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

In summary, the conversation discusses the calculation of the position where a 74 g mass should be suspended on a uniform meter stick to achieve equilibrium when a 103 g mass is suspended at 28 cm from the left end. The equation for this is Xcm = (m1D1 + m2D2)/(m1 + m2). The correct answer is D2 = 0.806 m or 80.6 cm from the left end of the meter stick.
  • #1
smillphysics
28
0
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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  • #2
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.
 
  • #3
I'm not quite sure what you are saying here? I solved for D2 above.
 
  • #4
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.
 
  • #5
(.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg)=.5cm
D2=.806m ?
 
  • #6
smillphysics said:
(.103kg*.28m+ .074kg*D2)/ (.103kg+.074kg)=.5cm
D2=.806m ?
Exactly!

(Be sure to give the answer in cm as requested.)
 
  • #7
You're the best!
 

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

1. What is the center of mass position?

The center of mass position is the point at which the mass of an object or system is evenly distributed in all directions. It is often referred to as the balance point or the center of gravity.

2. How is the center of mass position calculated?

The center of mass position can be calculated by finding the weighted average of the positions of all the individual masses in the object or system. It is given by the formula: xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where x is the position and m is the mass of each individual component.

3. Why is the center of mass position important?

The center of mass position is important because it helps us understand the overall motion and stability of an object or system. It also plays a crucial role in determining the effects of external forces on an object, such as torque and rotational motion.

4. Can the center of mass position be outside of an object?

Yes, the center of mass position can be located outside of an object. This is often the case with irregularly shaped objects or objects with varying densities. However, the center of mass position will always lie somewhere along the line of symmetry of the object.

5. How does the center of mass position affect an object's stability?

The lower the center of mass position, the more stable an object will be. This is because a lower center of mass means that there is less chance for the object to topple over due to external forces. Objects with a higher center of mass position are more prone to tipping over.

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