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

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In the pulley system lab, the focus was on predicting the weight of a known unknown mass using its y-component, as the system was in equilibrium. The x-component was not necessary for calculations because the horizontal forces balanced each other out, resulting in no net force in that direction. The tension forces T1 and T2 were resolved into their vertical components, which directly related to the weight of the unknown mass. The discussion emphasized that the unknown mass's weight could be determined by summing the vertical components of the tensions. Understanding these principles clarifies why the x-component was not needed in this specific scenario.
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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