Masses on strings and pulleys (iWTSE.org)

In summary: Please, see:https://technologystudent.com/gears1/geardex1.htmIn summary, the website states that two points, P and Q, have been set with values for acceleration. Two constraints were established, A=4 m/s² and B=0 m/s². If P moves up by 1 unit, then Q moves up by 1 unit as well. If B were stationary, we would have to assume that A wouldn't move either. However, because P rotates as well as moves, the constraints are maintained.
  • #1
TomK
69
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Homework Statement
The answer is 'C = 0 m/s²'.
Relevant Equations
Connected particles?
Pulley Dynamics 1.jpg

Pulley Dynamics 1 (Working).jpg


I don't understand the working shown on the website (seen above).

Firstly, they establish 'up' as the positive direction for acceleration, so A is +4 and D is -4.

They create two points, P and Q, and set acceleration values for them. I understand that connected particles have the same magnitude of acceleration, but I don't understand where they got '2' from.

The question states that B does not accelerate 'relative to the global coordinate system'. I looked-up this phrase, but I don't understand what it means.

This is what they say on the website: "The left-hand side of the system has two constraints: 'A = 4 m/s²' and 'B = 0 m/s²'. Looking at point P, these constraints mean that, for every 2 units that A rises in height by, point P rises by 1 unit." I don't know how they came to this conclusion.

Then, they say: "hence, the absolute acceleration of the connecting string is 2 m/s², so P = 2 m/s²." Again, why is it 2? I understand that Q must then be -2, but I don't get why P = 2m/s² in the first place.

I don't understand this question at all. Please help.
 
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  • #2
TomK said:
Looking at point P, these constraints mean that, for every 2 units that A rises in height by, point P rises by 1 unit.
Think about the length of the string connecting A and B. If P moves up 1 unit, how much more string is on the B side? How much less is on the A side?
TomK said:
Homework Statement:: The answer is 'C = 0 m/s²'.
Relevant Equations:: Connected particles?

relative to the global coordinate system
Here it just means relative to the support at the top of the apparatus.
 
  • #3
haruspex said:
Think about the length of the string connecting A and B. If P moves up 1 unit, how much more string is on the B side? How much less is on the A side?

Here it just means relative to the support at the top of the apparatus.

Sorry, I don't understand what you mean at all. Why does movement of the pulley affect movement of the mass blocks?

I am beginning to think that my application to university next year was a terrible mistake.
 
  • #4
TomK said:
Why does movement of the pulley affect movement of the mass blocks?
Actually, I should have been a bit more careful in my wording. We are not given that B is stationary, only that it is not accelerating.
But bear with me, suppose for the moment that it is stationary. If the pulley P moves up 1 unit, what happens to the distance PB? If the string does not break, what happens to the distance PA?
 
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  • #5
haruspex said:
Actually, I should have been a bit more careful in my wording. We are not given that B is stationary, only that it is not accelerating.
But bear with me, suppose for the moment that it is stationary. If the pulley P moves up 1 unit, what happens to the distance PB? If the string does not break, what happens to the distance PA?

If B were stationary, wouldn't we have to assume that A wouldn't move either? In the actual question, if A has a=4, why doesn't B have a=-4, if they are connected particles? I have the same confusion with C and D. I don't understand how this system works.
 
  • #6
TomK said:
If B were stationary, wouldn't we have to assume that A wouldn't move either?
No, you have to assume that pulley P rotates on its axis as well as moving up. The constraint is that the total length of each string is constant.
Try thinking of it the way around I described in post #4: if B is fixed in place but P is raised, what happens to A?
If it isn't obvious, do the experiment: tie one end of a string to the ground, the other to a small weight, and pass the string over a horizontal pencil. When you raise the pencil, what happens?
 
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  • #8
TomK said:
I am beginning to think that my application to university next year was a terrible mistake.
Relax, I have the same questions as you for this system and a general trouble understanding it , though I have a degree and master in Math (back from the distant 1999 when i did my master thesis which is on algorithms for numerical solutions to differential equations).
This problem the way i see it involves multiple frame of references (some of which are accelerating ) that's why it confuses me. Each moving pulley is a frame of reference according to my understanding. Whenever we consider multiple frames of reference I get confused. But that's just me.

