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Coefficient of Static Friciton

  1. Oct 13, 2005 #1
    Hello, I deeply need help on this problem. ANY help would be appreciated.

    Suppose a hanging 1.0 kb lab mass is attached to a 4.0 kg block on the table.
    A) If the coefficent of kinetic friction, uk is 0.20, what is the accereration of the block?
    B) What would be the minimun value of the coefficient of static friction, in order for the block to remain motionless?

    I have my force diagrams made, equations also. What I am getting confused on is do I add the remaning unbalanced forces, Ff and Fg or is the Ff negative? Any help please!!!

    Thank You!
     
  2. jcsd
  3. Oct 13, 2005 #2

    hotvette

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    The forces involved can be positive or negative depending on how you choose your coordinate system. For example, if you choose +y to be down, then mg is positive because it acts in the same direction as +y. The key to friction force is that it always acts in a direction opposite to motion. So, if you chose +x to be the direction the block on the table moves, then friction will be in the -x direction. The key is to be consistent with the coordinate system you decide to use.
     
  4. Oct 13, 2005 #3
    Oh I see. I have my axis as a traditional axis. So, I add my unbalanced forces according to the axis postive or negative values? I'm still stuck on the part b problem though, so any tad bit of help would help.
     
  5. Oct 13, 2005 #4

    Fermat

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    Part b) is a statics problem. All the forces balance.

    Have you got all the forces on the 4kg block ?
     
  6. Oct 13, 2005 #5
    Yes, I have the Fn y+, Fgy-, Ftx+, and Ffx-....
     
  7. Oct 13, 2005 #6
    Would I use the coefficient of kinetic friction from the last problem for the Ff in part b? I am really confused! Someone help please!!
     
  8. Oct 13, 2005 #7

    Fermat

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    It's now a statics problem! So the coefft of kinetic friction doesn't get involved. There's no movement.

    What you have is limiting friction.
    The block is static. When a force is applied to a static block, then there will always be static friction equal and opposite to the applied force on the block.
    As the force increases, the static friction increases until the limiting condition is reached. The static friction reaches a maximum value and the block is just about ready to move. The ratio of this friction/applied force to the normal reaction is called the coefft of static friction.

    This is the state that you have reached with the block. There is an applied force on the block. There is a static friction force opposing this. There is a normal reaction. That's all you need.
     
  9. Oct 13, 2005 #8
    I got it! Thanks!
     
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