Explain Force & Acceleration for Tension in Rope Around Cylinder Drum

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The discussion revolves around calculating the tension in a rope tied to a 55kg mass falling with an acceleration of 1.0 m/s². The tension is derived from the difference between gravitational acceleration (9.8 m/s²) and the given acceleration, leading to the formula T = mg - ma. Participants clarify that the mass is not accelerating faster than gravity but rather is subject to a net acceleration of 1.0 m/s² due to the opposing force of tension. The confusion stems from understanding how to account for both gravitational force and the resulting acceleration in the tension calculation. Ultimately, the tension in the rope is confirmed to be 484N based on this analysis.
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Homework Statement



A 55kg mass is tied to a massless rope wrapped around a solid cylindrical drum. The drum is mounted on a frictionless horizontal axle. When the mass is released, it falls with acceleration 1.0m/s^2 .

Find the tension in the rope.

my answer : 484N

the way I got is by doing this : 55(9.8-1.0).

I did this because my friend told me this but am confused on
why I have to subtract g from acceleration. Please explain in details.
 
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tnutty said:

Homework Statement



A 55kg mass is tied to a massless rope wrapped around a solid cylindrical drum. The drum is mounted on a frictionless horizontal axle. When the mass is released, it falls with acceleration 1.0m/s^2 .

Find the tension in the rope.

my answer : 484N

the way I got is by doing this : 55(9.8-1.0).

I did this because my friend told me this but am confused on
why I have to subtract g from acceleration. Please explain in details.

If there was no retarding force then the acceleration from gravity would have been unaffected. So for acceleration to be just 1, then there must be a counteracting force that relates to 9.8 - 1 ... in order to end with 1 as your acceleration.
 
This statement confuses me :
When the mass is released, it falls with acceleration 1.0m/s^2 .

Its falling down at 1 meters per second per second. This makes me think that
with gravity counted, it falls at 1 m/s/s.
 
tnutty said:
This statement confuses me :
When the mass is released, it falls with acceleration 1.0m/s^2 .

Its falling down at 1 meters per second per second. This makes me think that
with gravity counted, it falls at 1 m/s/s.

Right it does mean that.

It means that gravity is accelerating it at 9.8 and the drum is decelerating it at 8.8. The 8.8 then is the Tension in the cable.

Otherwise if it was free falling it would be slack.
 
Why should it not be decelerating at 1m/s/s instead of accelerating at 1m/s/s.
It seems like if if its accelerating then we should add 1.0 instead of subtract.
 
Ftotal = ma …

and Ftotal = mg - T :wink:
 
Ya but why is tension m*8.8 instead of m*10.8. After all it is falling 1.0m/s/s with the
gravity? It just seems counter intuitive to me.
 
tnutty said:
Ya but why is tension m*8.8 instead of m*10.8. After all it is falling 1.0m/s/s with the
gravity? It just seems counter intuitive to me.

No. It's not falling 1m/s2 faster than gravity.

It's falling at just 1m/s2

If it was static what would the Tension be? m*g right? And it's not moving.

So as TinyTim pointed out the tension will be m*g less the net acceleration of 1 m/s2
 
So the net acceleration is 1m/s/s.

If so then why isn't m*a = m*(1). Isn't a the net acceleration?
 
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tnutty said:
So the net acceleration is 1m/s/s.

If so then why isn't m*a = m*(1). Isn't a the net acceleration?

Yes …
tiny-tim said:
Ftotal = ma …

and Ftotal = mg - T :wink:

ma = mg - T,

so m*(1) = mg - T, and T = mg - m*(1).

No? :smile:
 
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