Painter (another Equilibirum problem)

  • Thread starter Thread starter Bryon
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on a physics problem involving equilibrium and torque calculations. A student weighing 95 kg constructs a platform using a 50 kg pine board, measuring 8.5 meters in length, supported by saw horses. The student experiments by repositioning one support and calculating the necessary mass of paint cans (1.59 kg each) required for balance. The correct torque equation derived is 1/3*95 - 1/2*50 = 2/3*m, leading to the solution for the total mass of cans needed.

PREREQUISITES
  • Understanding of torque and equilibrium principles in physics
  • Knowledge of basic mechanics and forces
  • Ability to perform algebraic manipulations and solve equations
  • Familiarity with gravitational force calculations (g = 9.81 m/s²)
NEXT STEPS
  • Study the concept of torque in rotational dynamics
  • Learn how to apply equilibrium equations in practical scenarios
  • Explore the effects of varying mass distributions on balance
  • Investigate real-world applications of equilibrium in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and equilibrium problems, as well as educators seeking practical examples for teaching rotational dynamics.

Bryon
Messages
98
Reaction score
0

Homework Statement


A physics student of mass mm = 95 kg gets a summer job painting houses. On his first project, he builds a platform using large pine board with a mass of mb = 50 kg The board has an overall length of L = 8.5 meters and set on two 1 meter tall saw horses so that l = 2.25 meters overhangs each side as shown above.


Later that day, after thinking about how cool rotational dynamics really is, the student decides to conduct an experiment. He removes one of the supports and places the other one 1/3 of the way from the left edge. Standing at the end of the board, he has his girl friend place paint cans, each of mass mc = 1.59 kg, on the opposite end. How many cans will the girl have to place on the board to provide the best balance? (You may neglect the small length of the board that both the man and the cans occupy. Assume both are points at the ends of the board.)

https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-12-Equilibrium/painter/painter_3.jpg


Homework Equations



torque = rFsin(theta)


The Attempt at a Solution


torque(A) = (1/3)mass(man)g - (2/3)*g*mass(cans) - (1/2)*g*mass(board) = 0
torque(A) = (1/3)*95*9.81 - (2/3)*9.81*m(cans) - (1/2)*9.81*50 = 0

I think for the mass of the cans I got 9.94 (finding the number of cans is easy), and I don't think its correct. I think I'm missing something, any ideas?
 
Last edited by a moderator:
Physics news on Phys.org
Check your calculation.
You can rewrite the second equation by canceling 9.81. It becomes
1/3*95 - 1/2*50 = 2/3*m
Now solve for m.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
26
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 10 ·
Replies
10
Views
8K