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Homework Help: Find the mass of the meter stick

  1. Apr 17, 2005 #1
    A uniform meter stick of mass M has a half-filled can of fruit juice of mass m attached to one end. The meter stick and the can balance at a point 12.5 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.79 N. Find the mass of the meter stick. Find the mass of the can of juice.

    I divided the scale reading 2.79 N by 9.81 m(gravity) to find the total mass. I got 0.2844036697. I did this because the scale reading should be equal to the gravitation force (so I think).
    Then I took the equation m1x1=m2x2. I used 0.875m for x1 and .125m for x2. I solved for m2. Then I used m1 + m1(7) = 0.2844036697 to find m1. Which I got 0.0305 which isn't right. Please tell me where I went or if I"m on the right track. If not please give me the formulas you used. Thank You.
  2. jcsd
  3. Apr 17, 2005 #2


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    The mass of the meter stick is distributed along its entire length. You cannot use the distance from the pivot to the end of the stick in your equation. You can treat all of the mass of the stick on each side of the pivot as if it were located at the center of each side. Think in terms of three masses and their three lengths. One is the can, and two are parts of the stick.


    After finding the force holding up the stick as you have done, think in terms of rotation around the end of the stick where the can is hanging. That force is at a given distance from the can end, and the weight of the meter stick can be considered to be in the middle of the stick. Either approach should get you to the same answer.


    Just an added comment to assure you there is no disagreement between the two replies you have gotten so far- this one and the one following. There are many different approaches to getting the final answer. The first one I gave you I thought would be the easiest for you to visualize, the second is computationally easier. The one you see in the next message is equivalent to the first one I gave you, except that all of the mass of the stick was considered to be at its center instead of thinking of the stick being in two parts. From a different point of view, the approach in the next reply is equivalent to my second appraoach except that the point about which the moments are calculated is the point of suspension instead of the end of the stick. You can use any point about which to calculate the moments. They must add up to zero no matter how you do it.
    Last edited: Apr 18, 2005
  4. Apr 17, 2005 #3
    You have a kind problem setter, who sets problems that can be solved without a calculator.

    Just draw the force diagram,
    at pos 0 - F_can pointing down
    at pos 12.5 - 2.79 pointing upwards
    at pos 50,0 - F_meterstick downwards (centre of mass of meter stick)

    therfore two equations need to be solved
    1.) 12.5 X F_can = .....
    2.) F_can + F_meterstick = ...
    the equations you can complete and solve.
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