Where Should a 150N Weight Be Placed on a Ladder to Achieve Equilibrium?

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Homework Help Overview

The problem involves a ladder weighing 200N, with a 150N weight that needs to be positioned to achieve equilibrium. The scenario includes a man acting as a pivot and applying an upward force. The goal is to determine the placement of the weight such that the force does not need to counteract the ladder's weight.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the concept of moments and equilibrium, discussing different interpretations of the pivot point and the calculations involved in determining the weight's placement.

Discussion Status

The discussion reflects a collaborative exploration of the problem, with participants confirming each other's calculations and sharing different methods of reasoning. There is no explicit consensus, but multiple approaches have led to similar conclusions.

Contextual Notes

Participants note potential ambiguities in the problem setup, such as the exact location of the pivot and the interpretation of the ladder's center of mass. These factors influence their calculations and understanding of the problem.

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Homework Statement


A ladder weighing 200N is being held by a man. The man shoulder acts as a pivot. His arm is pushing an upward force F.

http://i.imgur.com/dMrcRtK.jpg

Where could a weight of 150N be hang upon, so that force F do not need to counteract the weight of the ladder

Homework Equations


Moment= F x Perpendicular Distance

The Attempt at a Solution


I am not sure if I am understanding it correctly. If I am not wrong, the question is asking me, where can I hang a weight of 150N, so that force F no need to counteract the weight of the ladder. Meaning, assume force F = 0N.

Let the distance be x
Total clockwise moment = total anti clockwise moment
150x = 200*0.5
150x = 100
x = 2/3
= 0.667m.

Therefore it needs to be hang 0.667m away from the pivot or at 3.167m of the ladder.

Am I understanding this correctly?
 
Last edited:
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From your picture, I thought the pivot was at 0.25m (that's where "F" is drawn) but I guess that 0.25m is just the length of his arm?

Anyway, if the pivot is at 0.5m then I believe you've done it correctly, because I did it a different way and got the same answer. (I could have made a mistake, but as far as I can tell it is correct.)
 
Yes the pivot is at 2.5 mark of the ladder from the left. Sorry my picture wasn't clear.

Is my understanding correct? Thank you!
 
It appears to be, from your math. But there are usually multiple ways to understand something.

For example, you used the center of mass of the entire ladder to solve the problem, whereas I solved it using the center of mass of only the first meter (from the left) of the ladder.

The reason is, in my eyes, the first meter (from the left) is the only important part (because the other 3 meters are symmetrical around the pivot and therefore cancel out). So I took the mass of the first meter (50 Newtons, since it's 1/4 of the entire ladder) and multiplied it by the distance from the pivot (2 meters) and then I set it equal to 150xWe created our equations based on different understandings, yet they still led to the same answer.
 

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