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Homework Help: Rotational Equilibrium problem

  1. Feb 20, 2006 #1
    One end of a uniform 4.00-m-long rod of weight w (Fg) is supported by a cable. The other end rests against the wall, where it is held by friction. The coefficient of static friction between the wall and the rod = 0.500. Determine the minimum distance x from point A at which an additional weight w (the same as the weight of the rod) can be hung without causing the rod to slip at point A.

    Not sure how to tackle this one. Seperated tension force and friction force into x and y components. Just unsure of what to do with torques.
  2. jcsd
  3. Feb 20, 2006 #2
    Well, if it isn't rotating the sum of the torques is zero.

    Where's the point x?
  4. Feb 20, 2006 #3
    Hello sorry for being so sloppy, even left out the angle. It's kind of situated like this but not quite. http://hep6.physics.wayne.edu/harr/courses/2130/P8.17.jpg [Broken] Except the angle is at the bottom right and is 37 degrees above horizontal (North of west I guess?) Point A is all the way to the bottom left where the rod hits the wall. Point B is all the way at the right end of the rod. X is the distance from A to the hanging weight.
    Last edited by a moderator: May 2, 2017
  5. Feb 20, 2006 #4
    The bad thing about rotational equilibrium problems is the number of unknowns. The nice thing is we can take any axis we like as an axis of rotation. :cool:

    My advice is to do Newton's law in the x and y directions first and count the number of new equations you need. (If you've got "n" unknowns, you need "n" equations.) Then all you need to do is find enough axes of "rotation" to get the rest of the equations you need. To use the torque equation you need to pick an axis of rotation. If nothing is rotating we can pick any point in space we want for a rotation axis. Good points to consider:

    1) CM. Always a good choice.
    2) If there is a rod in the problem, either end of the rod.
    3) ANY point at which an unknown force acts on. The reasoning here is that this force has no moment-arm at this axis, so it drops out of the torque equation.
    4) Any other point at which a force is acting on. Same argument as above, but here we are just trying to find a new equation.

    The problem with choosing axes of rotation is that some axes give an equation that you already know. The above 4 choices of axis usually give you new equations to work with. (I make no promises, it depends on the physical system. But you should get at least SOME independent equations out of them.)

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