Explain Force & Acceleration for Tension in Rope Around Cylinder Drum

  • Thread starter Thread starter tnutty
  • Start date Start date
  • Tags Tags
    Explain
Click For Summary

Homework Help Overview

The discussion revolves around the dynamics of a mass tied to a rope wrapped around a solid cylindrical drum, focusing on the concepts of force, tension, and acceleration. The original poster presents a scenario where a 55kg mass falls with an acceleration of 1.0 m/s², seeking to understand the relationship between gravitational force and the resulting tension in the rope.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the reasoning behind subtracting the acceleration from gravitational force to find tension, questioning the implications of the mass's acceleration and its relationship to gravitational acceleration.

Discussion Status

The discussion is active, with participants expressing confusion over the calculations and the underlying physics principles. Some guidance has been offered regarding the relationship between net acceleration and tension, but multiple interpretations of the problem are still being explored.

Contextual Notes

Participants are grappling with the definitions of acceleration and the effects of forces acting on the mass, including the tension in the rope and gravitational force. There is a lack of consensus on the correct interpretation of the forces involved.

tnutty
Messages
324
Reaction score
1

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.
 
Physics news on Phys.org
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?
 
  • #10
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:
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
8K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 10 ·
Replies
10
Views
6K