Science is not always an easy way, sometimes there are obstacles, we must have patience and courage to overcome the obstacles.
 
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  • #9
haruspex said:
No, you have to assume that pulley P rotates on its axis as well as moving up. The constraint is that the total length of each string is constant.
Try thinking of it the way around I described in post #4: if B is fixed in place but P is raised, what happens to A?
If it isn't obvious, do the experiment: tie one end of a string to the ground, the other to a small weight, and pass the string over a horizontal pencil. When you raise the pencil, what happens?

I see what you mean by the pulley rotating about its centre, since it has the weight forces of both masses acting on opposite ends of its circumference.

If A has a=+4 (going upwards), does that mean B has to fall, therefore meaning that mass B is heavier than mass A, and the pulley is rotating clockwise?

How do they determine that point P is moving up and point Q is moving down?
 
  • #10
Delta2 said:
Relax, I have the same questions as you for this system and a general trouble understanding it , though I have a degree and master in Math (back from the distant 1999 when i did my master thesis which is on algorithms for numerical solutions to differential equations).
This problem the way i see it involves multiple frame of references (some of which are accelerating ) that's why it confuses me. Each moving pulley is a frame of reference according to my understanding. Whenever we consider multiple frames of reference I get confused. But that's just me.

Science is not always an easy way, sometimes there are obstacles, we must have patience and courage to overcome the obstacles.

I just think that I'm nothing exceptional, regardless of what it may be. Then, I ask myself, if I don't have any special skills, why would the world benefit from me studying engineering, in order to develop technology that makes lives better? There are many people on this forum alone that could take my place in life.
 
  • #11
TomK said:
I just think that I'm nothing exceptional, regardless of what it may be. Then, I ask myself, if I don't have any special skills, why would the world benefit from me studying engineering, in order to develop technology that makes lives better? There are many people on this forum alone that could take my place in life.
Hmm, you sound depressed for some reason.
Ok, so you want to be something exceptional, like Einstein for example. If you think Einstein was exceptional then you will encompass what he said that "genius is 99% hard work". It seems to me that you are on the right track as long as you study hard, come in this forum (and maybe other forums in internet) to discuss about the problems, and generally work hard toward your goal which is to become an exceptional engineer.

And don't worry too much if you will achieve your goal at the end, according to a Greek poet (K.P. Kavafis if you are interested) its not the final destination but the journey that counts. Its not so important if you will achieve your goal (of course as I said we must work hard towards our goal) but it is all the adventure you will have in your journey trying to achieve your goals. That's the thing that counts, and that's my view too.
 
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  • #12
TomK said:
I just think that I'm nothing exceptional, regardless of what it may be. Then, I ask myself, if I don't have any special skills, why would the world benefit from me studying engineering, in order to develop technology that makes lives better? There are many people on this forum alone that could take my place in life.
"I have no special talent. I am only passionately curious."

"It's not that I'm so smart, it's just that I stay with problems longer."

"The gift of fantasy has meant more to me than my talent for absorbing positive knowledge."

- Albert Einstein
 
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  • #13
TomK said:
I see what you mean by the pulley rotating about its centre, since it has the weight forces of both masses acting on opposite ends of its circumference.

If A has a=+4 (going upwards), does that mean B has to fall, therefore meaning that mass B is heavier than mass A, and the pulley is rotating clockwise?

How do they determine that point P is moving up and point Q is moving down?
Before you ask why again, let's observe together what is happening:

B can't fall because the problem is telling us so.
We also know that C is not moving, according to the known answer.
Then, we could draw an imaginary triangle that connects B, C and the fixed support of the big pulley, which is not moving either.
Neither mass B, the center of the big pulley or mass C are moving respect to each other.

Only masses A and D are moving up and down.
The three strings are keeping its respectives lenghts while unfolding around the pulleys.
Forced by the friction of the strings, the three pulleys are rotating clockwise because C+D are heavier than A+B.

pulley-dynamics-1-working-jpg.jpg
 
  • #14
Lnewqban said:
Before you ask why again, let's observe together what is happening:

B can't fall because the problem is telling us so.
We also know that C is not moving, according to the known answer.
Then, we could draw an imaginary triangle that connects B, C and the fixed support of the big pulley, which is not moving either.
Neither mass B, the center of the big pulley or mass C are moving respect to each other.

