Two masses, two pulleys and an inclined plane

[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 $a_1.$ $$2a_1s=v^2-v_0^2\implies a_1=\frac{\left(v^2-v_0^2\right)}{2s}=\frac{m_2/2- m_1~(\sin\!\theta+\mu_k~\cos\!\theta) }{\left(m_1+m_2/4\right)}g.$$