Tension of a rope in a pulley system

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Homework Help Overview

The discussion revolves around a pulley system involving a 1.0 kg mass and a 4.0 kg mass connected by a cord over a frictionless pulley. Participants are exploring the acceleration of the larger mass and the tension in the cord.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the acceleration and tension using Newton's second law, expressing confusion over which mass to use for the tension calculation.
  • Some participants suggest drawing force diagrams to clarify the forces acting on each mass.
  • There is a focus on whether to use the effective mass (3 kg) or the total mass (5 kg) in the tension formula.

Discussion Status

Participants are actively engaging with the problem, with one providing a diagram to illustrate the system. Guidance has been offered regarding the use of different masses for calculating tension, but no consensus has been reached on the best approach.

Contextual Notes

Participants are navigating the complexities of the problem setup, including the implications of using different masses in their calculations. The discussion reflects an emphasis on understanding the relationships between forces in the system.

winthos
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Homework Statement


A 1.0 kg mass and a 4.0 kg mass are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. What is the acceleration of the larger mass? What is the tension on the cord?

Homework Equations



F(net force)=F(tension)-weight

F=ma


The Attempt at a Solution


I found the acceleration using F=ma. The system has 3 kg of mass going down (4kg-1kg), which is equal to 29.4N of force (found by multiplying 3kg by 9.8m/s^2). The whole system has a mass of 5 kg, so I substituted into F=ma.

29.4N=5.0 kg + a

The acceleration is 5.88 m/s^2

Now comes the part I am confused about. The tension of the cord should be
F(net force)=tension-weight, or rearranged to be:

F(tension)=F(net force)+weight or (Tension=ma+mg)

I can substitute in what i know:
Tension =m(5.88m/s^2)+m(9.8m/s^2)

I am at a loss of which mass to use. Do i use the mass that i found in working on acceleration (3 kg), or the mass of the whole system (5 kg)?
 

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can u draw the force diagrams of each mass? that will help u a lot :)
 
I attached a quick diagram, i don't know if that's detailed enough, but that is basically what's going on. (i'm bad at drawing with a mouse, so it probably doesn't look that spectacular)- the 4kg mass is pulling the whole system down at 5.88m/s^2, which is great since that is the first half of the question.

My problem is which mass do i use to find the tension? the mass of 3 kg (because of the pulley 4kg-1kg=3kg, the 3kg was then used to find the amount of downward force, which turned out to be 29.4N) or the mass of the whole system, 5kg?
 
Last edited:
winthos said:
I attached a quick diagram, i don't know if that's detailed enough, but that is basically what's going on. (i'm bad at drawing with a mouse, so it probably doesn't look that spectacular)- the 4kg mass is pulling the whole system down at 5.88m/s^2, which is great since that is the first half of the question.

My problem is which mass do i use to find the tension? the mass of 3 kg (because of the pulley 4kg-1kg=3kg, the 3kg was then used to find the amount of downward force, which turned out to be 29.4N) or the mass of the whole system, 5kg?

ok, u are done basically, u can use any of the masses actually, so let's say we concentrate on the heavier one

the heavy mass' equation of force will be:

-Mg+T=-Ma

-(4 kg)(9.8 m/s^2)+T=-(4kg)(5.88 m/S^2)

that's it, u've found it :)

can u find it now using the smaller mass? :)
 

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