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ANOTHER static coefficient question

  1. Mar 19, 2004 #1
    Problem reads: A package case of mass m is left 10 m up on a 30 degree ramp overnight where it is at rest. a) what is the minimum value of the static coefficient of friction for this to be ture? Prove in general that the coefficient of static friction is [tex]\mu_s = \tan \theta [/tex]. b) during the night itrains and water seeps under the case reducing the coefficient of friction to .10 (both static and kinetic). With what speed does the case reach the bottom of the ramp?

    Ok... can I move on with part a without mass of the object? I am trying to move forward with [tex] F_{s} max = \mu_s FN [/tex]

    Oh man, physics...
  2. jcsd
  3. Mar 19, 2004 #2
    Did you draw a diagram of the forces? You have mg, N, and the friction. If you want the object to stay still [tex]\Sigma F[/tex] on it needs to be 0, in both X and Y axis. I suggest (and so will your teacher) that you lay the X axis on the ramp's plane.

    From [tex]\Sigma F_y = 0[/tex] you will be able to find the normal force (N) as a function of the object's mass m. Once you have that, write down [tex]\Sigma F_x = 0[/tex] to find the magnitude of the friction force. Divide that by N and you will have the coefficient of friction. :smile:
  4. Mar 19, 2004 #3
    As for (b), once the object is moving the friction is constant. You can solve the question by finding [tex]\Sigma F_x[/tex] (it's no longer zero; X is the only axis in which there is movement). Of course that depends on the object's mass, which you don't have, but you only need the constant acceleration so you divide by m and get a real number. From there it's just using this equation:

    [tex]{V_f}^2 = {V_0}^2 + 2ax[/tex]

    (You can also use energies and work to find the final velocity, but I'm not sure if you already studied that.)
  5. Mar 19, 2004 #4
    Thank you Chen for your reply! I still have stupid questions however.
    I did draw a FBD and see that the sum of all forces must be 0. Im still confused. When I take [tex]\Sigma F_y = 0[/tex], as FN + (-mg cos 30) = 0.... not knowing the normal force or the mass of the object, I am left with gravity and the angle only. What am I missing? (other than brains)

  6. Mar 19, 2004 #5
    You can leave N as a function of m, and as you noticed:

    [tex]F_N = (g\cos 30) m[/tex]

    When you write down [tex]\Sigma F_x = 0[/tex] and replace [tex]f_s[/tex] with [tex]F_N\mu[/tex] you will find that the mass cancels. :)
  7. Mar 19, 2004 #6

    Doc Al

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    Staff: Mentor

    You are doing fine. Just keep going.

    When it looks like you need data that is not given, chances are you don't really need it. That's the case here.

    The equation for y, gives FN = mg cos 30. Now write the equation for the x-components. Then combine them, as Chen advised.
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