Solving Confusions on Pulling a Spool

  • Thread starter ChiralSuperfields
  • Start date
In summary: Thank you for your reply @erobz!I think the hand would have unwound the same amount since each part as the angular velocity.
  • #71
haruspex said:
Since ##F_{net}=T\hat i - F_f\hat i##, ##W_{net} =( T\hat i - F_f\hat i)\cdot (r+\frac I{mr})\theta\hat i=(T-F_f) (r+\frac I{mr})\theta##.

Btw, just noticed that there’s a false assumption we all may have been making implicitly, that the friction acts to the left. In fact, it will act to the right. But the equations are fine, it's just that the value of Ff will be negative.
Thank you for your reply @haruspex!

So from that equation static friction contributes to the net work (so dose do work actually?). Assuming that ## T > F_f## then the ## W_{net} > 0##

Many thanks!
 
Physics news on Phys.org
  • #72
Callumnc1 said:
Thank you for your reply @haruspex!

So from that equation static friction contributes to the net work (so dose do work actually?). Assuming that ## T > F_f## then the ## W_{net} > 0##

Many thanks!
No!! It does no work whichever way it acts, for all the reasons given before.
 
  • Like
Likes ChiralSuperfields
  • #73
haruspex said:
No, no! ##L+l## relative to the ground.
The spool moves ##L## relative to the ground and the hand moves #l## relative to the spool.
Thank you for your reply @haruspex!

How dose it the hand move ##L + l## relative to the ground?

haruspex said:
What it says. If there were no friction, the the spool would rotate too fast for rolling contact, so the force of friction from the ground acts in the forward direction (right, in the diagram).
How would the spool rotate if there was no friction? Oh I guess the tension would provide a torque and therefore an angular acceleration and there would be other torque to counter it I guess.

Many thanks!
 
  • #74
haruspex said:
No!! It does no work whichever way it acts, for all the reasons given before.
Thank you for your reply @haruspex!

How sorry, the force of static friction is in the net work equation?
##W_{net} =(T-F_f) (r+\frac I{mr})\theta##

Many thanks!
 
  • #75
If the force of friction is to the right then actually the net work equation should be

##W_{net} =(T + F_f) (r+\frac I{mr})\theta## (since both T and F_f act in positive x-direction)

Is this please correct @haruspex ?

Many thanks!
 
  • #76
Callumnc1 said:
How dose it the hand move L+l relative to the ground?
The centre of the spool moves ##L## and rotates an angle ##\theta##. So the top of the spool moves further than the centre. A length ##l## of string that was wrapped around the spool has been payed out and is now horizontal.
Callumnc1 said:
How sorry, the force of static friction is in the net work equation?
Suppose you want to lift up a heavy plank, weight W, at its middle. The plank lies on two supports height H. You position a block, height H, at X from its midpoint. You go round to the other side and insert a rod under the midpoint of the plank to rest on the block. You hold the rod Y from the plank and lift distance L. The plank rises R.
Work done =WR
Lift force = FL, normal force from block =FB.
FL+FB=W
FLY=FBX
R/X=L/(X+Y)
Which leads to, work done by you = LFL=WR
You did all the work, the block did none.
Yet, both your lifting force and the normal force from the block feature in the net force on the plank.
You will just have to accept that a force that contributes to the net force does not necessarily do work.
Callumnc1 said:
If the force of friction is to the right then actually the net work equation should be

##W_{net} =(T + F_f) (r+\frac I{mr})\theta## (since both T and F_f act in positive x-direction)
No, as I wrote, the equations are fine. The sign you put on a force in an equation only has to match the convention you are adopting for which way is positive for that force. When you wrote -F in your force balance equation that just meant you were taking left to be positive for that force. When you later discover that F has a negative value that tells you that it acts to the right. There is no need to change any algebra.
 
  • Like
Likes ChiralSuperfields
  • #77
@Callumnc1

Look, you have to take a minute to draw a diagram if you are going to learn to physics...

