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

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The discussion focuses on determining the tension in a cord supporting a mass system involving a pulley. After releasing the masses, the tension in the cord was calculated to be 17N, with an acceleration of approximately 4.45 m/s². The key point clarified is that since the pulley is massless, the tension in the supporting cord (T(C)) is simply twice the tension in the ropes (T), leading to the equation T(C) = 2T. The confusion regarding the gravitational force (mg) acting on the pulley was addressed, confirming that it is zero due to the pulley’s negligible mass. Overall, the tension in the cord is straightforwardly derived from the system's dynamics.
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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