How to find tension of a single object in a pulley system?

Click For Summary

Homework Help Overview

The discussion revolves around finding the tension in a pulley system involving multiple masses. The original poster attempts to apply Newton's second law to derive an expression for tension based on the forces acting on the system.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's approach to calculating tension, questioning the inclusion of acceleration in the final expression. There is an exploration of how to express acceleration in terms of mass and gravitational force.

Discussion Status

Some participants have provided feedback on the calculations, noting the need to express results solely in terms of mass and gravity. There is an ongoing examination of the assumptions regarding the system's acceleration and the total mass involved.

Contextual Notes

Participants are working under the constraint of needing to express their answers in terms of mass (m) and gravitational acceleration (g), as specified in the problem directions.

fissifizz
Messages
29
Reaction score
2

Homework Statement



LETTER d only [/B]
phsy.PNG


Homework Equations


Fnet = ma

The Attempt at a Solution


So I considered 2m independently and set its net force equal to its gravity minus its tension.

(2m)a = Fg - Ft = (2m)g - Ft

I then rearranged the equation so that Ft = (2m)g - (2m)a = 2m(g-a)

There's no answer key, so I'm not sure if this is correct. I appreciate any assistance with this problem! Thanks!
 
Physics news on Phys.org
The problem is that your answer contains a. The directions say to express your answers in terms of m and g.

So, you need to find an expression for a in terms of m and g.
 
Mister T said:
The problem is that your answer contains a. The directions say to express your answers in terms of m and g.

So, you need to find an expression for a in terms of m and g.

Ah, ok. Since all the blocks are moving in a system, I'm going to assume that the acceleration calculated in letter b will be used.

Fnet = (5m)a = 5mg - 4mg (I hope I'm calculating this correctly)
(5m)a = mg
a = g/5

So time to plug back in: Ft = 2m(g-a) = 2m(g-(g/5))
Ft = 2m(4g/5)
Ft = 8mg/5

It seems like that's it. Was my original methodology correct though?
 
fissifizz said:
Ah, ok. Since all the blocks are moving in a system, I'm going to assume that the acceleration calculated in letter b will be used.

What do you mean? The blocks are connected, and the magnitude of their acceleration is the same.

fissifizz said:
Fnet = (5m)a = 5mg - 4mg (I hope I'm calculating this correctly)

The net force is correct, mg, but the whole mass is not 5m.
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 22 ·
Replies
22
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K