Three masses suspended on pulley

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Homework Help Overview

The problem involves a massless, frictionless pulley system with three suspended masses: 3 kg, 2 kg, and 6 kg. The original poster seeks to determine the tension in the string between two blocks and the acceleration of the lower left-hand block.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to solve for the tension and acceleration using a system of equations derived from Newton's second law. They express confusion regarding the acceleration of the lower left-hand block and whether it differs from the calculated value.
  • Participants point out a potential signage error in the equations, prompting a discussion on the correct placement of signs in the equations.
  • There are questions about the specifics of how to adjust the equations based on the direction of forces acting on the masses.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's equations. There is an emphasis on clarifying the correct signs for the forces involved, but no consensus has been reached regarding the correct values for tension and acceleration.

Contextual Notes

Participants are addressing potential errors in the setup of the equations, particularly concerning the direction of forces and the resulting signs in the equations. The original poster expresses uncertainty about the acceleration of the blocks, indicating a need for further clarification.

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Homework Statement


A pulley is massless and frictionless. 3 kg, 2 kg, and 6 kg masses are suspended on a pulley.

(a)What is the tension T1 in the string between the two blocks on the left-hand side of the pulley?

(b)What is the magnitude of the acceleration of the lower left-hand block?



Homework Equations


I solved for (a) using the equations:
T2-m1g=m1a
m2g+T1-T2=m2a
m3g-T1=m3a


The Attempt at a Solution



the combining those equations:
-m1g+m2g+m3g=m1a+m2a+m3a

then I solved for a
-29.4+19.6+58.8=11a
a=4.45455 m/s^2

then plugged into T1=m3g-m3a and got T1=32.0727N (right

for part (b) I thought that I already had a since I needed it to slove for T1, but it is wrong...

So does the lower left-hand block have a different acceleration?


Also, I tried putting in -4.45455 m/s^2 since m3 is 6kg and ways more than m1 and m2 combined... also didn't work
 
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You have a signage error in your 2nd relevant equation...the weight acts down, so it should have a minus sign in front of it. Redo your calc for the acceleration and value of T1.
 
PhanthomJay said:
You have a signage error in your 2nd relevant equation...the weight acts down, so it should have a minus sign in front of it. Redo your calc for the acceleration and value of T1.

In front of the whole equation or just infront of all m2?
 
gap0063 said:
In front of the whole equation or just infront of all m2?

Well, just in front of m2...on the left side of the equation, the weight acts down, and T2 acts down , so both those terms are minus; T1 acts up, so that's plus; on the right side of the equation, the acceleration is up, so that's a plus sign.
 
PhanthomJay said:
Well, just in front of m2...on the left side of the equation, the weight acts down, and T2 acts down , so both those terms are minus; T1 acts up, so that's plus; on the right side of the equation, the acceleration is up, so that's a plus sign.

What about m2 in front of acceleration?
 
gap0063 said:
What about m2 in front of acceleration?

Repeat...positive.
 

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