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Friction and Force

  1. Nov 9, 2009 #1
    A block M=7.50kg is initially moving up an incline of 25.0degrees and is increasing sppeed with a= 4.44. The applied Force, F, is horizontal. The coefficients of friction between the block and incline are ms=0.443 and mk=0.312.

    a) Fcos(25.0)-73.5sin(25.0)=0 F=34.3 This makes sense to me but does friction play a part in the applied force?

    b) N=mgcos(25.0)+Fsin(25.0) N=81.1

    c) fs=0.553*N
     
  2. jcsd
  3. Nov 9, 2009 #2
    I don't quite understand what you are doing.In the direction of the incline the net force is not zero it is F-T=ma where T=(coefficient)*N
    N=mgcos25
     
  4. Nov 9, 2009 #3
    Sorry i have a mistake N=mgcos25+Fsin25
     
  5. Nov 9, 2009 #4
    The force has has nothing to do with friction whereas the frixtion depends on the force because N the normal contact force depends on F in this excercise
     
  6. Nov 9, 2009 #5
    Thank you for the help

    Now, I have the equations
    F-T=ma
    N=mgcos(25.0)+Fsin(25.0)
    T=Coefficient*N

    How do I solve them with all of the unknowns?
     
  7. Nov 9, 2009 #6
    If you replace T with coefficient*N in the fisrt equation you have two equations with two unknowns.1)F-cof*N=ma
    2)N=mgcos25+Fsin25
     
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