Find the maximum length of x that will maintain equilibrium

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Homework Help Overview

The discussion revolves around a physics problem involving two uniform planks, each with a mass of 10 kg and a length of 2 m, arranged in a way that requires finding the maximum length of one plank that can extend over the edge of another while maintaining equilibrium. The participants explore concepts related to moments, forces, and equilibrium conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the role of moments and pivots in determining equilibrium, with some suggesting that the edge of the lower plank acts as a pivot point. There are attempts to formulate equations based on the principles of moments, and questions arise regarding the assumptions made about the setup, such as the thickness of the boards and the fixed nature of the lower plank.

Discussion Status

There is a mix of confusion and exploration among participants, with some expressing uncertainty about how to proceed. A few participants have proposed equations and attempted to relate the lengths and masses involved, while others have pointed out potential errors in reasoning or formulation. The discussion is ongoing, with various interpretations and approaches being explored.

Contextual Notes

Participants note that the thickness of the boards is not specified, leading to assumptions that they can be treated as lines. There is also mention of gravity being assumed as 10 m/s² for calculations, and some participants question the necessity of substituting known values into equations before reaching a symbolic form.

  • #61
Richie Smash said:
Ah yes I see you're right, I understand now finally... wow..

So would you confirm 5x2/2 is onto something?
Yes.

So that gives us the [clockwise] torque due to gravity acting on the overhanging right end of the top board. Now we have to figure out the counter-clockwise torque from the remaining left half of the top board.

The mass of the left half is ___ ?
The position of the center of gravity of the left half is ___ ?
The torque due to gravity acting on the left half is then ___ ?
 
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  • #62
the mass of the left half would be 5(l-x) and the postion of the center of gravity would be (l-x)/2

So the Moment would be (5l2-10lx+5x2)/2
 
  • #63
Richie Smash said:
Ah yes I see you're right, I understand now finally... wow..

So would you confirm 5x2/2 is onto something?
Yes! Now you have the moment acting on the right side of the board, according to the diagram in terms of the length x of that side.
What is the length of the other side?
 
  • #64
tnich said:
Yes! Now you have the moment acting on the right side of the board, according to the diagram in terms of the length x of that side.
What is the length of the other side?
Oops, sorry, missed the last two posts. Yes, you have the right expression for the moment on the other side.
 
  • #65
Richie Smash said:
the mass of the left half would be 5(l-x) and the postion of the center of gravity would be (l-x)/2

So the Moment would be (5l2-10lx+5x2)/2
Good. That looks entirely correct.

The tipping point is when the torques from the two sides are equal. You have formulas for both torques. Can you write down an equation that states that they are equal?
 
  • #66
Yes
(5x2-10lx+5l2)/2=5x2/2

And i know l =2
So x ultimately is 1?
 
  • #67
Richie Smash said:
Yes
(5x2-10lx+5l2)/2=5x2/2

And i know l =2
So x ultimately is 1?
Yes.

One could phrase it more generally: ##x=\frac{l}{2}##. That is, regardless of what the board length is, the tipping point is when exactly half of the board hangs over.
 
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