a metre stick is found to balance at the 49.7cm mark when placed on a fulcrum. when a 50g mass is attached at the 10cm mark, the fulcrum must be moved to the 39.2cm mark for balance. what is the mass of the stick? ==> i have NO idea what to do with this quesiton... have tried various ways to go about it but none have proved successfull, if anyone has any hints they would be appreciated!! thanks
We're clearly talking about torque, yes? Given the information: 1) Can you tell me where the center of mass of the meter stick is when the weight is not attached? 2) Once the weight is attached, the meter stick's center of mass does not move - what moves is the center of mass of the combined system. Why does it move? How far does it move? Given where the combined center of mass now is, what can you tell me about the torques exerted by the 50 g mass and the mass of the meter stick? Does tht help?
Take a look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html Look under Center of Mass: Continuous
well the center of mass of the meter stick is a little off half way along the stick, but a meter stick can not be heavier on one half then on the other, and there are no masses attached to either end (originally) so i dont know how i am meant to find the mass of the stick with such limited information. i went to the link above, just now, and found the m1L1=m2L2 formula... worked it out to be (.05kg x .1m)=(m2 x .497m) which came out as .01kg, or 10g as the mass of the stick... is that what you were getting at???
First off, there's really no logical contradiction in the center of mass of the meter stick not being exactly at the center. It may just be poorly made, or perhaps there's an end cap on one end not on the other. Beyond that, I began to work out the problem and realized that we have a difficulty. If the meter stick balances at the 24.7 cm mark, then that's where the center of mass is. If you put a weight at the 10 cm mark, then the center of mass should move that way, but the problem indicates that it moves in the other direction, to the 39.2 cm mark. This is not possible. Is it possible that the 50 g mass is put 10 cm from the other end of the meter stick? Working on that assumption, I get the mass of the meter stick to be 267g. Note: I'm not sure about the significant digits because I'm not sure how accurately you recorded the data. You'll have to work that out on your own. Is that sufficient?
ahhhhhh yes you're quite right!!! the stick balances at the 49.7cm mark, for some silly reason i decided to write the 24.7cm mark... mmm!!!! i think i might just have to wait to find out how to do it in my tutorial, i have no idea what formulas to use or anything, thanks for your help, but i think i am a bit beyond helping!! haha.. thanks!! :)
im having a problem with the same question! Just to correct the wrong figures, the question is: "A meter stick is found to balance at the 49.7cm mark when placed on a fulcrum. When a 50.0 grams mass is attatched at the 10.0 cm mark, the fulcrum must be moved to the 39.2cm mark for balance. What is the mass of the meter stick?" Any help would b greatly appreciated!! this is the last question on my homework arrrg isn't that always the way.
Hint: pretend the mass of the meter stick is concentrated at the stick's center of mass. That is, mentally replace the meter stick with a massless rod (with the same length, of course) that has a mass attached to it at the location of the stick's center of mass.