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Friction problem of stationary block

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Figure 6-24 shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a)μs = 0.620 and μk = 0.530 and (b)μs = 0.430 and μk = 0.340?

    2. Relevant equations


    3. The attempt at a solution

    Problem seems unsolvable to me. No mass given, no force, etc. Have no idea what to do.
  2. jcsd
  3. Feb 17, 2009 #2
    There is a good chance that at some point during your calculations, the mass will cancel out leaving you with just g somewhere. Either that or they just want your answer in an algebraic form.
  4. Feb 17, 2009 #3
    no I need a final answer. I'm sure it will cancel out since it is impossible to solve for but I don't know where to even began for that. Where do I even begin?
  5. Feb 17, 2009 #4


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    Step Number ONE: Do not try to solve the problem BEFORE you understand the problem!!!!

    Don't even think about what to do until you draw yourself a good picture with the forces labeled. Put in a set of coordinate axes. Then identify the x and y components of the forces. Note that in the y direction you have gravity acting down, but there is also a component of the applied force acting up. Note that there is a frictional force acting horizontally in one direction and the x component of the applied force acting in the other (I'm imagining the x axis being horizontal).

    Now ask yourself a couple questions: (1) Is the vertical component of the applied force larger than mg? If yes then the block comes off the ground. If no then it doesn't. (2) Is the horizontal component of the applied force larger than the static frictional force? If yes then the block moves horizontally. If no then the block doesn't move.

    If the block moves then you can use

    [tex] \Sigma F_x = ma_x [/tex]
    [tex] \Sigma F_y = ma_y [/tex]

    as appropriate.

    (By the way, I assume that in your other post you were referring to the textbook by Halliday and Resnick. Thousands of students have learned physics from that text. My own copy dates back to 1966. Oh, and thousands of students have cussed at it too :mad: :smile:)
  6. Feb 17, 2009 #5
    Well I don't understand how to solve for the mass. any idea? once i have the mass i can solve for the x and y components and do whatever else i need to do.
  7. Feb 17, 2009 #6
    Ahh I think I'm on the right track. mass does end up canceling since im solving for acceleration. so i assume mass isnt calculated when solving acceleration?

    actually im confused again. I've got a static and a kinetic friction. so how do i determine which friction is used for the x and y component for acceleration to calculate its magnitude?

    edit: giving up on the homework for tonight, i'm tired of looking at this stupid problem for 4 hours and getting no where. thanks for the reply but it didn;t help me :(
    Last edited: Feb 17, 2009
  8. Feb 18, 2009 #7


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    Since I'm into handing out advice today, working 4 hours on a problem (even if it's not a continuous 4 hours) is a waste of time. One ends up going around in circles. It's better to put it aside and even sleep on it. You may not know it but your brain does a lot of processing while you're sleeping.

    Okay, about static and kinetic friction. Here static means "not moving" and kinetic means "moving". Go back to my first response. I said to check to see if the horizontal force was greater than the static frictional force. You compute the static frictional force using the coefficient of static friction. Here's a general rule: If the static frictional force is larger than the component of the applied force acting in the same direction, the object ain't gonna move. If, on the other hand the component of the applied force is greater than the static frictional force, the object will move AND then you have to use the coefficient of kinetic friction to compute the frictional force opposing the motion.

    Now for your problem there isn't going to be any y component to the acceleration. Why? because the vertical component of the applied force is not large enough to lift the block. The block is going to slide horizontally, period.

    One last thing. Your first clue that the mass would cancel out of the equations was that the applied force was given as .500mg.
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