What is the mass of the meter stick?

  • Context: Graduate 
  • Thread starter Thread starter izmeh
  • Start date Start date
  • Tags Tags
    Balance Mass
Click For Summary
SUMMARY

The mass of the meter stick is determined to be 169 grams based on the balance of torques around the fulcrum. Initially, mass A (100 grams or 0.98 N) is positioned at 12.3 cm, while mass B (1.02 N) is at 86.1 cm. After mass B is removed, the fulcrum shifts 14 cm to the left, resulting in a new fulcrum position at 36 cm. The calculated torque from mass A and the meter stick leads to the conclusion that the weight of the meter stick is 1.659 N, translating to a mass of 169 grams.

PREREQUISITES
  • Understanding of torque and its calculation
  • Familiarity with the concept of center of mass
  • Basic knowledge of weight and mass conversion (N to kg)
  • Experience with balance problems in physics
NEXT STEPS
  • Study the principles of torque and equilibrium in physics
  • Learn about the center of mass and its applications in mechanics
  • Explore problems involving balance bars and fulcrums
  • Investigate the relationship between weight and mass in different gravitational fields
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in understanding the principles of balance and torque in physical systems.

izmeh
Here's the situation proposed in a problem.
I have a balance bar that is a meter stick.
It has a fulcrum at 50cms. Mass A(100gms[.98nt]) is at 12.3cms and mass B(1.02nt) is located at 86.1cms.
Mass B has been removed and the fulcrum C shifts fo the left by 14cms.
I need to determine the mass of the meter stick. How do i do this?
 
Physics news on Phys.org
The first part "It has a fulcrum at 50cms. Mass A(100gms[.98nt]) is at 12.3cms and mass B(1.02nt) is located at 86.1cms." just tells you that the bar itself has center of mass located at its center (since A is 50-12.3= 37.7 cm from the center and B is 86.1-50= 36.1 cm from the center- 37.7*.98= 36.1*1.02 (well, close enough for experimental error)).

After B is moved, the fulcrum shifts "14 cm to the left" so it is now at 36 cm. We still have mass A at 12.3 cm, 23.7 cm from the fulcrum. Its torque is .98*23.7= 23.226 (actually 0.23226 N-m). The center of mass of the bar is still at 50, 14 cm from the fulcrum so, taking the weight of the bar to be W, the torque around the fulcrum is
14W. To be balanced we must have 14W= 23.226 or W= 1.659 N. That means that the mass of the bar is 1.659/9.81= 0.169 kg or 169 grams.
 
thank you
 

Similar threads

  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 8 ·
Replies
8
Views
12K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
9K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
23K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K