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New to force problems, need some help

  1. Oct 14, 2004 #1
    I know I just posted about a different problem, but I think if I can figure out this problem I'll be able to do the rest of my homework. I've been gone from class for a couple days so I don't quite get this stuff...

    Here's the problem:

    A 4.00kg block is pushed along the ceiling with a constant applied force of 85.0N that acts at an angle of 55 degrees with the horizontal. The block accelerates at 6m/s/s. Determine the coefficient of friction between the block and the ceiling.

    This is what I worked out, but I doubt it's right:

    For my free body diagram I had one arrow pointing away to the left as the Ff, on arrow pointing down as Fg, one arrow angling up at 55 degrees to the horizontal as Fa, and one arrow angling down at 55 degrees to the horizontal as N. (<-- I don't think I even did my free body diagram correctly...)

    xFnet=Fa-N+Ff=ma
    =4(6)
    Ff=24


    85cos(53)=51.15N (<-- this is the x-axis force of the block being pushed)

    Ff=MN (<-- where M is the coefficient of friction)
    24=M(51.15)
    24/51.15
    M=0.47


    So what am I doing wrong here--I'm sure there's a bunch of things, but if you could please help me out I'd really appreciate it...
     
  2. jcsd
  3. Oct 14, 2004 #2

    cepheid

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    Gold Member

    Your FBD sounds mostly right. You have the weight, the friction force, the applied force (at an angle), and the normal force. But don't draw the normal force at an angle, because then by definition, it isn't a normal force. :smile: The normal to a surface is a line perpendicular to it, so the normal force is the force acting in that direction i.e. directly downward in this case. That will change your expression for the sum of forces in the x direction.

    The block accelerates horizontally, there is no motion in the vertical direction. So we have:

    [tex] \sum{F_x} = ma [/tex]

    [tex] \sum{F_y} = 0 [/tex]


    x-direction:

    [tex] \sum{F_x} = F_a\cos{55^o} - F_f = ma [/tex]

    y-direction:

    [tex] \sum{F_y} = F_a\sin{55^o} - F_g - N = 0 [/tex]

    [itex] F_a [/itex] is known. [itex] a [/itex] is known. [itex] F_g [/itex] can be calculated. That leaves two unknowns: [itex] F_f [/itex] and [itex] N [/itex]. Solve for each from the equations and then use the formula:

    [tex] F_f = \mu N [/tex]

    [tex] \mu = ? [/tex]
     
  4. Oct 14, 2004 #3
    Thanks cepheid, that helped a lot.
     
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