Only masses A and D are moving up and down.
The three strings are keeping its respectives lenghts while unfolding around the pulleys.
Forced by the friction of the strings, the three pulleys are rotating clockwise because C+D are heavier than A+B.

View attachment 272621
We only know B and C are not 'accelerating'. That doesn't mean they're not moving position.

If the left-pulley rotates, one mass has got to go up and the other has got to move down, correct, since the string is being pulled down more to the left/right by the heavier mass?

I don't see how the moving pairs of masses (i.e. A + B or C + D) can make the smaller pulleys move up/down (and the string connecting them go to the left/right). From the small pulley's point of view, if the same total weight force is acting on it, how can the string connecting the small pulleys move up/down?
 
  • #15
TomK said:
We only know B and C are not 'accelerating'. That doesn't mean they're not moving position.
The string linkage makes B share the condition of A: if A is acelerating, B must be either accelerating as well or not accelerating at all, B can't move position in any other way.
The same happens for C-D system.
 
  • #16
TomK said:
If the left-pulley rotates, one mass has got to go up and the other has got to move down, correct, since the string is being pulled down more to the left/right by the heavier mass?
...
That would be absolutely correct if you, as an observer, are standing at the small pulley.
Let's swith the observer's position, now you are standing at the fixed base of the big pulley.
You will see that A is moving and B is not, instead the small pulley is moving.
The string on the B-side grows in length exactly the same amount as the string on the A-side shrinks.

That tells you that:
1) The total length of the A-B string remains the same.
2) A is moving respect to both, the small and the big pulley.
3) B is moving respect to the small pulley.
4) The small pulley is moving respect to the big pulley.
5) B is not moving respect to the big pulley, as originally stated by the problem.
 
  • #17
TomK said:
I don't see how the moving pairs of masses (i.e. A + B or C + D) can make the smaller pulleys move up/down (and the string connecting them go to the left/right). From the small pulley's point of view, if the same total weight force is acting on it, how can the string connecting the small pulleys move up/down?
Can you visualize it if we assume the following?:
Weight of A = 10 Newtons
Weight of B = 20 Newtons
Weight of C = 30 Newtons
Weight of D = 40 Newtons
 
  • #18
Lnewqban said:
The string linkage makes B share the condition of A: if A is acelerating, B must be either accelerating as well or not accelerating at all, B can't move position in any other way.
The same happens for C-D system.
Then, why is acceleration of B zero, when acceleration of A is 4?
 
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  • #19
TomK said:
Then, why is acceleration of B zero, when acceleration of A is 4?
Because tension of string attached to mass B equals its weight.
##T=m_Bg##
The force pulling B up is as strong as force from gravity; therefore, they cancel each other and nothing motivates B to accelerate (##a=F{total}/m##).
 
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  • #20
Lnewqban said:
Because tension of string attached to mass B equals its weight.
And -how on earth- we know that? I don't seem to be able to infer it from the given data..

Btw the original statement doesn't say anything if all masses are equal or something @TomK is the original statement complete or is there something you didn't tell us
 
  • #21
I have my own explanation of this, i ll try to expose it with as much detail as possible.

We can break the acceleration of A (and B) into two components:
  1. an acceleration ##a_{sA}## which is the acceleration relative to the left small pulley ( that is if we take the left small pulley like it is not moving)
  2. an acceleration ##a_P## which is the acceleration with which the left small pulley is moving relative to the big center pulley, that is relative to the global reference frame. This acceleration is the same as the acceleration of point P that's why I chose that name.​
The total acceleration of A (relative to the global reference frame) is (by superposition principle) $$4=a_A=a_{sA}+a_P\text{ (1)}$$

The total acceleration of B is similarly ##a_B=a_{sB}+a_P## where the ##a_P ## term is the same cause B is also moving together with the small left pulley, as this pulley accelerates up.

Now it is crucial to notice that ##a_{sB}## and ##a_{sA}## are the accelerations that B and A would have if it was like the small left pulley wasn't moving so it will be ##a_{sB}=-a_{sA}##.

