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Frictional force word problem test tomorrow!

  1. Aug 6, 2014 #1
    these are technically not homework, i just need more help for my test tomorrow!
    I HAVE NO IDEA HOW TO DO THIS! can someone walk me through these and show me the steps! thank you :):confused:

    1.A sled is being pulled to the right with a force of 28 N. A frictional force is working opposite the applied force with a magnitude of 17 N. The mass of the sled is 4.5 kg. (a) What is the net force in the x direction? 28 – 17 = 11 N (b) What is the coefficient of friction between the sled’s blades and the surface? m = 0.385

    2. A football player with a mass of 86 kg, dives into the end zone and is slowed by the grass after he hits the ground by a force of friction of 580 N. What is µ between the player and the grass? m = 0.69

    3. A person pushes horizontally with a force of 350 N on a 68 kg chair to move it across a level floor. The coefficient of friction is 0.37. (a) What is the magnitude of the frictional force? 247 N (b) What is the acceleration of the chair? Fnet = 350 – 247 = 103 N, acceleration = 1.51 m/s2
     
  2. jcsd
  3. Aug 6, 2014 #2

    Nathanael

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    Sorry but you need to at least show an attempt.
    You need to try to understand what it is about the problem that you don't understand.

    This is copied from the thread at the top of this (Introductory Physics Homework) forum:

    This applies to your other question (about gravity) too.
     
  4. Aug 6, 2014 #3
    Sorry I did my work on appear and was too lazy to write them all online. For the coefficient question do I do 11N/28N
     
  5. Aug 6, 2014 #4

    Nathanael

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    No that's not right. What is coefficient of friction? Doesn't the force of friction depend on how hard two objects are pressing against each other?
     
  6. Aug 6, 2014 #5

    So is Ff 11N and FN 28N? Can you tell me how to identify the Ff and FN if I did it wrong
     
  7. Aug 6, 2014 #6

    Nathanael

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    [itex]F_{friction}[/itex] is said to be 17N in the problem
    ("A frictional force is working ... with a magnitude of 17N")

    (By "FN" you mean the normal force, right?)

    [itex]F_{normal}[/itex] is the force that the ground is pushing up on the sled
    (which is equal to the force with which the sled is pushing on the ground)



    P.S.
    Your answers for problem 3 are correct.
     
  8. Aug 6, 2014 #7
    so it would be 17/11
    which is 1.54
     
  9. Aug 6, 2014 #8
    is net force the same as normal force
     
  10. Aug 6, 2014 #9
    and for number 3a, the answer was online but i didn't get how to do it. i used Ff=(μ)(FN) and i got Ff=(.37)(350N) BUT my answer was 129.5 when the answer should be 247N
     
  11. Aug 6, 2014 #10

    Nathanael

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    Nope. Right now, gravity is applying a force on you, but you're not accelerating. The reason is because there is another force acting on you (equal in size and opposite in direction) from the ground (or from a chair or something) and it is called the normal force.

    To find the normal force, you note that the sled is not vertically accelerating, so there must be a net force of zero in the vertical direction.
    (So [itex]F_{gravity}+F_{normal}=0[/itex])
     
  12. Aug 6, 2014 #11
    so i have to do fnormal=fnet-fgravity which is 9.8N
     
  13. Aug 6, 2014 #12

    Nathanael

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    Well that's because you used the wrong value for the normal force.

    Kinetic friction does not depend on how hard you're pulling on the object, it depends on how hard the two object and the ground are pressed together (the normal force).
     
  14. Aug 6, 2014 #13

    Nathanael

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    So you're saying Fgravity is (negative) 9.8N?

    But isn't Fgravity=mg?

    The mass of the sled is not 1kg
     
  15. Aug 6, 2014 #14
    i got it! can u explain 3.a please? THANKS
     
  16. Aug 6, 2014 #15

    Nathanael

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    You had the right idea you just used the wrong normal force:
    You're given the mass of the chair so what is the normal force?
     
  17. Aug 6, 2014 #16
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