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Frictional force and largest angles

  1. Oct 5, 2005 #1
    A 17 kg block is at rest on an incline with angle 33. The coefficients of static and kinetic friction are 0.7 and0 0.59, respectively. The acceleration of gravity is 9.8m/s^2.

    1) What is the frictional force actin on the 17kg mass?
    2) What is the larges angle the incline can have so that the mass does not slide down the incline?

    for 1, I thought Ff was just the coefficient of static friction (0.7) times the normal force, but that apparently isn't right.

    for 2, I don't have a clue....
     
  2. jcsd
  3. Oct 5, 2005 #2

    Tom Mattson

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    Yes, it is right. How did you calculate it?
     
  4. Oct 5, 2005 #3
    I did Ff=(0.7)(17)(9.8)cos 33=97.806; however, when I put the answer into our homework server, it says the answer is wrong.
     
  5. Oct 5, 2005 #4

    Tom Mattson

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    Oh, good grief, I was wrong. You have [itex]F_f=\mu_SN\cos(\theta)[/itex] only when the static frictional force is a maximum. Sorry about that.


    Since you can't tell a priori that [itex]F_f[/itex] is equal to its maximum value here, you have to draw a free body diagram, sum the forces, and apply Newton's 2nd law to find [itex]F_f[/itex].
     
  6. Oct 5, 2005 #5
    Ok, I found the frictional force by using Ff=mg*sin33...it gave me 90.73686 (the right answer-yay!). So, how do I find the max angle? I think I have to use tangent, but I don't know on what or where or anything!
     
  7. Oct 5, 2005 #6
    You will have three vectors that must sum to be zero. Those are Ff+N+Fg=ma=0.

    If you write out each component individually, you should come up with a system that is easy to optimize :smile:

    Alex
     
  8. Oct 5, 2005 #7
    Have you managed to find the maximum angle? A simple three-step process:
    1. Set the friction force in terms of cos theta equal to the antiparallel aspect of the weight in terms of sin theta.
    2. Divide both sides by cos theta and rearrange to find tan theta.
    3. Take the inverse tan of both sides.
     
  9. Oct 5, 2005 #8
    :confused: antiparallel aspect???
     
  10. Oct 5, 2005 #9
    Wow! I wonder why I've never been taught this! (I hope it's not wrong...) The maximum angle possible for an incline before a block placed on it will slide is just the inverse tan of the coefficient of static friction. This is independent of the mass of the block or even the mass of the planet you are on. I'm surprised I've never been taught that (maybe I was snoozing).
     
  11. Oct 5, 2005 #10
    Parallel but in the opposite direction.

    If you set the x-axis parallel to the incline, then the friction force is also parallel to the x-axis. The x-component of the weight is then antiparallel to this friction force - parallel to the x-axis, but in the opposite direction to the friction.
     
  12. Oct 5, 2005 #11
    you rock!! Thanks soooooo much!!
     
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