1677300330262.png
 
  • Like
Likes ChiralSuperfields
  • #78
haruspex said:
The centre of the spool moves ##L## and rotates an angle ##\theta##. So the top of the spool moves further than the centre. A length ##l## of string that was wrapped around the spool has been payed out and is now horizontal.

Suppose you want to lift up a heavy plank, weight W, at its middle. The plank lies on two supports height H. You position a block, height H, at X from its midpoint. You go round to the other side and insert a rod under the midpoint of the plank to rest on the block. You hold the rod Y from the plank and lift distance L. The plank rises R.
Work done =WR
Lift force = FL, normal force from block =FB.
FL+FB=W
FLY=FBX
R/X=L/(X+Y)
Which leads to, work done by you = LFL=WR
You did all the work, the block did none.
Yet, both your lifting force and the normal force from the block feature in the net force on the plank.
You will just have to accept that a force that contributes to the net force does not necessarily do work.

No, as I wrote, the equations are fine. The sign you put on a force in an equation only has to match the convention you are adopting for which way is positive for that force. When you wrote -F in your force balance equation that just meant you were taking left to be positive for that force. When you later discover that F has a negative value that tells you that it acts to the right. There is no need to change any algebra.
Thank you very much for your reply @haruspex !

I'll ponder over this for a while before replying :)

Many thanks!
 
  • #79
erobz said:
@Callumnc1

Look, you have to take a minute to draw a diagram if you are going to learn to physics...

View attachment 322838
Thank you very much for your diagram @erobz !

That is very helpful :) I was not quite sure how to go about drawing the diagram

Many thanks!
 
  • #80
Callumnc1 said:
Thank you very much for your diagram @erobz !

That is very helpful :) I was not quite sure how to go about drawing the diagram

Many thanks!
Well, it's something that take some practice...If you're not even going to try, stumble, and maybe fail... you aren't going to get better at it. Sometimes a picture is worth a thousand words. This thread kind of proves that point.
 
Last edited:
  • Like
Likes ChiralSuperfields
  • #81
erobz said:
Well, it's something that take some practice...If you're not even going to try, stumble, and maybe fail though you aren't going to get better at it. Sometimes a picture is worth a thousand words...This thread kind of proves that point.
Thank you for your reply @erobz !

I agree with everything you said, I will try to draw a diagram next time. Perhaps I should draw my own diagram for the spool?

Many thanks!
 
  • #82
Callumnc1 said:
Thank you for your reply @erobz !

I agree with everything you said, I will try to draw a diagram next time. Perhaps I should draw my own diagram for the spool?

Many thanks!
Almost always there is a useful diagram that can be drawn for a problem.

Have a good night(or day).
 
  • Like
Likes ChiralSuperfields
  • #83
erobz said:
Almost always there is a useful diagram that can be drawn for a problem.
Thank you @erobz , I agree!

Many thanks!
 
  • Like
Likes erobz
  • #84
erobz said:
Have a good night(or day).
You too! Thanks for your help!
 
  • Like
Likes erobz
  • #85
haruspex said:
The centre of the spool moves ##L## and rotates an angle ##\theta##. So the top of the spool moves further than the centre. A length ##l## of string that was wrapped around the spool has been payed out and is now horizontal.
Thank you for your reply @haruspex !

I guess it makes sense intuitively that the distance the hand has moved with respect to the ground is the distance moved by the CM (L) + the amount of string unraveled (l)

Many thanks!
 
  • #86
haruspex said:
No, as I wrote, the equations are fine. The sign you put on a force in an equation only has to match the convention you are adopting for which way is positive for that force. When you wrote -F in your force balance equation that just meant you were taking left to be positive for that force. When you later discover that F has a negative value that tells you that it acts to the right. There is no need to change any algebra.
Thank you for your reply @haruspex !

Ok but if I was taking left to be positive in this equation ##W_{net} =(T - F_f) (r+\frac I{mr})\theta## then tension should be ##-T \hat i## correct?

However, if we choose right to be positive then ##F_{net} = T \hat i + F_f \hat i## and ##W_{net} =(T + F_f) (r+\frac I{mr})\theta##

Is that please correct?

Many thanks!
 
