What is the tension in the suspended cord when masses are released?

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Homework Help Overview

The problem involves determining the tension in a cord supporting a pulley system with two masses. The setup includes two masses, one weighing 1.2 kg and the other 3.2 kg, with the pulley and cords considered to have negligible mass.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve for the tensions in the ropes and questions the appropriate equation for the tension in the supporting cord. Some participants suggest that the analysis of forces on the pulley is simpler than initially thought.

Discussion Status

Participants are exploring the relationship between the tensions in the ropes and the tension in the supporting cord. There is a suggestion that the tension in the cord is simply twice the tension in the ropes, and some clarification is provided regarding the absence of a weight term for the pulley.

Contextual Notes

There is an ongoing discussion about the assumptions made regarding the mass of the pulley and the implications for the equations used in the analysis.

Ltcellis
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Homework Statement



Suppose the pulley is suspended by a cord C

Determine the tension in this cord after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords.

GIANCOLI.ch04.p54.jpg


Mass 1 : 1.2kg
Mass 2 : 3.2kg
Pulley and String mass is negligible

Homework Equations



T-m1g = -m1a
T-m2g = m2a
Tension of pulley = T(C) -mg - 2T = ma(?)

The Attempt at a Solution



So I solved for the tensions in both ropes. Since they're equal I got T=17N
For acceleraton I got 4.4545455m/s. I'm just not too sure on the equation on the Tension of Cord C. Would it be the equation I mentioned above? When I use that I get an answer around 96. Is that right?
 
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No, it's much simpler than that. Since the pulley is massless, the sum of the forces on it must add to zero. (Note further that there is no mg term acting on the pulley.)
 
So it's just T(C) = 2T?

because I was thinking mg was referring to the total weight.
 
Ltcellis said:
So it's just T(C) = 2T?
Yep, that's all it is.
because I was thinking mg was referring to the total weight.
You are analyzing the pulley, so mg can only refer to the weight of the pulley, which is zero.
 
Thanks a bunch
 

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