Tension downward acceleration problems

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SUMMARY

The discussion focuses on calculating tensions T1 and T2 in a system of two masses (m1 and m2) under downward acceleration of g/2. Participants emphasize the importance of drawing free body diagrams (FBDs) for each mass to establish the relationships between the forces acting on them. Key equations derived include T2 = 0.5(m2)g and T1 = 0.5(m1 + m2)g, which are confirmed as correct. The conversation highlights the necessity of understanding the dynamics of the system, particularly when considering massless and frictionless pulleys.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Ability to draw and interpret free body diagrams (FBDs)
  • Knowledge of tension in strings and pulleys
  • Familiarity with basic gravitational force calculations (g = 9.8 m/s²)
NEXT STEPS
  • Study the principles of dynamics involving multiple masses and pulleys
  • Learn how to derive equations of motion for systems with accelerating components
  • Explore advanced topics in tension calculations in non-static systems
  • Review examples of free body diagrams in physics problems
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Physics students, educators, and anyone involved in mechanics or engineering who seeks to deepen their understanding of tension in dynamic systems.

dkgojackets
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Im supposed to find T1 and T2 using m1, m2, and g. The whole system is accelerating downwards with magnitude g/2.

Maybe if I get this I can figure out the more complex ones. Thanks.
 
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dkgojackets said:
phys.jpg


Im supposed to find T1 and T2 using m1, m2, and g. The whole system is accelerating downwards with magnitude g/2.

Maybe if I get this I can figure out the more complex ones. Thanks.
Draw free body diagrams for both the masses and see if you can work it out. We will help you if you show us your attempt.
 
Its tough to put in paint, but for m1 I have T1 going up, T2 going down, and .5(m1)g going down. On m2 I have T2 up and .5(m2)g down.

Looking at m2, I have the sum of forces being T2 - .5(m2)g

I'm not used to having the system accelerate, but right now I can only think of making T2 = .5(m2)g. T1 would be .5(m1 + m2)g then. I just don't think that is correct.
 
Nevermind, it is :)
 
OK here's the one I am really stuck on. I am just not completely sure what's going on. I'm looking for T. The units are kg for mass of the blocks, and I am assuming massless/frictionless pulleys.

threepulley.jpg


I drew the FBD for each mass, I am just not sure how to connect it all.
 
So far I am thinking about trying to balance the force on each side of that middle pulley. I have the total mass on the right side that its carrying as 7+2.5? (since half of that 5 is attached the the "floor".) For the left side, I am thinking the mass there now is .5 (half of the 1 is attached to the "ceiling", so I need to get another 9 kgs (and 9 x 9.8 = 88.2 N as T).

Is this completely wrong thinking?
 
dkgojackets said:
OK here's the one I am really stuck on. I am just not completely sure what's going on. I'm looking for T. The units are kg for mass of the blocks, and I am assuming massless/frictionless pulleys.
You have two strings and therefore two tensions. You need to know how far each mass moves in relation to the other masses. With that information, your FBDs should get you a solvable system of equations.
 
Still stuck. I think I am missing the effect that the masses have when they are attached to the pulley, not the actual string.
 
dkgojackets said:
Still stuck. I think I am missing the effect that the masses have when they are attached to the pulley, not the actual string.
I think I misinterpreted the problem. Nothing is moving here? Is that correct?

It is so frustrating when this system freezes up like this. I’ve been frozen out for a long time.
In case I cannot get back in, here is what I have assuming nothing is moving.

TL is the tension in the upper left string
TR is the tension in the lower right string.

T + 1kg*g = 2TL
TL = 7kg*g + 2TR
TR = 5kg*g
 
Last edited:
  • #10
Thank you. That makes sense now.
 

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