How to Balance a Meter Stick with Unequal Masses at Different Positions?

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Homework Help Overview

The discussion revolves around balancing a meter stick with unequal masses positioned at different points. The original poster presents a scenario involving a meter stick pivoted at its center, with a specific mass at a known position and a second mass whose position needs to be determined for equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the center of mass equation to find the position for the second mass. There are attempts to derive the position D2 based on the given mass and distance of the first mass. Some participants express confusion about the calculations and seek clarification on the equations used.

Discussion Status

The discussion includes various attempts to calculate D2, with some participants confirming the use of the center of mass approach. There is a mix of interpretations regarding the equations and the final positioning of the second mass, with no clear consensus reached on the correct answer yet.

Contextual Notes

Participants are working under the constraints of a homework problem, which may include specific requirements for how answers should be presented. There is an emphasis on ensuring that the final answer is given in centimeters, as requested in the problem statement.

smillphysics
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1. A uniform meter stick is pivoted at its center. The meter stick has a M1 = 103 g mass suspended at the D1 = 28 cm position. Which is measured from the left end. At what position should a M2 = 74 g mass be suspended to put the system in equilibrium? Give your answer D2 in cm from the left end of the meter stick.



2. Because D2 is measured from the left D2= D2+.5m for the final answer.
Xcm= (m1D1+m2D2)/m1+m2


3. (.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg)
I get d2=.389m +.5m= .889 m which is wrong. How am I getting the math wrong or am I doing the problem incorrectly.
 
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You could use the center of mass equation to solve this problem. Where must the center of mass of M1 + M2 be for the stick to be in equilibrium about the pivot? (Measure all distances from the left end of the stick.) Set that center of mass equal to what it must be then solve for D2.
 
I'm not quite sure what you are saying here? I solved for D2 above.
 
smillphysics said:
I'm not quite sure what you are saying here? I solved for D2 above.
Show me the complete equation that you used. All I see is a formula for Xcm.
 
(.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg)=.5cm
D2=.806m ?
 
smillphysics said:
(.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg)=.5cm
D2=.806m ?
Exactly!

(Be sure to give the answer in cm as requested.)
 
You're the best!
 

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