Hanging mass on a massless pulley

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The discussion revolves around the dynamics of two blocks connected by a massless pulley, focusing on the equations of motion for each block. The user questions why assuming the net force (Fnet) for the hanging block (mass 2) is negative yields the correct answer, while setting the forces of friction equal does not. It highlights the importance of consistent sign conventions for acceleration when analyzing the system, as both blocks share the same acceleration magnitude but may differ in direction. Additionally, the user seeks clarification on how tension and weight can have different values in this scenario, especially when considering the impact of fixing mass 1 to the table. The conversation emphasizes understanding the relationship between forces acting on each block in a pulley system.
kasnay
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Homework Statement
A 3.5kg block is on a tabletop and is attached by a string to hanging block of mass 2.8 kg. The blocks are released from rest and allowed to move freely. If the table has a coefficient of friction of 0.4

A) Find the Acceleration
Relevant Equations
fnet x and fnet y
I have a conceptual question about this problem.
I can write the 3.5 kg block equation as Fnet(block 1)=(Force of tension)-(Force of friction)=m1a
I can write the 2.8 kg block as Fnet(block 2)=(Force of tension)-(Force of gravity2)=m2a

My question is this
If I set the forces of friction equal I get the wrong answer. However If I assume the fnet of block 2 is negative (because its going to fall) I get the correct answer.
Why do I need to assume the second block fnet is negative? Shouldnt the math already account for what needs to happen?
 
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You used the same symbol ##a## for the acceleration of both blocks. That means you are assuming the two blocks have the same acceleration - including the sign of their acceleration. If you take the positive direction for block 1 to be in the direction of the tension force acting on block 1, then what must be the positive direction for block 2 (downward or upward)?
 
kasnay said:
I can write the 2.8 kg block as Fnet(block 2)=(Force of tension)-(Force of gravity2)=m2a

My question is this
If I set the forces of friction equal I get the wrong answer.
How can tension and weight of mass 2 have different values?
How those two compare if you screw mass 1 to the table?

You mention forces of friction, but there is only one, if mass 2 is freely hanging from the pulley.
 
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