# Homework Help: Help on a kinematics (possibly) question ?

1. Apr 25, 2012

### Mark Rice

1. The problem statement, all variables and given/known data 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.

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

3. 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 ?

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Last edited: Apr 25, 2012
2. Apr 25, 2012

### dikmikkel

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?)

3. Apr 25, 2012

### Mark Rice

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.

4. Apr 25, 2012

### Mark Rice

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.

5. Apr 25, 2012

### dikmikkel

I think that that method has no physical justification. Try to pretend the boxes dont have that 100N first. The acceleration downward of the heavy mass must equal the acceleration upward of the smaller(The string cant 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 $a_{small} = -a_{big}$

6. Apr 25, 2012

### Mark Rice

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 ?

7. Apr 25, 2012

### dikmikkel

Yes think how a string which can't stretch connects two objects, one falling and one rising, how could these accelerations' magnitudes be different?

8. Apr 25, 2012

### Mark Rice

Yeah, this is the exact same thing that i was thinking when my teacher gave me the problem.

9. Apr 25, 2012

### dikmikkel

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. Apr 25, 2012

### Mark Rice

Well i just tried the above method ^, (finding system acceleration and adding etc) but i didn't really get the right answer

11. Apr 25, 2012

### dikmikkel

Looks to me like the small mass' acceleration is double my result. I think it is an error?

12. Apr 25, 2012

### Mark Rice

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. Apr 25, 2012

### dikmikkel

Flattering though huh :)
I get the accelerations to be:
Big mass: 0.27 m/s^2
Small mass: 8.67 m/s^2

14. Apr 25, 2012

### Mark Rice

Hmmmm, thanks very much for your help, i'll ask my teacher tomorrow if there is an error in the question :)

15. Apr 25, 2012

### dikmikkel

I think there is, but would you mind posting if there isn't?

16. Apr 25, 2012

### Mark Rice

No not at all :)