units. why is this bad: You can't add things with different units together!

    then you have this:
    "acceleration will be 2.75N"
    is that right in terms of units?

    Note: units will help you a LOT! So will a free body diagram. It looks, however, that you're headed the right direction. :smile:
  4. Nov 5, 2008 #3
    Ah I see, so that's where I went wrong on that part, I've drawn a free body diagram but I always thought I could just add the 2N onto it as it is and this is the first time I encountered a problem. (Obviously I made a big mistake when I said the acceleration was measured in N xD)

    EDIT: Can't seem to work out the force... Not sure how to turn the 2N force into the same unit... Really confused now :S
    Last edited: Nov 5, 2008
  5. Nov 5, 2008 #4
    Wow I solved the coefficient of friction...
    It was (g sin16.3 - 2 cos 16.3)/(-2 sin 16.3 - g cos 16.3) = 0.083

    No idea how to work out acceleration still though.
  6. Nov 5, 2008 #5
    Use F=ma... Force (in N) relates to mass (in kg) times acceleration (in m/s^2).

    When you have your free body diagram, you need to have the weight force parallel to the slope as m*g*sin theta... you left out the mass.

    Always check units!: in this case: N = kg*m/s^2
  7. Nov 5, 2008 #6
    Well as mass is 1kg, I can basically ignore it I think. I am correct in thinking that? Still not sure how the units apply here...
  8. Nov 5, 2008 #7
    Sigh.. still no luck after a few hours. I dunno how many hours I've spent in total on this...
  9. Nov 8, 2008 #8
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