How far to the right can a person walk before the plank will

  • Thread starter Thread starter QaH
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a uniform plank supported at two points, with a person standing on it. The scenario requires determining how far the person can walk to the right before the plank tips, considering the forces acting on both the plank and the person.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the center of mass of the plank and its relation to the position of the person. Questions arise regarding specific values used in the calculations, such as the origin of the 3 m measurement.

Discussion Status

Some participants are clarifying the steps taken in the calculations, while others are confirming the correctness of the expressions derived. There is an ongoing exploration of the assumptions and interpretations of the problem setup.

Contextual Notes

The problem setup includes specific measurements and forces, which may influence the interpretation of how the person’s position affects the stability of the plank. The discussion reflects a need for clarity on these values and their implications.

QaH
Messages
6
Reaction score
0

Homework Statement


A uniform plank 6.1 m long rests on two supports, 2.5 m apart (see figure below). The gravitational force on the plank is 130 N. The left end of the plank is 1.5 m to the left of the left support, so the plank is not centered on the supports. A person is standing on the plank one tenth of a meter to the right of the right support. The gravitational force on this person is 814 N. How far to the right can the person walk before the plank begins to tip?

Homework Equations

The Attempt at a Solution



The center of mass, since the plank is uniform, is 6.1m/2=3.05m
This means that the center of mass is 4m-3.05m=0.95m from the right support.
Our person is standing 0.1 meters to the right of the right support.
0.95m(130N)=(0.1m+x)814N
solving for x I got
[0.95m(130N)/814N]-0.1m=x
 
Last edited:
Physics news on Phys.org
Where did the 3m come from?
It would help if you explain what you did.
 
I changed what I did a bit in the description.
 
QaH said:
[0.95m(130N)/814N]-0.1m=x
Looks right.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
6K
Replies
14
Views
5K
  • · Replies 4 ·
Replies
4
Views
15K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
1
Views
2K
Replies
5
Views
10K