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Newton's laws - studying for a test

  1. Sep 19, 2011 #1
    Hello guys. I have a test in a couple of days, so I am hoping to get some practice. I'll post any questions I have here, I would appreciate if you guys can help me clear up any confusions.


    1. A 400-N block is dragged along a horizontal surface by an applied force F as shown. The coefficient of kinetic friction is uk = 0.4 and the block moves at constant velocity. The magnitude of F is:

    attempted solution:

    Since the block moves with constant velocity: [tex]\sum F = 0[/tex]
    Since
    [tex]\sum F = ma[/tex]
    [tex]F_y - mg = ma = 0 [/tex]
    [tex]F_x - u_kmg = ma = 0 [/tex]

    The correct answer is 150 N, but obviously the components above are far greater because mg = 400 N. I must be missing something conceptually simple... Please help.
     
  2. jcsd
  3. Sep 19, 2011 #2

    PhanthomJay

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    What is the direction of the applied force "as shown"? What answer do you get?
     
  4. Sep 19, 2011 #3
    oops. Sorry, here is a picture:

    [PLAIN]http://dl.dropbox.com/u/42149615/phq.jpg [Broken]

    Solving the simple equations I came up with above, I would get the components of the force as:

    Fy = mg
    Fx = Ukmg

    The magnitude of F will then of course be far greater then 150 N.
     
    Last edited by a moderator: May 5, 2017
  5. Sep 19, 2011 #4
    draw a free body diagram. in the vertical direction, there are three forces acting. normal reaction by the floor , (3/5)F , both of which are upwards and mg , which is downwards.
    since there is no net force in the vertical direction,

    [tex]N+\frac{3}{5}F=mg[/tex]

    [tex]\therefore N=mg-\frac{3}{5}F[/tex]

    so your force of friction which is opposing the applied force is

    [tex]F_{fr}=\mu_k N=\mu_k (mg-\frac{3}{5}F)[/tex]
     
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