How Far Should a 5.6 kg Box Be Placed for Balance?

Click For Summary
SUMMARY

The discussion focuses on the physics problem of balancing a board with a 2.0 kg box at one end and determining the distance a 5.6 kg box must be placed from the opposite end for equilibrium. Using the equation mass1 * (0.5 m) * (2.0 m) = mass2 * d, the calculated distance for the 5.6 kg box is 0.357 meters. This solution confirms the application of the principle of moments in static equilibrium scenarios.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with the concept of moments in physics
  • Basic algebra for solving equations
  • Knowledge of mass and distance relationships in balancing systems
NEXT STEPS
  • Study the principle of moments in greater detail
  • Explore static equilibrium problems involving multiple forces
  • Learn about torque and its applications in physics
  • Investigate real-world applications of balance in engineering
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding balance and equilibrium in physical systems.

skandol
Messages
3
Reaction score
0
A 1.2 kg board 2.0 m long is pivoted exactly at its center. A 2.0 kg box is at one extreme end. For the board to remain balanced, a 5.6 kg box should be placed how far away from the other end


Homework Equations



mass1(.5)(2.0)= mass2 (d)

The Attempt at a Solution



.357 meters
 
Physics news on Phys.org
Welcome to Pf.

That looks ok.
 

Similar threads

Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 10 ·
Replies
10
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 15 ·
Replies
15
Views
6K
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
8K