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

[smillphysics]

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.

 

[Doc Al]

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.

 

[smillphysics]

I'm not quite sure what you are saying here? I solved for D2 above.

 

[Doc Al]

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.

 

[smillphysics]

(.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg)=.5cm
D2=.806m ?

 

[Doc Al]

(.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg)=.5cm
D2=.806m ?

Exactly!

(Be sure to give the answer in cm as requested.)

 

[smillphysics]

You're the best!