Two masses, two pulleys and an inclined plane

[ThEmptyTree]

I was comparing it to a plane that is moving on the runway. Passengers inside could be walking relative to the plane, but Newton's second law could be applied to the plane using the total mass of passengers and plane as the m term in F=ma equation.

I know perfectly what you're talking about, but here it simply doesn't work and I kinda told you why.

The problem wouldn't be called challenging if it were that easy.

 

[vcsharp2003]

I know perfectly what you're talking about, but here it simply doesn't work and I kinda told you why.

The problem wouldn't be called challenging if it were that easy.

I think if we take a system of multiple parts then the a in F = ma needs to be the acceleration of the center of mass. In the application I did, I took the acceleration of the pulley P when it should have been the acceleration of the center of mass of P and moving masses 2 and 3. That's why my application of Newton's second law to the system was incorrect. Right?

 

[ThEmptyTree]

I think if we take a system of multiple parts then the a in F = ma needs to be the acceleration of the center of mass. In the application I did, I took the acceleration of the pulley P when it should have been the acceleration of the center of mass of P and moving masses 2 and 3. That's why my application of Newton's second law to the system was incorrect. Right?

Yes I think you are right here, I didn't think of the center of mass thing. Personally I avoid grouping objects because I think it's safer to take care of each object in particular.

 

[vcsharp2003]

Yes I think you are right here, I didn't think of the center of mass thing. Personally I avoid grouping objects because I think it's safer to take care of each object in particular.

If the parts of a system are at rest relative to each other then the acceleration of any part is also the acceleration of the center of mass of this system and then it's valid to apply F= ma with "m" being total mass of system and "a" being acceleration of any part of the system.

 

[vcsharp2003]

I have attached a PDF with the full solution.

Another way of solving this challenging problem could be to convert the non-inertial frame of reference to an inertial frame of reference.

In this case the pulley P is a non-inertial frame of reference and we can apply a pseudo force on masses $m_2$ and $m_3$ in a downward direction if pulley P is assumed to accelerate upwards. By doing this the accelerations of masses $m_2$ and $m_3$ will be as if the pulley P was fixed (not moving up or down) i.e. the masses will have same accelerations in magnitude but in opposite directions like in a typical Atwood machine.

 

[AzimD]

In post #10, @haruspex confirms the answers to parts A and B and that is also what I got, however, when it comes to part C (the OP's original question), I get something completely different and I'm unsure why.

I used T=(g*m_2)/2 for parts A and B and got them right, I also used them on Part C but according to OP, I can't use that on part C and rather should use T = ((m_2*g- m_2*a_2)/2 )... Why?

Edit: Here's Part C for clarification:
Now assume that the block on the incline plane is sliding upward. The coefficient of kinetic friction between the block and the incline surface is µk. Find the magnitude of the acceleration of the block on the inclined plane, a1. Express your answers in terms of some or all of the variables m1, m2, θ, µk and the acceleration of gravity g.

 

[haruspex]

should use T = ((m_2*g- m_2*a_2)/2 )... Why?

Draw a FBD for m₂. What are the forces on it? What equation does that give for its acceleration?

 

[AzimD]

I get that if I were to go the F=ma route and set the Forces to find the T I'd get F=ma such that m_2*g-2T=ma_2 so that T=((ma_2-m_2*g)/2).

I guess better worded my question would be: why am I setting 2T equal to m_2*g in parts A and B whereas I'm tying in the component of acceleration to the equation in Part C?

 

[kuruman]

I get that if I were to go the F=ma route and set the Forces to find the T I'd get F=ma such that m_2*g-2T=ma_2 so that T=((ma_2-m_2*g)/2).

I guess better worded my question would be: why am I setting 2T equal to m_2*g in parts A and B whereas I'm tying in the component of acceleration to the equation in Part C?

Look at the FBDs below for part (c) where $m_1$ is moving up the incline and $m_2$ is descending.

1. What is Newton's second law in the direction along the incline for $m_1$?
2. What is Newton's second law in the vertical direction for $m_2$?

 

[haruspex]

I get that if I were to go the F=ma route and set the Forces to find the T I'd get F=ma such that m_2*g-2T=ma_2 so that T=((ma_2-m_2*g)/2).

I guess better worded my question would be: why am I setting 2T equal to m_2*g in parts A and B whereas I'm tying in the component of acceleration to the equation in Part C?

