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Min. downward force to keep object from slipping

  1. Sep 30, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the min. downward force on the box that will keep it from slipping?
    Coefficients are:
    static= .36
    kinetic= .26

    the box is 30kg and there is a rope pointing -----> with 125N
    2. Relevant equations

    Ms n
    Mk n

    3. The attempt at a solution

    a= (Ms mg + Mk mg)/m = 6.1 m/s^2

    T= 125N
    mg= -294N
    n= 294N
    Fs= .36
    Fk= .26
    a= 6.1m/s^2

    F=ma -> F=125-.36-.26 = 124.4 ma= 30kg(6.1)= 183N

    I am not sure if any of this is correct and am stuck moving on from here. If this is correct, would I divide F/ma? Thanks for any help or clarification you can offer
  2. jcsd
  3. Sep 30, 2010 #2
    If the box isn't slipping that means it's not moving. What is the acceleration of a stationary box?
  4. Sep 30, 2010 #3
    The answer need to be in units of force and I tried two wrong answers; a= F/m = 4.2 m/s^2 and 0N. Here is the copied question:

    "What is the minimum downward force on the box in the figure that will keep it from slipping? The coefficients of static and kinetic friction between the box and the floor are 0.36 and 0.26, respectively."
  5. Sep 30, 2010 #4
    Well then you need to solve for a downward force that will make the force of friction equal to the force applied. You need to use the coefficient of static friction, as the box isn't moving.

    Can you tell me why the stationary box has a different coefficient of friction than one that is moving?

    As for solving, remember your net force equation. Since the box can't be accelerating Fnet=0.

    Three steps: Free body diagram, net force equation, substitution.

    Fnet= Ftension + Ffricton = 0
    You can rewrite this as: Ftension=Ffriction

    Don't forget that the force of friction is equal to the normal force applied by the surfaces it is in contact with, in this case the floor multiplied by the coefficient. Try writing the equation out that way and see if it helps. As I said, start with your free body diagram and label all the individual forces acting on it.
  6. Sep 30, 2010 #5
    Static friction is harder to overcome due to bonds formed between the box and the surface.

    Thanks, ill attempt it again
  7. Sep 30, 2010 #6
    Alright, so your saying that the FBD has a -w, +n, +T. W= 294 and normal = 294+.36 = 294.36, and T= 125?
  8. Sep 30, 2010 #7
    Sorry, it took me a bit to figure out what was going on with that last statement.

    From what I interpret that's mostly right. I'll tell you how this is setup, because you seem not fully grasp what the diagram is for. I'll try describe it. Draw a box. Now given your setup you have four forces acting on it. You have the force of tension going in one direction, and the force of friction opposite to that (right and left). Vertically you have the normal force pushing up. Going down you both have the force of gravity, and the additional downward force.

    Draw that out. This is what you use to setup your net force equation in each vector. Remember that you need to define a direction as positive. In this case up and right will be positive.

    In your X direction you have: Fnet=Ff+Ft=0, which you can rewrite as Ff=Ft. Since in this setup Ff is negative. In the Y axis you have Fnet=0 as well, so Fn=Fg+Fa.

    Now remember that Ff=(coefficient of friction)Fn. So lets put this together as Ft=(coefficient of function)Fn, and you have another equation that says Fn=Fg+Fa. That's plenty of hints. Put your algebra to use and solve for Fa.

    Now you can see that Fn = Fg + Fa (force applied).
  9. Oct 1, 2010 #8
    Thanks for the detailed response. I figured it out shortly before you responded:

    .36(294+F)= 125
  10. Oct 1, 2010 #9
    That's what I got.
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