How Does a Painter Balance on a Board While Painting?

  • Thread starter Thread starter turandorf
  • Start date Start date
  • Tags Tags
    Equilibrium
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a painter balancing on a board while painting. The scenario includes a painter of mass 100 kg standing on a 60 kg board that is supported by saw horses, with specific measurements provided for the board's length and overhang. The problem requires analyzing forces and torques to determine the balance conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the forces acting on the board and the painter, questioning how to calculate the forces at the supports and the conditions for tipping. They discuss the application of torque and the distribution of weight between the supports.

Discussion Status

Participants are actively engaging with the problem, offering insights into the calculations needed for torque and balance. Some have provided numerical values and attempted to clarify the relationships between the forces and distances involved. There is an ongoing exploration of the conditions for tipping and the role of the center of mass.

Contextual Notes

There is some confusion regarding the center of mass and how it is affected by the positions of the painter and the board. Participants are also considering the implications of removing one of the supports and the resulting balance conditions with paint cans being added to the board.

turandorf
Messages
18
Reaction score
0

Homework Statement


A physics student of mass mm = 100 kg gets a summer job painting houses. On his first project, he builds a platform using large pine board with a mass of mb = 60 kg The board has an overall length of L = 8.5 meters and set on two 1.1 meter tall saw horses so that l = 2.25 meters overhangs each side.
(a) the student stands over the support at point B, calculate the force exerted by the beam on the support at A__.
(b) b)How far from the left end of the beam can the painter stand before the board (and painter) begin to tip over?
(c) He now 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 = 1.75 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.)


Homework Equations



Torque=F(r)
F=ma

The Attempt at a Solution


I thought part (a) would be mg*r but that didn't work.
 
Physics news on Phys.org
In a) if he is standing over the support, then what weights are acting through that point?

For b) the sum of the torques (or moments if you like) will need to be 0, or there will be an incident and some paint to clean off the ground.

Where is the center of mass acting relative to a saw horse?

At tipping then won't his distance from the say horse times his weight need to be greater than the center of mass times its distance from the same point?
 
For (a) the weight is (100+60)9.81=1569.6N (man+board) Is this just split between the two points? Or do I have to use Torque=F(r)? (in which case r would be 4 meters [distance between B and A] right?)
 
turandorf said:
For (a) the weight is (100+60)9.81=1569.6N (man+board) Is this just split between the two points? Or do I have to use Torque=F(r)? (in which case r would be 4 meters [distance between B and A] right?)

Isn't just half the weight of the board acting through each support? Then the weight of the man is additionally acting over the support at one end?
 
Oh wow yeah can't see why I couldn't figure that out. (It equaled 294.3N) Thanks for the insight! So in (b), the torque of the man can't exceed 294.3(4+x), right? (Where x is the distance from the support to the painter and the 4 comes from the distance from point A to B.) Then if the torque pf the man equals Mg*x you can just set the two equations equal and sole for x. Let me know if there are any flaws in this. Thanks!
 
turandorf said:
Oh wow yeah can't see why I couldn't figure that out. (It equaled 294.3N) Thanks for the insight! So in (b), the torque of the man can't exceed 294.3(4+x), right? (Where x is the distance from the support to the painter and the 4 comes from the distance from point A to B.) Then if the torque pf the man equals Mg*x you can just set the two equations equal and sole for x. Let me know if there are any flaws in this. Thanks!

In b) you are dealing with a balance. First choose the point that it will pivot about ... like the saw horse at whatever end he stands at. He can move outside that only so long as the weight at the center of mass acting at guess where ... the center of mass ... times its distance from the pivot is not exceeded by his distance away to the outside times his weight.
 
OK so i pivot around point B. So the weight of the painter times the distance past B has to equal the center of mass times the distance to B which would be 2 m right? Or is the center of mass including the painter as well as the board? I really am confused, sorry.
 
turandorf said:
OK so i pivot around point B. So the weight of the painter times the distance past B has to equal the center of mass times the distance to B which would be 2 m right?

Yes. That's right.

Consider the elements acting separately is fine.

What you are determining is the condition at which the center of mass of the system, painter and board, lays no farther than the support point at the saw horse.
 

Similar threads

Replies
6
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
8K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 15 ·
Replies
15
Views
6K
Replies
5
Views
2K
Replies
5
Views
4K