Why didn't we need to calculate the x-component in our pulley system lab?

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Homework Help Overview

The discussion revolves around a lab experiment involving a pulley system where participants are tasked with predicting the mass of an unknown weight. The focus is on understanding the necessity of calculating the x-component of forces in the context of the system's equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the roles of tension in the pulley system and question why the x-component of forces was not calculated. There are attempts to resolve the tension into its components and explore the implications of equilibrium on the forces acting at a specific point in the system.

Discussion Status

Some participants have provided guidance on resolving forces and have prompted others to consider the mathematical expressions related to the components of tension. There is an ongoing exploration of the relationship between the forces and the unknown mass, with no explicit consensus reached yet.

Contextual Notes

Participants note the absence of a diagram that could clarify the setup of the pulley system, which may affect the understanding of the problem. The discussion is framed within the constraints of a lab assignment, emphasizing the experimental nature of the inquiry.

santoki
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We had a lab 2 weeks ago that dealt with predicting the "known unknown" masses (we know it's 250g and we have to experiment to try and get to 250g as close as possible) within a pulley system.

All three trials we had for part 1, where the known unknown was placed in the middle, we only needed to calculate for its y-component. I was just wondering why didn't we need to calculate the x-component?
 
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Hi santoki. Seems you forget to attach the diagram showing the pulley & rope system you were using.
 
NascentOxygen said:
Hi santoki. Seems you forget to attach the diagram showing the pulley & rope system you were using.

Oh sorry. Here is the system for trial 1:

qyVspeb.jpg
 
Let's label as point A that point on the M-shaped rope where the unknown mass is hanging from.

First step, resolve tension T1 into its horizontal and vertical components.
Repeat this for T2

Since the system is in equilibrium, what can you say about the sum at A of the horizontal components of all the forces acting there? See whether you can form this into a mathematical expression.

And about the sum of the vertical components at A? Do the same for that.

See any clues yet towards answering your original question?
 
NascentOxygen said:
Let's label as point A that point on the M-shaped rope where the unknown mass is hanging from.

First step, resolve tension T1 into its horizontal and vertical components.
Repeat this for T2

Since the system is in equilibrium, what can you say about the sum at A of the horizontal components of all the forces acting there? See whether you can form this into a mathematical expression.

And about the sum of the vertical components at A? Do the same for that.

See any clues yet towards answering your original question?

I was thinking that the masses hung at the sides served only to change the direction of the force applied and the unknown force is the equilibrant force which is equal in magnitude but of opposite direction to the resultant force. So if I add T1y and T2y together, the magnitude of this resultant vector will be the weight of the unknown mass.

Am I thinking of it correctly?
 
santoki said:
So if I add T1y and T2y together, the magnitude of this resultant vector will be the weight of the unknown mass.

Am I thinking of it correctly?
That's it in words, so can you express this mathematically?
 

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