  • #87
Callumnc1 said:
Thank you for your reply @haruspex !

Ok but if I was taking left to be positive in this equation ##W_{net} =(T - F_f) (r+\frac I{mr})\theta## then tension should be ##-T \hat i## correct?

However, if we choose right to be positive then ##F_{net} = T \hat i + F_f \hat i## and ##W_{net} =(T + F_f) (r+\frac I{mr})\theta##

Is that please correct?

Many thanks!
You took right as positive for Fnet and T, left as positive for Ff: ##F_{net}=T-F_f##, and with clockwise as positive for torque about ground level, ##\tau_{net}=2rT##.
With right as positive for all forces, ##F_{net}=T+F_f##, torque as before.
With left as positive for all forces, ##F_{net}=T+F_f##, and with clockwise still as positive for torque about ground level, ##\tau_{net}=-2rT##.
 
  • Like
Likes ChiralSuperfields
  • #88
haruspex said:
Suppose you want to lift up a heavy plank, weight W, at its middle. The plank lies on two supports height H. You position a block, height H, at X from its midpoint. You go round to the other side and insert a rod under the midpoint of the plank to rest on the block. You hold the rod Y from the plank and lift distance L. The plank rises R.
Work done =WR
Lift force = FL, normal force from block =FB.
FL+FB=W
FLY=FBX
R/X=L/(X+Y)
Which leads to, work done by you = LFL=WR
You did all the work, the block did none.
Yet, both your lifting force and the normal force from the block feature in the net force on the plank.
You will just have to accept that a force that contributes to the net force does not necessarily do work.
Thank you for your reply @haruspex !

That seems like the most non-intuitive physics fact I've ever heard of!

Do you know where to learn that fact from?

BTW, I am going to draw a diagram and post it sometime so that I make sure I understand your analogy correctly

Many thanks!
 
  • #89
haruspex said:
You took right as positive for Fnet and T, left as positive for Ff: ##F_{net}=T-F_f##, and with clockwise as positive for torque about ground level, ##\tau_{net}=2rT##.
With right as positive for all forces, ##F_{net}=T+F_f##, torque as before.
With left as positive for all forces, ##F_{net}=T+F_f##, and with clockwise still as positive for torque about ground level, ##\tau_{net}=-2rT##.
Thank you for your reply @haruspex !

What I did was wrong, correct?

I am not allowed to take right as positive for one force and left as positive for another force as you have to define one coordinate system and use it to for determining the direction of the forces

Many thanks!
 
  • #90
Callumnc1 said:
What I did was wrong, correct?
Yes,
Callumnc1 said:
I am not allowed to take right as positive for one force and left as positive for another force as you have to define one coordinate system and use it to for determining the direction of the forces
No, you are free to define positive separately for each force, each displacement, each acceleration. Just be clear about it and do it consistently through the equations.

If all such variables are along the perpendicular axes, it is a good idea to pick one convention for them all. If not, you could decompose each oblique variable into such components or define the conventions individually.

Many prefer to guess which way each will act and define positive accordingly. That's ok as long as you handle each consistently and accept that some may turn out to have negative values.

That said, there may be cases where the equation does change if you guessed wrongly, but I can only think of quite contrived ones.
 
  • Like
Likes ChiralSuperfields
  • #91
haruspex said:
Yes,

No, you are free to define positive separately for each force, each displacement, each acceleration. Just be clear about it and do it consistently through the equations.

If all such variables are along the perpendicular axes, it is a good idea to pick one convention for them all. If not, you could decompose each oblique variable into such components or define the conventions individually.

Many prefer to guess which way each will act and define positive accordingly. That's ok as long as you handle each consistently and accept that some may turn out to have negative values.

That said, there may be cases where the equation does change if you guessed wrongly, but I can only think of quite contrived ones.
Thank you for your reply @haruspex! That is very interesting!

Many thanks!
 

Similar threads

  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
734
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
294
  • Introductory Physics Homework Help
Replies
3
Views
672
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
922
  • Introductory Physics Homework Help
Replies
13
Views
4K
Back
Top