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Force and Coefficient of Friction Problem

  1. Oct 16, 2005 #1
    Okay I'm really stumped on this question and I would be really glad if someone can take me step by step on how to do this question... I have the answer but I have no clue how to obtain it!

    The Question is:
    I obtained the answer for the force of friction by just using Fhorizontal = cos55 * 100 = 58 N (answer is 57 N)

    Now.. to find the coefficient.. we need to know two things right? The force of friction which we have and the normal force. But how can we know what the normal force is if we are not given mass?



    However I'm not sure if that is the correct way to do it... if someone could explain how to get both answers step-by-step it would be great thanks!
     
  2. jcsd
  3. Oct 16, 2005 #2

    Päällikkö

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    Homework Helper

    Assume a massless shovel (it is, after all, described as light), what would the normal force be?
     
  4. Oct 16, 2005 #3
    normal force would be 0 then as there is no object for there to be a gravitational force on? but that would make everything (being coefficient of friction and force of friction.. according to the equation Ffriction = mew * Fnormal) equal to 0 :grumpy:
     
  5. Oct 16, 2005 #4
    Normal force would be the ground acting against the shovel, else the shovel would push the ground down.
     
  6. Oct 16, 2005 #5
    yup but in this case, the normal force is = to the force of gravity on the shovel

    if we assume if the shovel is massless... it means theres no force of gravity acting on it and therefore no normal force... correct?

    i'm really confused.. so do we have to find the vertical force that she is exerting on the handle?
     
  7. Oct 16, 2005 #6

    HallsofIvy

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    NO, it's not- that's the whole point!

    You were told the girl pushed the shovel "with a force of 100 N". You correctly (approximately) calculated the horizontal component of force to be "cos55 * 100 = 58 N (answer is 57 N)". What do you think happened to the rest of the force? Where there is a "horizontal component", don't you thing there is likely to be a "vertical component"?

    By the way, how did you get "58 N"? What exactly did your calculator say?
     
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