So we will have $$0=a_B=a_{sB}+a_P=-a_{sA}+a_P\text{ (2)}$$

(1),(2) are a simple system which we solve we find ##a_{sA}=a_P=2##.

I hope this helps. If you have questions please be free to ask. And remember K.P Kavafis: It doesn't matter so much if at the end you 'll understand this its all about the mini adventure coming to this forum and asking questions and communicating with us and the multinational scientific community of Physics Forums!.
 
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  • #22
Delta2 said:
And -how on earth- we know that? I don't seem to be able to infer it from the given data..

Btw the original statement doesn't say anything if all masses are equal or something @TomK is the original statement complete or is there something you didn't tell us

The question is complete, taken straight from here: https://i-want-to-study-engineering.org/q/pulley_dynamics_1/
 
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  • #23
Delta2 said:
And -how on earth- we know that? I don't seem to be able to infer it from the given data..
Because those are the only two forces on B and we are told it is not accelerating.
 
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  • #24
TomK said:
The question is complete, taken straight from here: https://i-want-to-study-engineering.org/q/pulley_dynamics_1/
Did you try a simpler problem among those before attempting solving the one we have been discussing?
This one is basically half as complex and the site shows a detailed solution:
https://i-want-to-study-engineering.org/q/pulley_dynamics_2/

It is important that you can see that both, velocity and acceleration of those pulleys, must be different and the reason behing that fact.
One (top-left) only redirects the tension of the string, while the other (bottom-right) has certain mechanical advantage (whatever it loses in force it gains in displacement and vice-verse).

pulley_dynamics_2.png

cub_simple_lesson05_fig5.jpg


cub_simple_lesson05_fig6.jpg
 
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  • #25
I have attempted the original question from the beginning. It spans multiple pages of solving simultaneous equations, so this post will only show the equations without their derivation. This is what each variable seen in the equations refers to:

T1 = tension in string connecting block A and block B
T2 = tension in string connecting left and right smaller pulleys
T3 = tension in string connecting block C and block D

mA = mass of block A
mB = mass of block B
mC = mass of block C
mD = mass of block D

aC = acceleration of block C (this is our solution)

g = gravitational acceleration (= 9.81)

6 unique equations:

Equation 0 (equation of motion of block B): mB.g = T1

Equation 1 (equation of motion of block A, T1 from equation 0 is substituted into here): 4.mA = mB.g - mA.g

Equation 2 (equation of motion of left pulley): T2 = mA.g + mB.g

Equation 3 (equation of motion of right pulley): T2 = mC.g + mD.g

Equation 4 (equation of motion of block D): T3 = 4.mD + mD.g

Equation 5 (equation of motion of block C): T3 = mC.aC + mC.gNew equations formed through equating any two previous equations:

Equation 6 (T2 equated from equations 2 and 3): mA + mB = mC + mD

Equation 7 (T3 equated from equations 4 and 5): 4.mD + mD.g = mC.aC + mC.g

Equation 8 (mA equated from equations 1 and 6): [mB.g / g + 4] = mC + mD - mB

Equation 9 (mD equated from equations 7 and 8): aC = [(mB / mC).(2g + 4)] - 2g - 4I am now questioning what I should do with equation 9. Do I have enough information from the other equations to find aC? This is where I am stuck.
 
  • #26
TomK said:
I have attempted the original question from the beginning. It spans multiple pages of solving simultaneous equations, so this post will only show the equations without their derivation. This is what each variable seen in the equations refers to:

T1 = tension in string connecting block A and block B
T2 = tension in string connecting left and right smaller pulleys
T3 = tension in string connecting block C and block D

mA = mass of block A
mB = mass of block B
mC = mass of block C
mD = mass of block D

aC = acceleration of block C (this is our solution)

g = gravitational acceleration (= 9.81)

6 unique equations:

Equation 0 (equation of motion of block B): mB.g = T1

Equation 1 (equation of motion of block A, T1 from equation 0 is substituted into here): 4.mA = mB.g - mA.g

