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Static Friction of a string

  1. Feb 17, 2004 #1
    I missed the section on static friction and I can't get this question. Can someone help me out?

    What mass can you hang onto the string so that the 0.8kg box will just begin to move?

    The 0.8kg box is sitting on a flat surface with a coefficient of friction of 0.50. This box is attached to a string which extends in a straight line but then is attached to a wall 45 degrees above the horizontal. At the point where the string begins to rise a strand of string is attached which holds the second mass.
  2. jcsd
  3. Feb 17, 2004 #2
    The equation for static friction is [tex]F_f=\mu_s\\F_n[/tex]

    For the box (mass1) to start to move, Ft>Ff. So what you must do is find the Ff using the above equation (remember, Fn has the same magnitude as Fg since the box is at rest in the y-direction). Once you have calculated Ff, you know that any Ft with a greater magnitude than the calculated Ff will cause the Box to accelerate, as I said above. You then calculate the mass of the hanging mass using an Ft = Ff (the one you calculated), and your final answer will be that anything greater than this mass will accelerate the first box from rest.
  4. Feb 17, 2004 #3
    But does the string hanging at a 45 degree angle, on the other side affect the result.

    Calculating Ff of 0.8kg mass

    Fg = mg
    Fg = (0.8)( 9.8)
    Fg = 7.84N

    Fn=Fg So, Ff = uFn

    Ff = (0.5)(7.84)
    Ff = 3.92N

    After that I'm stumped. When the box starts to move Fnet on the 0.8kg box is 0 right?
    Last edited: Feb 17, 2004
  5. Feb 17, 2004 #4
    Yes, it does affect it. The string attaching the hanging mass to the wall exerts half of its force horizontally and half of its force vertically (If the tension inbetween the wall and second mass is Ft2, Ft2 sin 45 and Ft2 cos 45 are the same). We want Ft2 to be at the threshold between a system which is moving, and one which is not. So you can set the x-component of Ft2 to Ft, and then solve for Ft2. Since we know that mass2 is not moving up or down before mass1 starts moving, we can also assume that the y-component of Ft2 is the same as Fg2 (force of gravity on mass2). We can set them equal, and solve for the mass of mass2.
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