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Rock - Don't Roll!

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A climber of mass 72 kg can stand on a rock sloping at 40degrees to the horizontal. What can you deduce about the coefficient of static friction between his boots and the rock?


    I'm really lost on this one. His weight = mg so 72*9.8 = 705.6 N

    and thats about as far as I get. Anyone able to help make sense of what my next steps are and why? Thank you

    - Otis
  2. jcsd
  3. Jul 18, 2007 #2


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    If the man is not sliding down the hill, what does this tell you about the net force in the direction of the slope? What does this then tell you about the value of the friction force?
  4. Jul 18, 2007 #3

    Well, the friction force would equal the applied force because he's not moving anywhere. I'm not sure what it says about the net force. I'm supposed to show some sort of calculations with my answer as well, but not sure what all I'm able to calculate.
  5. Jul 18, 2007 #4


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    You'll want to start by splitting the vertical force mg into components normal to the slope and parallel to the slope.
  6. Jul 18, 2007 #5


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    Remember that there will be a component of gravity acting along the slope.

    Now the only two forces acting in the direction we care about, here the direction of the slope, are friction and the component of gravity. So, the total force in that direction is:

    [tex]\Sigma F_{along \ slope} = F_{gravity} + F_{friction}[/tex]

    Now, since the guy is not accelerating at all, what does Newton's First Law say the total force has to be equal to? Once you know this, you should be able to solve for the friction force and then the coefficient.
  7. Jul 18, 2007 #6

    An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

    He's at rest, so the total force must be equal to the friction force?

    I want to say the force is 0 but isn't there always some force?

    Dick says to split the vertical force mg into components normal to the slope and parallel to the slope.

    Can I go sin = o / h

    sin40 = 705.6 N / h

    h = 705.6 N / sin40 = 1097.7 N

    is that the force on the slope? what am I supposed to do with it?

    I was just starting to get this stuff without the angles and now the angle has me totally lost. :\
  8. Jul 18, 2007 #7


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    You are making an error I've seen before. There are separate similar triangles for the force diagram and the slope diagram. The hypotenuse of one is not the hypotenuse of the other. The hypotenuse of the force diagram is mg. So the force parallel to the slope is mg*sin(theta). Not mg/sin(theta). Try and straighten this out on your own for now. Because I've gotta go.
  9. Jul 19, 2007 #8
    If you have a slope of 40degrees to the horizontal and a vertical line coming straight down forming a triangle.. how can that vertical line possible be the hypotenuse? theres going to be a 90degree angle where the hypotenuse and horizontal meet up and hypotenuse will be opposite that angle.

    I wish I could draw a triangle in here but I've no idea how. I will attempt to explain what I'm seeing when I read what you said.

    take a 1, 2, sqrt3 triangle

    the angle opposite the 1 is 60

    the angle opposite the sqrt3 is 30

    the angle opposite the 2 is 90

    So sin(theta) = opposite / hypotenuse

    sin60 = 1 / 2

    when I read what you said it appears to me that the 1 is the hypotenuse. So I'm obviously totally missing something here. I can't see where any other triangles are formed that I have info for or can use.
  10. Jul 19, 2007 #9

    Doc Al

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    Whenever you use a right triangle to find the components of a vector, the full vector is always the hypotenuse. In this case you are trying to find the components of the weight, which is represented by a vertical line: so that vertical line is the hypotenuse of your right triangle. Since you want components parallel and perpendicular to the slope, the other sides of your triangle must be parallel and perpendicular to the slope.

    This attached diagram might help you visualize the components and how they might form a triangle.

    Attached Files:

  11. Jul 19, 2007 #10


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  12. Jul 19, 2007 #11

    Thanks for that link, it helped a lot. Starting to make sense now. I feel pretty confident I can get it now and with a little more practice have it down for tests.

    - Otis
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