Equation 2 (equation of motion of left pulley): T2 = mA.g + mB.g

Equation 3 (equation of motion of right pulley): T2 = mC.g + mD.g

Equation 4 (equation of motion of block D): T3 = 4.mD + mD.g

Equation 5 (equation of motion of block C): T3 = mC.aC + mC.gNew equations formed through equating any two previous equations:

Equation 6 (T2 equated from equations 2 and 3): mA + mB = mC + mD

Equation 7 (T3 equated from equations 4 and 5): 4.mD + mD.g = mC.aC + mC.g

Equation 8 (mA equated from equations 1 and 6): [mB.g / g + 4] = mC + mD - mB

Equation 9 (mD equated from equations 7 and 8): aC = [(mB / mC).(2g + 4)] - 2g - 4I am now questioning what I should do with equation 9. Do I have enough information from the other equations to find aC? This is where I am stuck.
In the problem in this thread, we are not given any masses or tensions, and it is a digression to try to work with forces at all.
It is purely a kinematics problem. If you assign acceleration variables to all of the points that can move independently (each mass, each pulley) then you can write equations for the 'accelerations' of the lengths of sections of string.

Let the lengths of the vertical sections of string above these points be LA for the section above A, LP for the section above P, etc.
What equation can you write relating ##\ddot{L_A}, \ddot{L_B}##? Hint: start with an equation relating LA, LB and the (constant) total length of the string, then differentiate twice.
Next, what equation can you write relating ##a_P, a_A, \ddot{L_A}##?
 
  • #27
Equation 2 and 3 of post #25 are wrong according to my opinion. The way you have written them is like you have assumed that the left and right pulleys have zero acceleration. I think the correct equations are:
$$T_2-(m_A+m_B)g=(m_A+m_B)a_L$$
$$T_2-(m_C+m_D)g=(m_C+m_D)a_R$$
where ##a_L,a_R## the accelerations of the left and right pulley respectively, for which we are not given that ##a_L=a_R=0##.
 
  • #28
Delta2 said:
Equation 2 and 3 of post #25 are wrong according to my opinion. The way you have written them is like you have assumed that the left and right pulleys have zero acceleration. I think the correct equations are:
$$T_2-(m_A+m_B)g=(m_A+m_B)a_L$$
$$T_2-(m_C+m_D)g=(m_C+m_D)a_R$$
where ##a_L,a_R## the accelerations of the left and right pulley respectively, for which we are not given that ##a_L=a_R=0##.
As I wrote in post #26, forces and masses are irrelevant to solving the problem given.
 
  • #29
haruspex said:
As I wrote in post #26, forces and masses are irrelevant to solving the problem given.
Yes ok, just saying those two equations are wrong, right?
Also what do you think about my post #21
 
  • #30
Delta2 said:
just saying those two equations are wrong, right?
Yes. Since the common mass centre of A and B is accelerating upwards, T2 must exceed the sum of their weights.
Delta2 said:
what do you think about my post #21
Yes, that is relevant and correct.
 
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  • #31
Delta2 said:
Equation 2 and 3 of post #25 are wrong according to my opinion. The way you have written them is like you have assumed that the left and right pulleys have zero acceleration. I think the correct equations are:
$$T_2-(m_A+m_B)g=(m_A+m_B)a_L$$
$$T_2-(m_C+m_D)g=(m_C+m_D)a_R$$
where ##a_L,a_R## the accelerations of the left and right pulley respectively, for which we are not given that ##a_L=a_R=0##.

Oh, I assumed ma = 0, since we are told the pulleys are light (negligible mass). If using forces is wrong, I guess it doesn't matter now that it's wrong.

I don't really understand how to solve the problem with kinematics. This problem is a mess.
 
  • #32
TomK said:
Oh, I assumed ma = 0, since we are told the pulleys are light (negligible mass)
Seems to me you got a point here. hmm for the moment i can't resolve this, let me think abit more...
TomK said:
I don't really understand how to solve the problem with kinematics. This problem is a mess
Please read my post #21. @haruspex says that it is correct...
 
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  • #33
Delta2 said:
Seems to me you got a point here. hmm for the moment i can't resolve this, let me think abit more...

Please read my post #21. @haruspex says that it is correct...