Because in parts A and B the acceleration was zero, maybe?

 

[AzimD]

Because in parts A and B the acceleration was zero, maybe?

I've attached the full question to this reply! There is no mention of acceleration being 0 from what I can tell!

 

[haruspex]

There is no mention of acceleration being 0

"barely slides", i.e. negligible acceleration.

 

[AzimD]

"barely slides", i.e. negligible acceleration.

Got it, so technically what we did with parts A and B is a F=ma analysis where 2T-m_2*g=m_2*a_2=0 and thus 2T=m_2*g?

 

[haruspex]

Got it, so technically what we did with parts A and B is a F=ma analysis where 2T-m_2*g=m_2*a_2=0 and thus 2T=m_2*g?

Yes.

 

[AzimD]

Yes.

So just to confirm: Conceptually, we ALWAYS have to do a F=ma analysis.
I've attached the new solution that I've come to. Does it look correct?

 

[kuruman]

So just to confirm: Conceptually, we ALWAYS have to do a F=ma analysis.
I've attached the new solution that I've come to. Does it look correct?

No, you do not ALWAYS have to do a F = ma analysis. You can get the answer by energy considerations.

Your solution for $a_1$ does not look correct because it has $a_2$ on the right-hand side. The problem asks you in part (c) to "express your answer in terms of some or all of the variables m₁, m₂, theta, µ_k and the acceleration of gravity g."

 

[AzimD]

I'm redoing this problem, and I have another question revolving around the Tensions... Would the tension provided by m2 on Pulley 2 be different than the two Tensions provided by the rope on Pulley 2? If so would that be 2T=T_2?
Is this even relevant?

 

[AzimD]

I'm redoing this problem, and I have another question revolving around the Tensions... Would the tension provided by m2 on Pulley 2 be different than the two Tensions provided by the rope on Pulley 2? If so would that be 2T=T_2?
Is this even relevant?

Never mind, I answered my own question. It does matter. I think I got it right this time, still not entirely sure though. Seems that the answer is a difficult answer to come across. I may have done something wrong.

 

[haruspex]

Never mind, I answered my own question. It does matter. I think I got it right this time, still not entirely sure though. Seems that the answer is a difficult answer to come across. I may have done something wrong.

Why the minus sign in $a_2=-a_1/2$?

 

[kuruman]

Never mind, I answered my own question. It does matter. I think I got it right this time, still not entirely sure though. Seems that the answer is a difficult answer to come across. I may have done something wrong.

You have a sign error in the equation for the hanging mass that is descending. The equation for $m_1$ is correctly written as
$$T-m_1g\sin\theta-\mu_kg\cos\theta=m_1a_1.$$ Vectors "up the incline" (tension and acceleration) are positive and down the incline (friction and gravity component) are negative. Specifically, the symbols $T$ and $a_1$ stand for positive numbers.

Now let's look at the hanging mass. Let's say that "down" is positive and "up" is negative. We write $$-2T+m_2g=m_2a_2.$$ Here the symbols $T$ and $a_2$ also stand for positive numbers.

So far so good. Now we know that the magnitude of $a_2$, a positive number, is half that of $a_1$ also a positive number. So we write $$a_2=\frac{1}{2}a_1.$$ There is no minus sign. It looks like you have to do the algebra one more time.

 

[AzimD]

So far so good. Now we know that the magnitude of $a_2$, a positive number, is half that of $a_1$ also a positive number. So we write $$a_2=\frac{1}{2}a_1.$$ There is no minus sign. It looks like you have to do the algebra one more time.

For the length of the rope, I get:
L=2y_2+y_1+constants so that d^2L = 0 = 2a_2+a_1 which means that a_2=(-a_1)/2

How is this wrong? The only way I can see there not to be a negative is if either both of the y_2 or the y_1 were either negative in length which doesn't make sense. Is it a directional thing?

@haruspex, this is why I have the negative sign in my accelerations! Not sure if it's a directional thing, but I can't see where that minus sign disappears.

 

[erobz]

For the length of the rope, I get:
L=2y_2+y_1+constants so that d^2L = 0 = 2a_2+a_1 which means that a_2=(-a_1)/2

The way you chose your coordinate for the bodies is not consistent with the equations you wrote.

