Problem with two pulleys and three masses

[erobz]

So the mass M is the same as m2 then?? M = 2kg?

Slow down. You need some more things here. Do a FBD of the lower pulley. What are the forces acting on it? Remember you are isolating the pulley when you do this.

 

[Erdi]

The force acting on it will be T2 upward and m2 and M downwards

 

[erobz]

The force acting on it will be T2 upward and m2 and M downwards

$m_2$ or $M$ aren't forces. You are to be "freeing" the pulley from all other bodies, severing the ropes. What are the forces acting to pull the pulley down. What is the force acting to pull the pulley upward? Are these forces balanced on the pulley?

 

[Erdi]

$m_2$ or $M$ aren't forces. You are to be "freeing" the pulley from all other bodies, severing the ropes. What are the forces acting to pull the pulley down. What is the force acting to pull the pulley upward? Are these forces balanced on the pulley?

If that's the case, forces acting on it downwards is the gravitational force. BUT i don't understand what you mean with the upward force. Because you said to isolate the pulley, but you agreed that it wasnt accelerating (vertically at least) so the upward force is the same as downward force then? which is the gravitatioal force ??

 

[erobz]

If that's the case, forces acting on it downwards is the gravitational force. BUT i don't understand what you mean with the upward force. Because you said to isolate the pulley, but you agreed that it wasnt accelerating (vertically at least) so the upward force is the same as downward force then? which is the gravitatioal force ??

We are freeing the pulley from the other bodies. Not from the forces. What are the internal forces inside the rope that we expose when we "free" the pulley.

 

[Erdi]

We are freeing the pulley from the other bodies. Not from the forces. What are the internal forces inside the rope that we expose when we "free" the pulley.

Im sorry, I am lost. I really don't have a clue.

 

[erobz]

Im sorry, I am lost. I really don't have a clue.

Which forces are associated with the ropes?

 

[Erdi]

Originally, the pulley B and m1

 

[erobz]

Originally, the pulley B and m1

Lets back up. What are the forces acting on $m_2$?

 

[Erdi]

That should be just the gravity and tension 1, tension 1= M*g

 

[erobz]

That should be just the gravity and tension 1, tension 1= M*g

$T_1 \neq Mg$ That would be the case if $M$ or $m_2$ were NOT accelerating, but they are.

What is the force balance on $m_2$

 

[Erdi]

its m2g

 

[erobz]

its m2g

Maybe I'm using confusing wording. That is just the force of weight. What is the sum of the forces acting on $m_2$. I think you have already written this in an earlier post.

 

[Erdi]

If that's not T - mg = ma or just T - mg , then i really don't understand at all

 

[erobz]

If that's not T - mg = ma or just T - mg , then i really don't understand at all

Ok good:

$$T_1 - mg = ma$$

Now for the other mass $M$?

 

[Erdi]

Thats also got to T1 - Mg = ma

 

[erobz]

Thats also got to T1 - Mg = ma

close. be careful about the coordinate system you chose. If $m2$ goes up is positive $a$ then what does that mean for the acceleration of $M$ here. Also, you used two different symbols for $M$

 

[Erdi]

T1 + mg = ma?

 

[erobz]

T1 + mg = ma?

You chose a coordinate system where up was positive in #46 whether you did so intentionally or not. That is upward accelerations are positive. Upward forces are positive, downward forces are negative, downward accelerations are negative. Try again. And use a capital $M$ leave the little $m$

 

[Erdi]

Well i think the formula is correct, so what you mean is:
T - Mg = M*(-a)

 

[erobz]

Well i think the formula is correct, so what you mean is:
T - mg = m*(-a)

Good. Please use capital $M$ though so you don't get confused about which hanging mass.

 

[Erdi]

Good. Please use capital $M$ though so you don't get confused about which hanging mass.

I know i edited it, guess it didnt go throug..

 

[erobz]

I know i edited it, guess it didnt go throug..

Also its $T_1$ you missed the subscript.

 

[erobz]

Now, of the forces on the LHS. Which one is supplied by the rope?

 

[Erdi]

Back to stage 1 again. That is m2g isn't it??

 

[erobz]

Back to stage 1 again. That is m2g isn't it??

The $m_2$ 's weight does not make $m_2$ go up. The other force i.e. $T_1$ (the tension) does. The tension is supplied by the rope.

I'm going to be honest here. It seems like you have some road ahead of you before you solve this problem. I got to get to bed, its midnight here. Don't lose hope, you'll figure it out. Someone will pick back up with you at some point. Goodnight.

In the meantime, it would be helpful if you learned to use Latex to format your equations. There is a guide on how to do this in the lower left corner of the new relpy box.

 

[Erdi]

Okay thanks for your help tho!

 

[Lnewqban]

I know that the tension from pulley B (T1) has to be equal the m*g of m1 for m1 to have acceleration = 0. But i can't figure how this works because the m2 is already heavier. And so the block(M) has to be negative weight?

Welcome, @Erdi !

I believe that that statement shows the misconception that is the root of your confusion.
Any mass is weightless in free fall (reason for which your guts feel funny during a roller coaster ride).

Please, see:
https://en.m.wikipedia.org/wiki/Weightlessness

The more you slowdown that free fall, the heavier the mass becomes.
If you are able to stop that mass completely (like Earth's surface and a bathroom scale do to your body), that mass "adquires" its natural or normally measurable weight when in repose.

If the same mass is suddely moved upwards (like your body in an elevator), its measurable weight increases, as weight is a force and an accelerated mass resists that acceleration showing a reaction force.

 

[Lnewqban]

Im sorry, I am lost. I really don't have a clue.

Just imagine that you are removing M and that you are grabing the rope while standing on the ground.
You keep a solid grip on that rope and m2, which is four times heavier that m1, starts moving down until m1 hits the top fixed pulley.
After that moment, your hand feels a force of m2g value pulling up.

If you slowly release your grip, the rope starts sliding up and m2 falls even lower.
Your hand starts feeling less force simultaneously as m2 is loosing weight.

If you fully release the rope, m2 becomes completely weightless and unable to exert any force on the movable pulley.
As a result, m1 is free to fall down, loosing weight at the same time.

There must be a balance point for which you keep some gripping force on the sliding rope in such a way that m2 continues falling, but m1 remains static.
At that point, your hand should feel a pulling upwards force, which value is the key to the solution of this problem.

 

[kuruman]

You might profit if you research "Atwood machine" on the web. Here is a compact treatment but there is plenty more including videos. Please read carefully and try to understand how one proceeds to solve such problems. The method is straightforward:

1. Choose a system, in this case one of the masses, say $M$.
2. Find the net force (sum of all the forces acting on it) $F_{\text{net,M}}.$
3. Set this sum equal to the mass of the system times its acceleration. This gives you one equation, $F_{\text{net,M}}=Ma$
4. Repeat for the other mass, $m_2$. Note that if $M$ accelerates up, $m_2$ must accelerate down. You should get a second equation $F_{\text{net},\text{m}_{2}}=-m_2a.$
5. Combine the two equations to find the acceleration.

See how this plan is executed in the link I provided. The link does not provide the tension, but you can find it easily with a little algebra from either $F_{\text{net,M}}=Ma$ or $F_{\text{net},\text{m}_{2}}=-m_2a$ if you have an expression for the acceleration $a$.