Physics: Finding the Unknown Weight in a Moment Problem

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A uniform beam weighing 50N and 3.0m long is supported by a pivot 1.0m from one end, and when a load W is added, the beam remains in equilibrium. To find W, it's essential to consider the forces acting on the beam and draw a free body diagram. The weight of the beam acts at its center, and the equilibrium condition requires that the moments around the pivot point sum to zero. The calculation shows that W equals 25N, which, along with the beam's weight, results in a total reaction force of 75N at the pivot. Understanding the balance of forces and moments is crucial for solving such equilibrium problems.
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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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