Could you please validate my answer?

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Homework Help Overview

The problem involves a uniform plank supported at two points, with a painter walking on an overhanging section. The goal is to determine how far the painter can walk before the plank tips over. The subject area includes concepts of static equilibrium and torque in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up an equation based on moments to find the tipping point. Some participants question the clarity and correctness of the original poster's mathematical expressions and reasoning.

Discussion Status

Participants are actively engaging with the original poster's work, seeking clarification on the steps taken and the variables used. There is a focus on understanding the setup of the problem and the application of torque principles, with no clear consensus reached on the correctness of the original poster's approach.

Contextual Notes

Some participants note potential confusion regarding the definitions of variables and the physical setup, including the distances involved and the forces acting on the plank. There are indications of missing explanations for certain terms and calculations.

qwerty11
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Homework Statement



A uniform plank of length 6.00m and mass 30.0 kg rests horizontally across two horizontal bars of a scaffold. the bars are 4.50 m apart, and 1.50m of the plank hangs over one side of the scaffold. How far can a painter of mass 70.0kg walk on the overhanging part of the plank before it tips?


Homework Equations



w1*d-w2*d=0

The Attempt at a Solution



70g*d*Na*0-30g+((6.2)-1.5)+Nb*4.5=0
70g*d-30g+((6/2)-1.5)=0
d=(30g*(6/2)-1.5)=0
d=0.6428m

Is my answer correct and is meters the right unit d should be in?
 
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Your explanation is incomprehensible to me. What are you attempting to do here?

AM
 
That is the distance the painter can walk on the overhang park of the plank before it tips.
 
qwerty11 said:
That is the distance the painter can walk on the overhang park of the plank before it tips.
I understand what the question is asking. I am having difficulty understanding your work. Can you explain the steps you are taking and your reasoning? You have to find the moments about the support nearest the overhang. I don't understand how you get this:

70g*d*Na*0-30g+((6.2)-1.5)+Nb*4.5=0

For example, why do you multiply by 0 in the first term? Why are there 4 terms here and only two in your formula? Why do you subtract 30g? What is Na and Nb? What does 6.2 represent?

AM
 
70g should be 70kg which is the mass of the painter. Na is the weight of the plank (again should be kg instead). 6.2 should be 6/2 which is the length of the board on each side of the pivot. -1.5 is subtracting the non-overhang part out. 4.5 is the distance the bars are apart from each other.
 
Bump.
 
Bump.
 
70g*d*Na*0-30g+((6.2)-1.5)+Nb*4.5=0
It is still difficult to make sense of this.

The torque due to the painter is (70 kg)*g*d that appears at the beginning of your expression. It makes no sense to multiply that by the weight Na of the plank or by zero, and you have done both.

(30 kg)*g is the weight of the plank. You are adding that to the moment arm of the plank's center-of-mass, "6/2 - 1.5", but you need to multiply them in order to get the torque due to the plank's weight.

Why does the plank's weight appear twice in the equation, as Na and as 30*g?

You never explained what Nb is.
 

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