# Forces question, a pedagogical machine

#### holezch

forces question, a "pedagogical machine"

1. Homework Statement

All surfaces are frictionless. How much force must be applied to M1 in order to keep M3 from rising or falling?

Picture:

http://www.slideshare.net/brigittperalta/sol-maquina-pedagogica-1546585

2. Homework Equations

F = Ma

3. The Attempt at a Solution

I think my understanding of these force questions is way off.. if the surfaces are frictionless, how will M1 affect the M2 on top of it? I think in order to keep M3 from falling, you need the tension to equal to the weight of M3..

thanks

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#### tiny-tim

Homework Helper
I think my understanding of these force questions is way off.. if the surfaces are frictionless, how will M1 affect the M2 on top of it? I think in order to keep M3 from falling, you need the tension to equal to the weight of M3.
Hi holezch!

Your diagram shows a block of mass m1 being pulled by a horizontal force F.

On the horizontal top of the block, a mass m2 is attached to a horizontal string which goes over a pulley to become vertical, and is attached to a mass m3 which is free to rise or fall in a vertical hole in the block.

ok … there is no vertical acceleration, so yes, T = m3g, and yes, the block (m1) cannot affect m2.

Hint: you need the acceleration of m2 to equal the acceleration of the block.

#### holezch

Re: forces question, a "pedagogical machine"

okay, would I have to pull M1 enough so that the whole vertical acceleration of the system is equal to the weight of M3? Since how M3 falls or risings is horizontal, vertical forces won't affect that

thanks a lot!!

#### tiny-tim

Homework Helper
okay, would I have to pull M1 enough so that the whole vertical acceleration of the system is equal to the weight of M3? Since how M3 falls or risings is horizontal, vertical forces won't affect that

thanks a lot!!
"the whole vertical acceleration of the system" doesn't make any sense …

the system doesn't move "as a whole".

and "since how M3 falls or rises is horizontal" doesn't make any sense either.

Just do Newton's second law three times, once for each of the three masses.​

#### holezch

Re: forces question, a "pedagogical machine"

"the whole vertical acceleration of the system" doesn't make any sense …

the system doesn't move "as a whole".

and "since how M3 falls or rises is horizontal" doesn't make any sense either.

Just do Newton's second law three times, once for each of the three masses.​
wouldn't it move together? if M2 sits on top of M1..

thanks

#### tiny-tim

Homework Helper
No, because all surfaces are frictionless …

if there's no friction, and if there was no tension from m3, there's no force, so (good ol' Newtons first law) m2 ain't going nowhere

#### holezch

Re: forces question, a "pedagogical machine"

No, because all surfaces are frictionless …

if there's no friction, and if there was no tension from m3, there's no force, so (good ol' Newtons first law) m2 ain't going nowhere
okay, so I guess the big m1 block has to push m3 which pulls the cord and pulls m2.. so then you have to push m3 until the acceleration of m2 is equal to m1? Is that so the tension of the string is balanced out?

thanks

#### tiny-tim

Homework Helper
okay, so I guess the big m1 block has to push m3 which pulls the cord and pulls m2.. so then you have to push m3 until the acceleration of m2 is equal to m1? Is that so the tension of the string is balanced out?

thanks
(just got up :zzz: …)

You need to think about each mass separately … what are the forces on it? … what is its acceleration?

Concentrate on m3 (the one in the vertical hole) …

what are the horizontal forces on m3? … what is its horizontal acceleration?

what are the vertical forces on m3? … what is its vertical acceleration?

#### holezch

Re: forces question, a "pedagogical machine"

(just got up :zzz: …)

You need to think about each mass separately … what are the forces on it? … what is its acceleration?

Concentrate on m3 (the one in the vertical hole) …

what are the horizontal forces on m3? … what is its horizontal acceleration?

what are the vertical forces on m3? … what is its vertical acceleration?
thanks tiny-tim, I drew my free body diagram focusing individually on masses.. for M3, I have that ma1 is pushing it sideways, mg is pulling it down, and so the total force is diagonal m(a1+g) and this is what is pulling the cord as well. Then the cord pulls on M2 by m(a1+g).. and then by newton's 3rd law, M2 should pull back by m(a1+g) as well.. but then that would cause an equilibrium in the pulley system for any force you apply to M1 ?? that's definitely not right...

#### holezch

Re: forces question, a "pedagogical machine"

my concept of tension is obviously wrong.. if I imagined a case when M2 is not moving and M3 is just pulling down, I'll get an equilibrium!

If M3 is pulling on the cord, wouldn't that become the tension?

#### tiny-tim

Homework Helper
for M3, I have that ma1 is pushing it sideways, mg is pulling it down, and so the total force is diagonal m(a1+g) and this is what is pulling the cord as well.
Sorry, but that's completely wrong.

A cord (string, rope, etc) only experiences a force (tension) along its length.

The length is vertical, so the force is vertical.

(Yes, I agree m3 is also getting a horizontal acceleration, but that's from being pushed horizontally by the wall of the hole … it has nothing to do with the cord.)

Start again, calculating the tension T only from the vertical part of F = ma for m3.

#### holezch

Re: forces question, a "pedagogical machine"

Sorry, but that's completely wrong.

A cord (string, rope, etc) only experiences a force (tension) along its length.

The length is vertical, so the force is vertical.

(Yes, I agree m3 is also getting a horizontal acceleration, but that's from being pushed horizontally by the wall of the hole … it has nothing to do with the cord.)

