Help on a kinematics (possibly) question ?

  • Thread starter Thread starter Mark Rice
  • Start date Start date
  • Tags Tags
    Kinematics
Click For Summary

Homework Help Overview

The discussion revolves around a kinematics problem involving a pulley system with a 100N force applied upwards. The original poster is tasked with proving values related to the system, which includes a 5kg mass and a 2kg mass. There is confusion regarding the application of forces and the resulting accelerations in the system.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the masses and the system's acceleration. There are attempts to derive equations based on the forces and to understand the implications of the 100N force. Questions arise about the validity of splitting the force between the masses and the physical justification for various approaches.

Discussion Status

Participants are actively engaging with the problem, suggesting different methods to analyze the system. Some have proposed equations to relate the forces and accelerations, while others express uncertainty about their calculations and the assumptions made. There is no explicit consensus on the correct approach, but several productive lines of reasoning are being explored.

Contextual Notes

There is mention of the problem being intentionally challenging, as it was given to the original poster alone. Participants are also considering the implications of the string connecting the masses and how that affects their accelerations.

Mark Rice
Messages
37
Reaction score
0

Homework Statement

See attached picture please. It is supposed to be a pully system with 100N being applied upwards and i need to prove the values given in the diagram i drew and my teacher gave me it as a question to do since we have finished the scottish higher course.

Homework Equations

Obviously it is using Fun=m * a but i am thrown off by the idea of the 100 N force upwords.

The Attempt at a Solution

My whole class was revising for exams etc and my teacher gave me this question to see if i could do it. In the 50 minute period i was really confuzed and had no idea how to do it and it is really irritating me. I obviously know that the unbalanced force on the 5kg mass must be 1N to produce that acceleration and the unbalanced force on the 2kg mass must be 30.4N. But i have no idea how to prove it ?
 

Attachments

  • Physics question.png
    Physics question.png
    10.7 KB · Views: 498
Last edited:
Physics news on Phys.org
The whole system has an upward acceleration(possibly)
100N - m_1g - m_2g = m_1a_{sys} + m_2a_{sys} = (m_1+m_2)a_sys
The heaviest will have an acceleration less than the system and the lightest will have a bigger one(do you follow?)
 
dikmikkel said:
The whole system has an upward acceleration(possibly)
100N - m_1g - m_2g = m_1a_{sys} + m_2a_{sys} = (m_1+m_2)a_sys
The heaviest will have an acceleration less than the system and the lightest will have a bigger one(do you follow?)

Yeah i tried this and didn't really know where to go with it, (ie working out the upward unbalanced force of the whole system). PS I do higher physics at high school and i think my teacher gave me this question to try challenge me - additional information.
 
One person pointed out that if i just split ( ? ) the 100N, 50 N between each mass, (ie upward force of 50N on box 2kg and 19.6N downwards = 30.4N Fun = 15.2 m/s/s but i thought that might just have been a lucky guess and that there was no physics behind the idea of splitting of the 100 N into 50N on each mass. Am i right in assuming this is incorrect physics and if that is in fact the reason why; why would you split the 50N.
 
I think that that method has no physical justification. Try to pretend the boxes don't have that 100N first. The acceleration downward of the heavy mass must equal the acceleration upward of the smaller(The string can't be stretched).
Then you could solve the eq.s assuming the string is massless:
m_{small}a_{small} = T - m_{small}g
-m_{big}a_{big} = m_{big}a_{small} = T - m_{big}g
Then you could add the acceleration of the system to the found acceleration(sign sensitive)
But you use that a_{small} = -a_{big}
 
Hmmmm, I'll try that ^ and see if that works, so i just add the accelerations together (being careful with + and -) Would this be a physically valid method of proving it ?
 
Yes think how a string which can't stretch connects two objects, one falling and one rising, how could these accelerations' magnitudes be different?
 
dikmikkel said:
Yes think how a string which can't stretch connects two objects, one falling and one rising, how could these accelerations' magnitudes be different?

Yeah, this is the exact same thing that i was thinking when my teacher gave me the problem.
 
And also the system accelerates with some magnitude upward and the smaller mass accelerates with greater magnitude upward because of the connection to the greater mass, isn't it intuitive?
 
  • #10
Well i just tried the above method ^, (finding system acceleration and adding etc) but i didn't really get the right answer
 
  • #11
Looks to me like the small mass' acceleration is double my result. I think it is an error?
 
  • #12
dikmikkel said:
Looks to me like the small mass' acceleration is double my result. I think it is an error?

Sorry was otherwise occupied, but I'm really not sure what to do then. I think this question was supposed to be intentially difficult since i was the only one she gave it to. It isn't compulsary but it's just irritating not knowing why.
 
  • #13
Flattering though huh :)
I get the accelerations to be:
Big mass: 0.27 m/s^2
Small mass: 8.67 m/s^2
 
  • #14
Hmmmm, thanks very much for your help, i'll ask my teacher tomorrow if there is an error in the question :)
 
  • #15
I think there is, but would you mind posting if there isn't?
 
  • #16
dikmikkel said:
I think there is, but would you mind posting if there isn't?
No not at all :)
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 6 ·
Replies
6
Views
12K
  • · Replies 36 ·
2
Replies
36
Views
10K
  • · Replies 16 ·
Replies
16
Views
4K