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Rotational equilibrium formula

  1. Jan 6, 2007 #1
    I have a homework question for physics. i have read the chapter in the book on it and have studied my notes and everything. i have been trying different ways for hours but i am unable to get any where. if you could help me get started on this problem that would be wonderful.

    i thought that the .100kg and the .700kg is the mass and the gravity is 9.81 m/s squared and the force is 19.6N. and i drew a picture of my paper but i do not have any idea of how to use this information to get the answer or what kind of formula to use. This is the problem below. Please help me if you can.

    A 0.100kg meterstick is supported at its 40.0cm mark by a string attached to the ceiling. A 0.700kg mass hangs vertically from the 5.00cm mark. a mass is attached somewher on the meterstick to keep it horizontal and in both rotational and translational equilibrium. if the force applied by the string attaching the meterstick to the ceiling is 19.6 N, determine the following:
    a) the value of the unknown mass
    b) the point where the mass attaches to the stick.

    -Confussed on Physics problem:|
  2. jcsd
  3. Jan 6, 2007 #2


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    Staff Emeritus
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    The point of this question is to use moments (also known as torques). Pick one point as a pivot, and equate the moments to ensure no rotation. Then pick another point and do the same. If you pick the pivots wisely, then this will give you two equations involving the two unknowns (namely, the unknown mass, and position of the mass), which you will be able to solve.
  4. Jan 7, 2007 #3


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    also, you need to take into account the mass of the meter stick as it's given in the problem. you can do this by using the middle of a section as the distance of a section of the ms from the middle. for the mass you use the fact the the ms is uniform so 0.5 of the stick is 0.5 * the mass, 0.2m is 0.2 * the mass...
    for example: from 0 to 0.4meters:
    mass = 0.4*0.1kg
    radius from pivot(0.4meters):
    0.4/2 = 0.2meters
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