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Homework Help: Block being slid up and inward

  1. Mar 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A 0.5kg block is being slid up a chalkboard with an applied force of 6.0N upward and 2.0N inward towards the board. If the coefficient of kinetic friction is 0.4: find the force of friction and the acceleration of the block.

    2. Relevant equations
    coefficient of kinetic friction=force of friction/force normal

    force of gravity=mass*acceleration

    3. The attempt at a solution
    for force of friction:

    0.4=force of friction/0.5

    for acceleration:


    Is the above method and answer correct? need help.:confused:
  2. jcsd
  3. Mar 25, 2007 #2
    You state that the normal force is 0.5? Why?

    *Nice username*
  4. Mar 25, 2007 #3
    Isnt force normal same as the weight
  5. Mar 25, 2007 #4
    The normal force is a contact force, it is not the same as the weight. You must always solve for the normal force by using newton's second law.
    The block has a force of 2.0 N into the board. Does the object accelerate into the board?
  6. Mar 25, 2007 #5
    From the question..........do u assume that it does??.............i am guessing it dosent........only upwards.

    Thanks for replying
  7. Mar 25, 2007 #6
    You should picture the motion. It is reasonable to assume that the only acceleration is upwards. So, if the acceleration into the board is zero, what does that imply for the net force into the board?
  8. Mar 25, 2007 #7
    Sorry we took long to reply.........we didnt no u posted it...........but ne ways........ummmmmm........does the net force into the board equal the applied force????
  9. Mar 25, 2007 #8
  10. Mar 25, 2007 #9
    Remember Newtons second law: [tex]F_net = ma[/tex].
    So if a is zero in one direction, what can you say about the net force in that direction?
  11. Mar 25, 2007 #10
    then the net force is 0 isnt it?
  12. Mar 25, 2007 #11
    Yes, good.
    So what forces act on the block, in the direction of into or out of the board?
  13. Mar 25, 2007 #12
    out of the board??????
  14. Mar 25, 2007 #13
    There is a force acting on the block into the board as mentioned in the problem. You said that the a is zero into the board, hence the net force is zero into the board. If there is one force into the board and the net force is zero...
  15. Mar 25, 2007 #14
    so then we do not take the inward force into consideration??/

    im confused
  16. Mar 25, 2007 #15
    Net force means all the forces. There is stated a force into the board, you said the net force into the board is zero, what does that imply? (are there other forces?)
  17. Mar 25, 2007 #16
    isnt there an upward force of 6N though
  18. Mar 25, 2007 #17
    Yes, let us just consider the forces into and out of the board for now. Newtons second law is a vector equation which mean you need to only deal with one direction at a time. (Right now we are working on getting the frictional force correct.)
  19. Mar 25, 2007 #18
    ok yeah i understand so it means we solve it in two componenets right?
  20. Mar 25, 2007 #19
    Yes, an upward/downward direction and an into/out of direction.
    You have a force into the board presumably due to the person pushing on the block, are there any other forces in this direction? (or in a direction directly opposite this, which a physicist will just say as the same direction)
  21. Mar 25, 2007 #20
    every action there is an equal and opposite reaction?
  22. Mar 25, 2007 #21
    So if I push on the block (against the wall) that is a force on the block by me. The block doesn't accelerate into the wall, we can see that physically, so there must be another force on the block (at least one). What could be providing this force?
  23. Mar 25, 2007 #22
    The board pushes backwords on us??
  24. Mar 25, 2007 #23
    Close, the board does not push back on me, I am not in contact with it, but it does push on something, what is it and how much must it be pushing?
  25. Mar 25, 2007 #24
  26. Mar 25, 2007 #25
    Yes, very nice. So this is the normal force that arises in you friction equation.
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