For Block 1 ( on the incline ) you chose $\nearrow^+$, and for block 2(hanging mass) you chose $\downarrow^+$ ( presumably referenced at the pulley at the top of the incline), thus:

$$ L = - y_1 + 2y_2 - \rm{const.} $$

You can imagine if the pulley is accelerating up the incline (positive acceleration by the convention you chose), the hanging mass is acceleration downward ( also positive acceleration by the convention you chose), and visa-versa.

Your result is telling you that if block 2 (hanging mass) is accelerating up ( negative acceleration ), then block 1 is also acceleration up the incline. Your spider senses better be tingling at that notion.

 

[AzimD]

The way you chose your coordinate for the bodies is not consistent with the equations you wrote.

For Block 1 ( on the incline ) you chose $\nearrow^+$, and for block 2(hanging mass) you chose $\downarrow^+$ ( presumably referenced at the pulley at the top of the incline), thus:

$$ L = - y_1 + 2y_2 - \rm{const.} $$

You can imagine if the pulley is accelerating up the incline (positive acceleration by the convention you chose), the hanging mass is acceleration downward ( also positive acceleration by the convention you chose), and visa-versa.

Your result is telling you that if block 2 (hanging mass) is accelerating up ( negative acceleration ), then block 1 is also acceleration up the incline. Your spider senses better be tingling at that notion.

Oh okay, so basically the way I should conceptualize it is if Y_2 increases, then Y_1 has to decrease and thus in that regard, I must make it -Y_1 in my equation. Understood. This would also fix the acceleration issue!

 

[kuruman]

Oh okay, so basically the way I should conceptualize it is if Y_2 increases, then Y_1 has to decrease and thus in that regard, I must make it -Y_1 in my equation. Understood. This would also fix the acceleration issue!

I'm glad you sorted that out. If you post your solution with your answer, I will post mine that shows how you can do this using the work-energy theorem as an alternate method.

 

[erobz]

Oh okay, so basically the way I should conceptualize it is if Y_2 increases, then Y_1 has to decrease and thus in that regard, I must make it -Y_1 in my equation.

I don't know If there is a clear conceptualization to be had without some mathematical funny business. The idea is that you have to fix a coordinate system, and lengths are measured relative to it in the typical fashion ( i.e. positions on one side of it are positive and positions on the other side negative).

Not to pull you back into the weeds, but there is a question that I have. While what I did gets the proper result (and I know it to be commonly used to solve these inextensible rope pulley system problems), what is the idea that lets us say the constraint $L$ the length of the rope - something which is always positive as far as I can tell can be added in this way. I mean, if we added up the RHS we really aren't going to get $L$. So, I feel like it is still a bit hand wavy?

I feel like it should be:

$$ L = |-y_1| + 2|y_2| + \sum \rm{const}$$

How does the math work?

 

[AzimD]

I don't know If there is a clear conceptualization to be had without some mathematical funny business. The idea is that you have to fix a coordinate system, and lengths are measured relative to it in the typical fashion ( i.e. positions on one side of it are positive and positions on the other side negative).

Not to pull you back into the weeds, but there is a question that I have. While what I did gets the proper result (and I know it to be commonly used to solve these inextensible rope pulley system problems), what is the idea that lets us say the constraint $L$ the length of the rope - something which is always positive as far as I can tell can be added in this way. I mean, if we added up the RHS we really aren't going to get $L$. So, I feel like it is still a bit hand wavy?

I feel like it should be:

$$ L = |-y_1| + 2|y_2| + \sum \rm{const}$$

I'm not sure if I fully understand your question... was the question rhetorical lol

 

[erobz]

I'm not sure if I fully understand your question... was the question rhetorical lol

I have some "doubts" whether what I told you is mathematically consistent! I'm asking the pro's around here to clarify.

 

[AzimD]

I'm glad you sorted that out. If you post your solution with your answer, I will post mine that shows how you can do this using the work-energy theorem as an alternate method.

I believe I've gotten it right this time.

 

[AzimD]

I have some "doubts" whether what I told you is mathematically consistent! I'm asking the pro's around here to clarify.

Got it. This is where I learned the constraint condition method. Perhaps you can use it to answer your question! https://ocw.mit.edu/courses/8-01sc-...-pulley-problem-part-ii-constraint-condition/

 

[erobz]

Got it. This is where I learned the constraint condition method. Perhaps you can use it to answer your question! https://ocw.mit.edu/courses/8-01sc-...-pulley-problem-part-ii-constraint-condition/

Doesn't quite answer it, but thanks trying.

