# HW Lab PHYS101: Net torque of pulley system and meterstick position balance

1. Oct 31, 2012

### pmalayavech

THANKS, I ONLY NEED HELP #2 NOW, PLEASE
10/31/12 @ 9:10 pm
bump

1. The problem statement, all variables and given/known data
1)If the fulcrum of a 0.1 kg meter stick was placed at the 40 cm mark and a 200 g mass at the 0 cm mark, would it be possible to balance the meter stick with a 200 g mass on the long side of the meter stick? What would the position of the mass have to be to balance the meter stick?

2) What is the torque acting on the pulley system below

2. Relevant equations

LA1=LA2

3. The attempt at a solution

FROM DIAGRAM
#2, this is my first time working on this type of problem. how should I combine the other "counterclockwise" 2 pulleys? value correct? 10N x radius #2 of 1m= 10Nm

and the other T1=15N x radius#1 of 2m= 30Nm??

How do i combine these 2 forces? Are they negative?

Last edited: Oct 31, 2012
2. Oct 31, 2012

### tiny-tim

welcome to pf!

hi pmalayavech! welcome to pf!
yes, technically torque is a vector

but in a 2D case like this, all you need to say is "clockwise" or "anti-clockwise"

(I'm not clear … are you asking about question 1) also? )

3. Oct 31, 2012

### pmalayavech

Re: welcome to pf!

Tim!, hey thanks

that is correct, so it would be

(20N)(2m)+(15N)(1M)cc=(30N)(2m)cw

(45Joules)cc = (60Joules)cw

there for the net force is a Fnet of 15 Joules counterclockwise

YES, i need help on the first one as well, any idea, please.

4. Oct 31, 2012

### tiny-tim

hey pmalayavech!
i] are we using the same diagram?

ii] this shouldn't be an equation!!

you should simply be adding the torques (some positive, some negative)!
what have you done so far?

5. Oct 31, 2012

### pmalayavech

i) yes, same diagram

ii) ok, no equations, I was thinking about conservation, where net force will equal to zero.

ok, cc is negative and cw is positive? -40Nm-15Nm+60Nm= 15Nm

from the pulley, I am looking at the left side as negatives and the right side positives. I hope I had multiplied the radius to the right "T"

6. Oct 31, 2012

### tiny-tim

+60 is right, but i don't see how you got the other two

7. Oct 31, 2012

### pmalayavech

Alright, for #1

for Larm 1
I did r(1)=0-40
=-40cm
Tcc= -40 x 200
=-800 gcm for m1La2

now for m2LA2
i am stuck

8. Oct 31, 2012

### tiny-tim

let's see …
ok, now you need the torque for the 500g stick, at the 50 cm mark, and for the other 200g mass, at x cm

9. Oct 31, 2012

### pmalayavech

#1,

for Tcc= -8000gcm

the fact that fulcrum only moved to the left 10cm (at 40cm), where it ought to be balance at 50cm (center), if we had 2 equal weights on both side, it would balance "cc" at 0cm and "cw" at 100cm, agree?

so, since fulcrum moved to the left 10cm(40cm), weight at 0cm, on the "cw" side, the weight of that should moved 10cm to the left as well to balance. so the "cw" 200g mass should be at 90cm!!

yes agree?

also for #2, this is my first time working on this type of problem. how should I combine the other 2 pulleys? value correct? 10N x radius #2 of 1m= 10Nm

and the other T1=15N x radius#1 of 2m= 60Nm??

do i have atleast these values correct?

Thanks again

10. Oct 31, 2012

### pmalayavech

need help with the second one. please

11. Oct 31, 2012

### pmalayavech

Just the last one

for #2, this is my first time working on this type of problem. how should I combine the other 2 pulleys? value correct? 10N x radius #2 of 1m= 10Nm

and the other T1=15N x radius#1 of 2m= 60Nm??

do i have atleast these values correct?

12. Oct 31, 2012

### pmalayavech

still trying, and no references to check answers with

13. Nov 1, 2012

### tiny-tim

hi pmalayavech!

(just got up :zzz:)
yes
no

i] it would have to be 20 cm, wouldn't it?

ii] anyway, you're not taking the weight of the ruler itself into account

write the equation out properly, or you'll never learn how to do these problems​
yes, that's correct now!

14. Nov 1, 2012

### pmalayavech

Thank you, I understand now. after working out the problem, it was common sense to have to have it at 20cm, to evenly distribute the weight.
8000cc=8000cw, duhh

And yes the "net" force is adding it all up i presume
130 Joules

I was able to turn this in complete now, even though late, its fine. I understand thanks
Timmy