Physics: Finding the Unknown Weight in a Moment Problem

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Homework Help Overview

The problem involves a uniform beam in equilibrium, with a known weight and length, supported at a pivot. The task is to determine the unknown weight that, when added to one end of the beam, maintains equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss identifying forces acting on the beam and the importance of drawing a free body diagram. There are questions about the placement of the beam's weight and the role of the pivot in the equilibrium condition.

Discussion Status

Participants are actively engaging with the problem, providing hints and guidance on how to approach the equilibrium condition. There is recognition of the need to account for all forces, including the reaction force at the pivot, and some participants express understanding of the concepts involved.

Contextual Notes

Some participants mention the importance of considering the distances from the pivot when calculating moments, indicating that assumptions about the beam's weight distribution and pivot location are being examined.

seiei
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Homework Statement


A uniform beam of weight 50N is 3.0m long and is supported on a pivot situated 1.0m from one end. When a load of weight W is hung from that end, the beam is in equilibrium. What is the value of W?

I know that the sum of all moments about any given point on an object in equilibrium is equal to zero but I'm not sure how to go about this, any help would be appreciated!
 
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Start by identifying the forces acting on the beam. Draw a diagram. Hint: Where does the weight of the beam act?
 
Start by drawing a free body diagram of the plank and a pivot point. Put in all the forces and the appropriate distance where it acts.

Hint: Uniform beam means that the weight acts at its geometrical centre.
 
At the centre of the beam the weight of 50N acts downwards. That means 1xW=0.5x50 which means W=25N. Is that right?
 
seiei said:
At the centre of the beam the weight of 50N acts downwards. That means 1xW=0.5x50 which means W=25N. Is that right?

You forgot a force, the pivot point provides a reaction!

Else your sum of forces in the vertical direction would not balance.
 
So the sum of the weight of the beam and the weight of the load = The reaction force? Sorry I'm new to this :(
 
seiei said:
So the sum of the weight of the beam and the weight of the load = The reaction force? Sorry I'm new to this :(


Yes that is right. You see on your free body diagram the beam's weight and the load act in the same direction?

For equilibrium, if the sum of the forces in the vertical direction weren't balance then the beam would move down. Agree?
 
Yes I see that, thanks! So the weight of the load (25N) and the weight of the beam (50N) are added together to make the reaction force on the pivot (75N)?
 
seiei said:
Yes I see that, thanks! So the weight of the load (25N) and the weight of the beam (50N) are added together to make the reaction force on the pivot (75N)?

Yes.


(normally you'd need to be careful where you take your moments in these types of things. Since you took moments about the pivot point, the distance of R from the pivot was 0, so it worked out fine)
 

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