Hanging mass on a massless pulley

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SUMMARY

The discussion centers on the dynamics of a two-block system connected by a massless pulley, specifically analyzing the forces acting on a 3.5 kg block and a 2.8 kg block. The equations of motion are established as Fnet(block 1) = (Force of tension) - (Force of friction) = m1a and Fnet(block 2) = (Force of tension) - (Force of gravity) = m2a. A key insight is that the sign of the net force for the second block must be negative when it is falling, which resolves discrepancies in calculations. This highlights the importance of consistent directional assumptions in analyzing the system.

PREREQUISITES
  • Understanding Newton's Second Law of Motion
  • Familiarity with tension in a pulley system
  • Knowledge of free-body diagrams
  • Basic concepts of friction and gravitational forces
NEXT STEPS
  • Study the effects of different masses on pulley systems
  • Learn about free-body diagram construction for multi-block systems
  • Explore the role of friction in dynamic systems
  • Investigate the implications of massless pulleys in mechanical systems
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Physics students, educators, and anyone interested in understanding the mechanics of pulley systems and the dynamics of connected masses.

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)?
 
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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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