Newton's laws - Pedagogical machine

In summary: Hey, thanks for checking this out .M2 and M3 are connected via a rope so,supossing it doesn't expand, those 2 masses should have the same acceleration,right ?No, not with respect to the Earth reference frame. (They would have the same acceleration relative to M1.)For example, what would happen to M3 if M2 were held fixed relative to the Earth while M1 moves to the left? What if M2 were released from rest? What if M2 were given an initial velocity to the left?If M2 were released from rest M3 would fall to the ground .If M2 were given an initial velocity to the left it would accelerate towards the point of attachment of the string .If
  • #1
Saitama
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Homework Statement


A "Pedagogical machine" is illustrated in the attachment. All surfaces are frictionless. What force F must be applied to ##M_1## to keep ##M_3## from rising or falling?

Ans. clue. For equal masses F=3Mg

Homework Equations





The Attempt at a Solution


I would like to do this problem in both inertial and non-inertial frames but first with the inertial frame.
The net force acting on the system is F. Hence,
[tex]F=(M_1+M_2+M_3)a \Rightarrow a=\frac{F}{M_1+M_2+M_3}[/tex]
Let a' be the acceleration ##M_2## and ##M_3## with respect to ##M_1##.
For ##M_1##, a normal force (at the point where it is in contact with ##M_3##) and a tension acts on it.
[tex]F-N-T=M_1a (*)[/tex]
For ##M_2##, ##T=M_2(a+a')##
For ##M_3##, ##M_3g-T=M_3a'## and ##N=M_3a##
Substituting T and N in the equation (*)
[tex]F=M_1a+M_3a+M_2a+M_2a'[/tex]
Solving, ##a'=0##. This is obviously wrong. :confused:

Any help is appreciated. Thanks!
 

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  • #2
Your problem stems from using ##F=(m_1+m_2+m_3)a##. This is only true if a'=0, so of course you derived a'=0 as a consequence.

Suppose the rope is cut. The mass m3 is going to fall to the bottom of the shaft, but while doing so the shaft keeps mass m3 from moving side-to-side with respect to mass m1. Thus from a horizontal perspective, if the rope has been cut then ##F=(m_1+m_3)\ddot x_1##, where ##\ddot x_1## is the horizontal component of the center of mass of mass m1. What about mass m2? It's not a part of the system anymore, at least with regard to horizontal motion. The surface is frictionless, and you don't care about the vertical component of the force on mass m1.

Now suppose the rope is intact. What horizontal forces, if any, does this add to the m1+m3 system? Hint: What does the pulley do?
 
  • #3
Why are you solving for a' ? a' is zero if F is such that M3 doesn't rise or fall.

Considering horizontal direction,for M2 →T=M2a

Considering vertical direction,for M3 →T=M3g

a=(M3g)/M2

For equal masses, a=g and F=3Mg
 
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  • #4
D H said:
What horizontal forces, if any, does this add to the m1+m3 system?

Tension. It would act in the direction to left. Correct?

Also, I would like to derive the acceleration of individual masses, not only for the case when M3 is not allowed to either rise or fall.
 
  • #5
Hi Pranav :)

What is bothering you in this problem ?

Tension will act leftwards on M1 and rightwards on M2.
 
  • #6
Tanya Sharma said:
Hi Pranav :)

What is bothering you in this problem ?

I don't know but I just can't reach the final answer.
I will write down the equations again.
For ##M_1##, ##F-N-T=M_1a##.
For ##M_2##, ##T=M_2(a+a')##
For ##M_3##, ##N=M_3a## and ##M_3g-T=M_3a'##

Oh well, looks like I get the answer by solving the above equations. I should have been a bit more careful. :redface:

Thanks for the help D H and Tanya! :smile:

Also, how do I solve this in non-inertial frame?
 
  • #7
Pranav-Arora said:
Also, how do I solve this in non-inertial frame?

First off, what frame do you think is suitable here?
 
  • #8
Pranav-Arora said:
Also, how do I solve this in non-inertial frame?

Apply a pseudo force in horizontal direction(opposite to M1) on M2 and M3 and solve just as you would do normally.
 
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  • #9
Consider the frame with acceleration ##a = \frac{F}{M_1 + M_2 + M_3}##. What is the condition that ##M_3## is neither falling nor rising in this frame? It's just that ##M_2## has to be at rest in this frame. Take it from there; it's thankfully very simple.
 
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  • #10
WannabeNewton said:
Consider the frame with acceleration ##a = \frac{F}{M_1 + M_2 + M_3}##. What is the condition that ##M_3## is neither falling nor rising in this frame? It's just that ##M_2## has to be at rest in this frame. Take it from there; it's thankfully very simple.

That worked. Thanks WBN! :smile:
 
  • #11
Does anyone know what's wrong with my answer for the 2nd part of this exercise?
It says F = 0 and wants the acceleration for M1 .

