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The sums of forces along x and y

  1. May 8, 2006 #1
    I have a problem I'm working on where the general premise is that there is a box being pushed along the ceiling at a constant speed. The force F is at some angle with respect to the vertical. There is a coefficient of kinetic friction between the box and the ceiling and the persons hands and the box. The latter of which keeps the persons hands from slipping.
    I am trying to be deliberately vague because I want help on the concept not the problem itself.

    A = angle
    [tex]F_a[/tex] = Force along the angle A
    [tex]\mu_k[/tex] = coefficient of friction along the ceiling
    [tex]\mu_{k2}[/tex] = coefficient of kinetic friction with persons hands
    My thinking is that
    [tex]\sum f_x = f_a*sinA - ff = 0 \Rightarrow f_a*cosA = \mu_k*N + \mu_{k2} *N[/tex]
    [tex]\sum f_y = f_a*cosA - mg - n = 0[/tex]

    Is there a flaw in my logic? Because when I put in all the information my result is unrealistic. I have attached the free body that goes along with my thinking.

    Attached Files:

    Last edited: May 8, 2006
  2. jcsd
  3. May 9, 2006 #2
    If the person's hands aren't slipping then ignore the coeffecient of friction between the hands and the box for now.

    Now resolve your forces:

    In the y direction:
    [tex]F_a*cos(A) - mg - N = 0 [/tex] N is normal reaction between box and ceiling
    [tex]N = F_a*cos(A) - mg[/tex]

    In the x direction:
    [tex]F_a*sin(A) - \mu*N = 0 [/tex]
    [tex]F_a*sin(A) - \mu(F_a*cos(A) - mg) = 0[/tex]

    Now as far as the coeffecient of friction between the hands and the box, this should be a limiting condition.

    [tex]F_a*sin(A) < F_a*cos(A)*\mu[/tex]
    [tex]A < Tan^-1(\mu)[/tex], so the above equations only hold true if this condition is met.
  4. May 9, 2006 #3
    The force along the x direction in your problem is incorrect because the friction between the box and the hands is important to the problem overall.

    However in bug fixing your problem I was able to see where I went wrong so thanks.
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