In post 21, what is this 'superposition principle' you mentioned? How do you know that the 'acceleration of the small pulley relative to big pulley' + 'acceleration of block relative to the small pulley' = 'absolute acceleration of the block'?
 
  • #34
TomK said:
In post 21, what is this 'superposition principle' you mentioned? How do you know that the 'acceleration of the small pulley relative to big pulley' + 'acceleration of block relative to the small pulley' = 'absolute acceleration of the block'?
It's a standard equation of relative motion.
If P moves up a distance yP, and A moves up a distance yAP relative to P (i.e. gets that much closer to P), then A must have moved up the sum of those two distances, yA=yP+yAP.
Differentiating produces a similar statement about relative velocities:
vA=vP+vAP
And again gives accelerations:
aA=aP+aAP.
 
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  • #35
haruspex said:
It's a standard equation of relative motion.
If P moves up a distance yP, and A moves up a distance yAP relative to P (i.e. gets that much closer to P), then A must have moved up the sum of those two distances, yA=yP+yAP.
Differentiating produces a similar statement about relative velocities:
vA=vP+vAP
And again gives accelerations:
aA=aP+aAP.

That makes sense. I'll try the question again with what you've suggested.
 
<h2>1. What is the principle behind masses on strings and pulleys?</h2><p>The principle behind masses on strings and pulleys is the conservation of energy and Newton's laws of motion. In this system, the tension in the string is equal throughout and the masses experience equal and opposite forces.</p><h2>2. How do you calculate the tension in a string in a system with multiple masses and pulleys?</h2><p>The tension in a string can be calculated using the equation T = mg, where T is the tension, m is the mass of the object, and g is the acceleration due to gravity. In a system with multiple masses and pulleys, the tension in the string will be equal throughout the system.</p><h2>3. What is the difference between a fixed pulley and a movable pulley in a mass and string system?</h2><p>In a fixed pulley, the pulley is attached to a fixed point and only changes the direction of the force. In a movable pulley, the pulley is attached to the object being lifted and changes both the direction and magnitude of the force. This results in a mechanical advantage, making it easier to lift heavy objects.</p><h2>4. How does the number of pulleys affect the mechanical advantage in a mass and string system?</h2><p>The more pulleys there are in a system, the greater the mechanical advantage. This is because each additional pulley reduces the amount of force needed to lift an object. For example, a system with four pulleys has a mechanical advantage of 4, meaning the force needed to lift the object is reduced by a factor of four.</p><h2>5. What are some real-world applications of masses on strings and pulleys?</h2><p>Masses on strings and pulleys are commonly used in everyday objects such as elevators, cranes, and weightlifting machines. They are also used in more complex systems, such as in sailboats to adjust the tension on the sails, and in exercise equipment to provide resistance for strength training.</p>

1. What is the principle behind masses on strings and pulleys?

The principle behind masses on strings and pulleys is the conservation of energy and Newton's laws of motion. In this system, the tension in the string is equal throughout and the masses experience equal and opposite forces.

2. How do you calculate the tension in a string in a system with multiple masses and pulleys?

The tension in a string can be calculated using the equation T = mg, where T is the tension, m is the mass of the object, and g is the acceleration due to gravity. In a system with multiple masses and pulleys, the tension in the string will be equal throughout the system.

3. What is the difference between a fixed pulley and a movable pulley in a mass and string system?

In a fixed pulley, the pulley is attached to a fixed point and only changes the direction of the force. In a movable pulley, the pulley is attached to the object being lifted and changes both the direction and magnitude of the force. This results in a mechanical advantage, making it easier to lift heavy objects.

4. How does the number of pulleys affect the mechanical advantage in a mass and string system?

The more pulleys there are in a system, the greater the mechanical advantage. This is because each additional pulley reduces the amount of force needed to lift an object. For example, a system with four pulleys has a mechanical advantage of 4, meaning the force needed to lift the object is reduced by a factor of four.

5. What are some real-world applications of masses on strings and pulleys?

Masses on strings and pulleys are commonly used in everyday objects such as elevators, cranes, and weightlifting machines. They are also used in more complex systems, such as in sailboats to adjust the tension on the sails, and in exercise equipment to provide resistance for strength training.

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