How Far Can a Person Walk on a Plank Before It Tips?

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Homework Help Overview

The problem involves a uniform plank of length 6 m and mass 30 kg, which is positioned horizontally on a scaffold with part of it overhanging. The question is how far a 70 kg person can walk on the overhanging section before the plank tips over.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of torque and the setup of the problem, including the choice of pivot point. There are attempts to clarify the definitions and calculations related to torque and forces acting on the plank.

Discussion Status

The discussion includes various interpretations of the problem setup and calculations. Some participants provide calculations while others seek clarification on specific terms and values used in the equations. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working with the assumption of uniform gravitational acceleration and discussing the implications of the mass distribution on the plank. There are also references to potential misunderstandings regarding terminology related to torque and lifting.

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The problem is a uniform plank of length 6 m and mass 30 kg rests horizontally on a scaffold, with 1.5 m of the plank hanging over one end of the scaffold. How far can a 70 kg person walk on the overhanging part of the plank before it tips?

Relevant equations
Sum of all the forces = 0
Sum of the torque must= 0Attempt-
(1) I make the pivot point at the point of overhang so 1.5 of of the board is over the pivot point and -4.5 is to the left of the pivot point.
(2) 1.5 is .25 of the length so it is .25 of the mass so the mass over the pivot point is 7.5. 4.5 is .75 of the length so it is .75 of the mass so the mass to the left of the pivot point is 22.5 kg.
(3) since the sum of all the torque is 0 then 22(-4.5)+7.5(1.5)+70(x)= 0
(4) solving for x I got 1.29 the correct answer is 0.643.
 
Last edited:
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First you calculate the torque you need to lift the bar
Torque=F*d then T=300 * 1.5 = 450 N*m
now you do sum of the torque must be = 0
-450 + 700 *x = 0
solve for x = 0,643
 
What do you mean by "lift" the bar?
 
Tip the plank down. my bad xD
 
Ok I understand. Where did you get the 300 in calculating the torque needed to lift the bar?
 
its 30kg * 10 m/s²
 
Ok thanks a lot!
 

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