?
Solved

forces in a pulley

Posted on 2010-08-24
13
Medium Priority
?
316 Views
Last Modified: 2012-05-10
Hi

There is a diagram below taken from a video. The video says that because the cord is assumed to be massless and frictionless the 2 tension forces are equal in magnitude. I don't understand why they are the same.

I also don't understand the implication of the cord being massless. Does it have to be massless so you can ignore its weight?

thanks
pulley.jpg
0
Comment
Question by:andieje
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 7
  • 5
13 Comments
 

Author Comment

by:andieje
ID: 33510436
Does it mean that the tension force on the right of block 1 is the same as the tension force pulling up on block 2 because these are the same force but have had their direction changed by the pulley?
0
 

Author Comment

by:andieje
ID: 33510514
Also, why is the acceleration the same for both blocks? And i don't understand why the formula chages from

Fx = m1.a1
Fy = Ft - m2.g = m2.a2

to Fy = -m2.a2 (note the minus sign)


p3.jpg
p1.jpg
0
 
LVL 27

Accepted Solution

by:
d-glitch earned 2000 total points
ID: 33510590
You said it was a video, so I assume the masses are moving.

You might set up an experiment like this to measure the coefficient of friction between the table and sliding mass.

If the cord is not massless, the tension in the cord increases with height/length.
You can hang a rope vertically with no load.  The free end of the cord has T=0.
But 100 feet of rope can wiegh 50 pounds.

Any friction in the pulley takes/subtracts tension from the horizontal cord.
You could clamp the rope between two block (very high friction) and support the entire load.
This would leave not tension in the horizontal run.
0
Technology Partners: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 

Author Comment

by:andieje
ID: 33510597
Is it just because the accelerations have opposite directions? e.g. if box 1 was accelerating in the positive x direction then box 2 would be acclerating in negative y direction, and if box 1 was acclerating in the negative x direction then box 2 would be accelerating in the positive y direction, so the accelerations always hav opposite signs?

I still dont see why magnitudes of accelerations would be equal though
0
 

Author Comment

by:andieje
ID: 33510605
The boxes are not moving in the video
0
 
LVL 27

Expert Comment

by:d-glitch
ID: 33510692
The acceleration must be the same for both blocks because they are tied together under tension.

If the blocks are acceleration, so is the cord.  
If the cord were not massless, the vertical and horizontal tension forces would differ by the force
required to accelerate the cord.
0
 
LVL 27

Expert Comment

by:d-glitch
ID: 33510838
>>  Is it just because the accelerations have opposite directions?

Pretty much.  In a straight massless cord, the tension forces are always equal in magnitude
and opposite in direction.  The pulley lets you change the direction, but there is still the constraint
that you have noted.
0
 
LVL 32

Expert Comment

by:phoffric
ID: 33513170
Conflicting ideas:
>> The boxes are not moving in the video
>> why is the acceleration the same for both blocks?

Are you sure the boxes are not moving? Maybe you are looking at a snapshot picture of moving boxes. If not moving, the velocities are constant 0; and then accelerations are constant 0 (i.e., no accelerations).
0
 

Author Comment

by:andieje
ID: 33513584
its probably a snapshot of moving boxes but they aren't moving in the video. Like you say they are talking about acceleration so some movement is implied
0
 

Author Comment

by:andieje
ID: 33516510
I'm still not fully convinced why the boxes are accelerating at the same rate

would it be possible for the boxes to accelerate at different rate whilst tied together under tension?
0
 
LVL 27

Expert Comment

by:d-glitch
ID: 33517153
No.

In the context of this problem, the cord can not stretch.
If m1 moves, m2 moves the same distance.

The displacements, the velocities, and the accelerations all have to match.

m2*g is pulling on the vertical end of the cord.
Friction proportional to m1*g is pulling on the horizontal end.
This guarantees tension will be maintained.
0
 
LVL 27

Expert Comment

by:d-glitch
ID: 33520547
>>  The displacements, the velocities, and the accelerations all have to match at every instant.


0
 

Author Closing Comment

by:andieje
ID: 33526230
thanks
0

Featured Post

Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! https://www.google.com/about/careers/students/guide-to-technical-development.html How to Know You are Making a Difference at EE In August, 2013, one …
Article by: Nicole
This is a research brief on the potential colonization of humans on Mars.
Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaac…
I've attached the XLSM Excel spreadsheet I used in the video and also text files containing the macros used below. https://filedb.experts-exchange.com/incoming/2017/03_w12/1151775/Permutations.txt https://filedb.experts-exchange.com/incoming/201…
Suggested Courses

752 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question