Tension of a rope in a pulley system

AI Thread Summary
In a pulley system involving a 1.0 kg and a 4.0 kg mass, the acceleration of the larger mass was calculated to be 5.88 m/s² using the net force equation. The tension in the cord is derived from the forces acting on the masses, leading to confusion about which mass to use for calculations. It was clarified that either mass can be used to find tension, but focusing on the heavier mass simplifies the process. The equation for the heavier mass shows how to calculate tension based on its weight and acceleration. Ultimately, both masses can be utilized to arrive at the correct tension value.
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? :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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