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

[pmalayavech]

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

Homework Statement

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

Homework Equations

LA1=LA2

Torque net= radius x force

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?I understand 60Nm is correct from the clockwise but not the other. Please help!

Please help -- Greatly Needed

 

[tiny-tim]

welcome to pf!

hi pmalayavech! welcome to pf!

I found the sum of the forces for each body (the pulley and the hanging weight (Newton)) but I don't know how to do the equation, it is a vector so it should be negative one end and positve on the other.

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? )

 

[pmalayavech]

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? )

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.

 

[tiny-tim]

hey pmalayavech!

Tim!, hey thanks

that is correct, so it would be

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

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)!

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

what have you done so far?

 

[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"

 

[tiny-tim]

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

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

 

[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

 

[tiny-tim]

let's see …

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?

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

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

 

[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

 

[pmalayavech]

need help with the second one. please

 

[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?

 

[pmalayavech]

still trying, and no references to check answers with

 

[tiny-tim]

hi pmalayavech!

(just got up :zzz:)

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?

yes

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!

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​

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?

yes, that's correct now!

 

[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