So here's my analysis, assuming the right be the positive direction for x'x axis and down be positive for y'y axis.

For M2 :

x'x axis : F2 = M2a2

F2 is the rope tension

y'y axis motion is irrelevant .

For M3 :

y'y axis : M3g-F2 = M3a2
x'x axis : -N = M3a1

For M1 :

x'x axis : N-F2 = M1a1
y'y is irrelevant .

N is the normal force acting on M3 by M1 "wall's".

By solving the system i end up with this :

a1 = -[ M1M3/(M1M2 + M1M3 + M2M3 + M3^2) ]*g

It's all good except for the term M2M3 . According to the solution it should be 2M2M3 .
I can't figure out where I'm wrong though . Do i miss something out or is it just the book ?
 
  • #12
Avaro667 said:
For M3 :

y'y axis : M3g-F2 = M3a2
Is it correct to take the y-component of acceleration of M3 to be equal to the acceleration of M2?

a1 = -[ M1M3/(M1M2 + M1M3 + M2M3 + M3^2) ]*g

It's all good except for the term M2M3 . According to the solution it should be 2M2M3 .
The numerator, M1M3, of your expression for a1 doesn't look right. As M1 goes to infinity, what should the value of a1 approach?
 
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  • #13
TSny said:
Is it correct to take the y-component of acceleration of M3 to be equal to the acceleration of M2?

The numerator, M1M3, of your expression for a1 doesn't look right. As M1 goes to infinity, what should the value of a1 approach?

Hey, thanks for checking this out .
M2 and M3 are connected via a rope so,supossing it doesn't expand, those 2 masses should have the same acceleration,right ?

About M1M3 you're right,i miscalculated this part actually,it turns out to be M2M3 .
I still can't get that 2 for that term in dominator though .
 
  • #14
Avaro667 said:
Hey, thanks for checking this out .
M2 and M3 are connected via a rope so,supossing it doesn't expand, those 2 masses should have the same acceleration,right ?
No, not with respect to the Earth reference frame. (They would have the same acceleration relative to M1.)

For example, what would happen to M3 if M2 were held fixed relative to the Earth while M1 moves to the left?
 
  • #15
TSny said:
No, not with respect to the Earth reference frame. (They would have the same acceleration relative to M1.)

For example, what would happen to M3 if M2 were held fixed relative to the Earth while M1 moves to the left?

So basically i subconsciously changed the reference frame during the analysis for these 2 bodies ?
I'm not sure how to proceed though,too many uknowns,but a few equations only .
 
  • #16
You have all the equations except for the correct relation between a1, a2, and a3y. See if you can discover that relation.
 
  • #17
TSny said:
You have all the equations except for the correct relation between a1, a2, and a3y. See if you can discover that relation.

I gave it a try and i think i worked this out .
The set of equations i managed to come up with are 5 of them :

F2=M2a2
M3g-F2=M3a3
N-F2=M1a1
-N = M3a1
a1-a2+a3 = 0

Solving gave the correct answer .
Sometimes I'm a bit confused whether 2 objects should have or not the same acceleration . I guess the safest way is to suppose they don't ,unless it's quite obvious,like in the case of M1's and M3's horizontal acceleration a1 ? Thank you so much for helping with this one,much appreciated :bow: .
 

1. What are Newton's three laws of motion?

Newton's three laws of motion are fundamental principles that describe the behavior of objects in motion. The first law, also known as the law of inertia, states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force. The second law, also known as the law of acceleration, states that the acceleration of an object is directly proportional to the force applied and inversely proportional to its mass. The third law, also known as the law of action and reaction, states that for every action, there is an equal and opposite reaction.

2. What is the Pedagogical machine in relation to Newton's laws?

The Pedagogical machine is a teaching tool that uses simple experiments and demonstrations to help students understand and visualize Newton's laws of motion. It allows students to see the principles of motion in action and provides a hands-on learning experience to deepen their understanding of these laws.

3. How does the Pedagogical machine demonstrate Newton's first law?

The Pedagogical machine can demonstrate Newton's first law by showing that an object at rest will remain at rest unless acted upon by an external force. For example, a ball placed on a flat surface will not move unless a force, such as pushing or pulling, is applied to it. This reinforces the concept of inertia and how objects tend to resist changes in their state of motion.

4. Can the Pedagogical machine be used to explain gravity and Newton's laws?

Yes, the Pedagogical machine can be used to demonstrate the effects of gravity and how it relates to Newton's laws. For example, by using different masses of objects and varying the angle of the ramp, students can observe how the force of gravity affects the acceleration of objects and how it follows Newton's second law.

5. How does the Pedagogical machine make learning about Newton's laws more engaging?

The Pedagogical machine makes learning about Newton's laws more engaging by providing a hands-on and interactive experience for students. It allows them to see the laws in action and encourages them to think critically and make connections between the experiments and the principles being taught. This active learning approach can help students better understand and remember the concepts of Newton's laws.

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