How do you calculate the length of a board using torque?

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Homework Help Overview

The problem involves calculating the length of a board using torque principles, specifically in the context of a cat walking on the board and causing it to lift off a support. The scenario includes a 7 kg cat and a 15 kg board, with the cat positioned at a specific point when the board begins to lift.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the balance of torque on both sides of the board and the implications of the board lifting off the support. Some question the interpretation of the problem, particularly regarding the position of the cat and the center of mass of the board. Others suggest using the relationship between forces and distances to set up equations.

Discussion Status

The discussion includes various interpretations of the problem setup and attempts to apply torque equations. Some participants have proposed specific equations and methods to approach the problem, while others are still seeking clarification on the setup and assumptions involved.

Contextual Notes

There is mention of a specific educational context (NCDES Physics 12) and references to resources that may not have covered the necessary concepts in detail. Participants also note the challenge of solving the problem without knowing the length of the board upfront.

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Homework Statement


A 7kg cat walks to the end of a 15 kg board resting on two saw horses as shown below.
When the cat is at the 1m position as shown, the board “just” lifts off of
the left hand support. How long is the board?



Homework Equations



T = D x F

The Attempt at a Solution


I know that the torque on both sides has to be equal, but I have no idea how to approach this without knowing the length!
 

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Perhaps I'm misreading the question but it does say that "A 7kg cat walks to the _end_ of a 15 kg board".

So you can calculate how far the centre of mass of the board is from the right hand horse (call it L).
 
Hello,

Sorry for grave-digging this thread, but I have the exact same problem as you. Perhaps you were also a NCDES Physics 12 student with the same workbook aha. I may have figured out the solution, for all the NCDES Physics 12 students googling this question aha. No where in our lessions were we shown how to do this with just T=Fd

Because the board is about to lift, there is now weight on the left hand side and that saw horse can be removed so there is just one pivot point.

http://hyperphysics.phy-astr.gsu.edu/hbase/cmms.html (m1gL1=m2gL2)

Plug in the values and solve for L2, then find the total distance.
 
CyberneticTitan said:
Hello,

Sorry for grave-digging this thread, but I have the exact same problem as you. Perhaps you were also a NCDES Physics 12 student with the same workbook aha. I may have figured out the solution, for all the NCDES Physics 12 students googling this question aha. No where in our lessions were we shown how to do this with just T=Fd

Because the board is about to lift, there is now weight on the left hand side and that saw horse can be removed so there is just one pivot point.

http://hyperphysics.phy-astr.gsu.edu/hbase/cmms.html (m1gL1=m2gL2)

Plug in the values and solve for L2, then find the total distance.
So what answer do you get?
 
haruspex said:
So what answer do you get?

Using m1gL1=m0gL2

(7kg)(9.8m/s2)(1m)=(15kg)(9.8m/s2)(L2)

Solve for L2 results in 0.4666...meters, which is the distance from the pivot to the center of gravity.

Add 1 meter to find the distance from the cat to the center of gravity, which is 1.4666...meters.

Double to find the total length of the board, which is 2.9333..meters before sigfigs.

This is right according to: http://bclearningnetwork.com/LOR/media/ph/learning_guides/PH12_2013/PH12_U2_notes.pdf (Page 19, answers in the back.)
 

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