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Force and a couple other problems

  1. May 28, 2007 #1
    Problem 1:
    Calculate the normal force exerted on a 2.84 kg book resting on a surface inclined at 32.9° above the horizontal.
    I figure u have to use 9.8 m/s but where does the 32.9 deg come into play?


    Problem 2:
    The coefficient of static friction between a block and a horizontal floor is 0.360, while the coefficient of kinetic friction is 0.125. The mass of the block is 4.74 kg. A horizontal force is applied to the block and slowly increased. What is the value of the applied horizontal force at the instant that the block starts to slide?
    What is the net force on the block after it starts to slide?

    Im not even sure what to start with for this one.
    Thanks for the Help :smile:
     
  2. jcsd
  3. May 28, 2007 #2
    1. If you draw a diagram you have the answer. What forces are acting? Fg pointing downwards and Fn perpendicular to the plane. You know how to do it now? :smile:
    2. You can start by calculating what you can. For example, what is the static friction force? What force will get the object moving? (Hint: An object that just starts to move in this case has no net force acting on it.)
     
  4. May 28, 2007 #3
    1. ok i drew the diagram but am still unaware what to do. do i need to get X,Y components?

    2. Im really not sure what the SF force is. Is there a formula i can use?
     
  5. May 28, 2007 #4
    ahh i think i got the first one is it:
    Fn= mg(cosx)?

    Still stuck on the second one tho...
     
  6. May 28, 2007 #5
    Ok so i got the F in the second problem at 16.7 but how do i calculate the net force for the second problem?
     
  7. May 28, 2007 #6
    any1 know for the net force?
     
  8. May 28, 2007 #7
    Yes, the first one is Fn= mg(cosx).
    And yes, for the second one the force is 16.7 N.
    Now to calculate the net force you need to see what forces are acting.
    There is force of gravity and normal force, which cancel each other out and we don't have to worry about them.
    Then there is the force of friction and the applied force, which we just found out to be 16.7 N.
    So we need to find the force of friction. It is different from the first case, because the object is moving now and the coefficient of kinetic friction is involved now, rather than the coefficient of static friction from before.
    So calculate friction force for when the object is moving.
     
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