Start again, calculating the tension T only from the vertical part of F = ma for m3.
but if you pushed the block sideways so it starts moving horizontally, wouldn't it pull the cord sideways as well? :S

#### tiny-tim

Homework Helper
but if you pushed the block sideways so it starts moving horizontally, wouldn't it pull the cord sideways as well? :S
You need to consider one mass at a time …

For m3, the cord is vertical, so the tension T3 acting on m3 is vertical.

For m2, the cord is horizontal, so the tension T2 acting on m2 is horizontal.

(and obviously in this case T2 = T3)

#### holezch

Re: forces question, a "pedagogical machine"

You need to consider one mass at a time …

For m3, the cord is vertical, so the tension T3 acting on m3 is vertical.

For m2, the cord is horizontal, so the tension T2 acting on m2 is horizontal.

(and obviously in this case T2 = T3)
thanks, but I meant for M3, if M1 pushes M3 horizontally, the cord will start getting pulled horizontally? :S

#### tiny-tim

Homework Helper
thanks, but I meant for M3, if M1 pushes M3 horizontally, the cord will start getting pulled horizontally? :S
If you mean that the cord will start moving horizontally, the answer is no, not necessarily.

If you mean that the cord will start getting a horizontal tension, that has nothing do to with the force of m1 on m3.

Why did you ask this?

Which of the three masses were you considering the forces on, and in which direction?

(btw, i'm going to bed now :zzz: …)

#### holezch

Re: forces question, a "pedagogical machine"

I was trying to focus on m3 and then m2.. :( Could you please show me a worked out solution so I know what to do for next time? I'd really appreciate it.. thanks

#### tiny-tim

Homework Helper
I was trying to focus on m3 and then m2.. :( Could you please show me a worked out solution so I know what to do for next time? I'd really appreciate it.. thanks
erm noooo

You have three unknowns: a, the acceleration of m1, b, the acceleration of m2 and m3, and T, the tension.

Do three separate F = ma equations for forces and acceleration:

i] for m1 in the horizontal direction

ii] for m2 in the horizontal direction

i] for m3 in the vertical direction​

That will give you three equations for three unknowns, which you can then solve.

#### holezch

Re: forces question, a "pedagogical machine"

okay, if we want m2 to have the same acceleration as m1, don't we only need to push m1 so that the total force is equal to mg/2 (if they had equal masses) ? since that's how m2 will be moving.. thanks

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#### holezch

Re: forces question, a "pedagogical machine"

erm noooo

You have three unknowns: a, the acceleration of m1, b, the acceleration of m2 and m3, and T, the tension.

Do three separate F = ma equations for forces and acceleration:

i] for m1 in the horizontal direction

ii] for m2 in the horizontal direction

i] for m3 in the vertical direction​

That will give you three equations for three unknowns, which you can then solve.
acceleration of m3 = mg/2
T = mg/2
acceleration of m2 = mg/2
acceleration of m1 = m1a1 (we can choose this force)

#### holezch

Re: forces question, a "pedagogical machine"

I still need to produce a force to counter balance the mg from m3.. I don't know what kind of force would do that :(

#### tiny-tim

Homework Helper
(just got up :zzz: …)
acceleration of m3 = mg/2
(did you copy that from some other problem? )

You mean the vertical acceleration?

NO, the vertical acceleration of m3 is zero.

Try again.​

#### holezch

Re: forces question, a "pedagogical machine"

(just got up :zzz: …)

(did you copy that from some other problem? )

You mean the vertical acceleration?

NO, the vertical acceleration of m3 is zero.

Try again.​
I'm confused :S if we suppose that the vertical acceleration of m3 is zero, how do we explain it? (there are no counterbalancing forces :S)
thanks

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#### cepheid

Staff Emeritus
Gold Member
Re: forces question, a "pedagogical machine"

I'm confused :S if we suppose that the vertical acceleration of m3 is zero, how do we explain it? (there are no counterbalancing forces :S)
thanks
You realize that that is the objective of this problem, right?

All surfaces are frictionless. How much force must be applied to M1 in order to keep M3 from rising or falling?
TWO forces act on m3: 1. gravity and 2. the force due to the tension in the rope.

#### holezch

Re: forces question, a "pedagogical machine"

You realize that that is the objective of this problem, right?

TWO forces act on m3: 1. gravity and 2. the force due to the tension in the rope.
yea :P I understand.. I just don't understand how the tension could balance out the weight :S there are no other forces on the top block going the other direction :S So I guess the question is, how can you push m1 so that you create a force on m2 that opposes the pull by the cord? I wouldn't know the answer to that :S thanks

edit: okay , I have some idea.. I think, if m1 is my reference frame, then m2 will be going in the opposite direction as well.. is this on the right track? without pushing m1, m2 would be pulling the cord by mg/2.. so would you pull m1 forward by mg? (a total force of mg/2) Then the total force on the cord by m2 would be mg, and it would cancel out the weight of m3 thanks

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#### tiny-tim

Homework Helper
I just don't understand how the tension could balance out the weight …
That's because you're not writing the equations out!!

For m3: tension + weight = m3*acceleration = 0.

For m2: tension = m2* acceleration.

For m1: force = m1* acceleration.

Write them all out, and solve them!
edit: okay , I have some idea.. I think, if m1 is my reference frame, then m2 will be going in the opposite direction as well.. is this on the right track? without pushing m1, m2 would be pulling the cord by mg/2.. so would you pull m1 forward by mg? (a total force of mg/2) Then the total force on the cord by m2 would be mg, and it would cancel out the weight of m3 thanks
A reference frame has to be inertial, that is, stationary or with constant velocity.

You can't use m1 as a frame, since it's accelerating.

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