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Forces - friction

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data

    A man is pulling a 40.0 kg crate across a level floor with a horizontal force, and the coefficient of kinetic friction is μk = 0.59 for the crate and the floor.

    14. What is the normal force acting on the box?
    a) 231 N
    b) 340 N
    c) 196 N
    d) 392 N

    15. What horizontal force must the man apply to get the crate to accelerate at 2.0 m/s2 ?
    a) 311 N
    b) 231 N
    c) 81 N
    d) 40 N

    2. The attempt at a solution

    For 14:
    N = mg = (40.0)(9.8) = 392N

    For 15:
    N-Fx = max
    392 - Fx = (40.0)(2.0)
    Fx = 312N

    but there's no 312N there and I didn't even incorporate the μk?
     
  2. jcsd
  3. Mar 31, 2014 #2
    Did you draw a free body diagram? Is N a horizontal force or a vertical force?

    Chet
     
  4. Mar 31, 2014 #3
    N is a horizontal force.
     
  5. Apr 1, 2014 #4

    haruspex

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    No, it's a vertical force, the vertical component of the reaction from the floor.
    What, then, is the frictional force?
    What is the relationship between the frictional force, the applied force, and the resulting acceleration? (Your equation was wrong.)
     
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