 

[erobz]

I think I worked out my issue. The inextensible rope constraint is actually this:

$$ L = |-y_1| + 2|y_2| + \sum \rm{const}$$

Then we can say that the following expression is equivalent:

$$ L = \sqrt{( -y_1)^2} + 2\sqrt{( y_2)^2}+ \sum \rm{const} $$

Which taking the derivative of both sides:

$$ 0 = \frac{1}{2}\frac{1}{\sqrt{(-y_1)^2}}( -2 y_1) \dot y_1 + \frac{1}{\sqrt{(y_2)^2}}( 2 y_2) \dot y_2 = - \dot y_1 + 2 \dot y_2 $$

and again taking the time derivative:

$$ 0 = -\ddot y_1 + 2\ddot y_2$$

And all seems to be right with the world again.

 

[haruspex]

I have some "doubts" whether what I told you is mathematically consistent! I'm asking the pro's around here to clarify.

If the lengths of the straight sections are initially $l_1, l_2=l_3$ then the total length is $L=l_1+2 l_2+c$.
If block 1 is displaced $y_1$ up the slope when block 2 descends $y_2$ then $L=l_1-y_1+2 l_2+2y_2+c$. Differentiate.
Is that you were looking for?

 

[erobz]

If the lengths of the straight sections are initially $l_1, l_2=l_3$ then the total length is $L=l_1+2 l_2+c$.
If block 1 is displaced $y_1$ up the slope when block 2 descends $y_2$ then $L=l_1-y_1+2 l_2+2y_2+c$. Differentiate.
Is that you were looking for?

I was thinking of coordinates. In #57 I suggested to apply the coordinates to the constraint, but that doesn't seem to be copasetic. Adding "lengths" and "coordinates" like that is crap that just happens to work because of what I found in post #66.

But yes, what you are suggesting appears to skirt that issue.

 

[kuruman]

I believe I've gotten it right this time.

Yes, you got it right. Here is a complete solution using the work-energy theorem.

Suppose mass $m_1$ slides up the incline a distance $s$. If the axle of pulley 2 were attached to the rope, mass $m_2$ would descend distance $s$. But that is not the case. The rope goes around pulley 2 in which case mass $m_2$ descends distance $s/2$. So in the same time that $m_1$ moves by some amount, mass $m_2$ moves by half that amount. This means that $v_2=\frac{1}{2}v_1$ and $a_2=\frac{1}{2}a_1.$

Now let's say that $m_1$ moves distance $s$ up the incline with initial speed $v_0$ and final speed $v$. The change in kinetic energy of the two-mass system is $$\Delta K=\frac{1}{2}m_1\left(v^2-v_0^2\right)+\frac{1}{2}m_2\left[\left(\frac{v}{2}\right)^2-\left(\frac{v_0}{2}\right)^2\right]=\frac{1}{2}\left(v^2-v_0^2\right)\left(m_1+\frac{m_2}{4}\right).$$The change in potential energy is $$\Delta U=m_1~g~s~\sin\!\theta-m_2~g~\frac{s}{2}.$$ The work done by friction is $$W_{\!f} = -\mu_k~N~s=-\mu_k ~m_1~g~s~\cos\!\theta.$$We apply the work-energy theorem to the two-mass system, $$\begin{align}
& \Delta K+\Delta U=W_{\!f} \nonumber \\
& \frac{1}{2}\left(v^2-v_0^2\right)\left(m_1+\frac{m_2}{4}\right)+m_1~g~s~\sin\!\theta-m_2~g~\frac{s}{2}=-\mu_k~m_1~g~\cos\!\theta \nonumber \\ & \frac{\left(v^2-v_0^2\right)}{2}=
\frac{m_2~g~s/2- m_1~g~s(\sin\!\theta+\mu_k~\cos\!\theta) }{\left(m_1+m_2/4\right)}.\nonumber \\ \end{align}$$ Since all forces are constant, all accelerations are constant. We can use the kinematic equation relating acceleration, displacement and speed squared to find $a_1.$ $$2a_1s=v^2-v_0^2\implies a_1=\frac{\left(v^2-v_0^2\right)}{2s}=\frac{m_2/2- m_1~(\sin\!\theta+\mu_k~\cos\!\theta) }{\left(m_1+m_2/4\right)}g.$$