(I.E. Irodov problem 1.72) Dynamics problem involving a pulley

[Adesh]

Homework Statement

Find the acceleration of the body 2 in the arrangement shown in the figure, if it mass is $n$ times as great as the mass of the body 2 and the angle of inclination plane forms with horizontal is equal to $\alpha$. The mass of the pulleys and the threads, as well as the friction are assumed to be negligible. Look into possible cases.

Relevant Equations

Newton's 2nd Law $\Sigma F = ma$

.

First thing that I notice is that the other part of string (to which body 2 is connected) is fixed. Therefore, I concluded that the body 2 is stationary with respect to the pulley 2 (let's agree that pulley to which body 1 is connected is pulley 1 and in same way pulley 2 is defined).

Forces on body 1:
$m_1 g \sin \alpha$
$T_1$
Let the acceleration of body 1 be $a$, then we have $T_1 - m_1 g \sin\alpha = m_1 a$.

The string which is going down from pulley 1 will have same acceleration as the body 1, so we can conclude that the acceleration of the body 2 will also be $a$.

But I'm unable to progress from here, please guide me.

Thanks.

 

[Lnewqban]

... body 2 is stationary with respect to the pulley 2...

That is not necessarily correct.
If mass of 1 is not too big, mass 2 can descend twice as fast as pulley 2 can.
If mass of 1 is huge, mass 2 can move up twice as fast as pulley 2 can.

Note the ratio between forces and the displacement of the thread is 2.
Please, look at this animation (upside-down view applies to your problem):

 

[Adesh]

@Lnewqban Your animation is extremely nice.

But to be honest, I need a little more explanation why body 2 will not be stationary to pulley 2.

 

[Lnewqban]

...But to be honest, I need a little more explanation why body 2 will not be stationary to pulley 2.

Imagine that the string which is going down from pulley 1 will have same acceleration in the up direction.
Based on that, consider an infinitesimal movement of the center of pulley 2, upwards and with same acceleration.

Please, describe to me how that displacement and movement is for left and right points that are tangent to string 2.
Could the left point of pulley 2 move up, if solidly tied by friction to the section of the string that is anchored to the wall?

Does not work that point, instantaneously, just like a fulcrum?
If it does, and the load is located by the axis of the pulley, could the tension of the string by the right point of the pulley be equal to that load?
What about its displacement, its velocity and its acceleration?

 

[haruspex]

more explanation why body 2 will not be stationary to pulley 2.

The length of string running over pulley 2 is constant. If pulley 2 moves vertically then the portion on the left of it changes, so the portion on the right changes. That changes the distance between it and mass 2.

 

[Adesh]

The length of string running over pulley 2 is constant. If pulley 2 moves vertically then the portion on the left of it changes, so the portion on the right changes. That changes the distance between it and mass 2.

Can't string become loose on the left side of pulley 2?

 

[Adesh]

Here is my Free Body Diagram for the problem:


For body 1 we have
$$
T_1- m_1 g \sin \alpha= m_1 a$$

For body 2 and with respect to pulley 2 we have :
$$
T_2 - m_2 g = m_2 a’$$
($a’$ is the acceleration of body 2 with respect to pulley 2).

What should I do next?

 

[haruspex]

Can't string become loose on the left side of pulley 2?

Only if mass 2 is in free fall, which is not going to happen here.

For body 2 and with respect to pulley 2 we have :
$$
T_2 - m_2 g = m_2 a’$$
($a’$ is the acceleration of body 2 with respect to pulley 2).

On the LHS you have all the forces acting on mass 2, no? Why does that not yield the acceleration in the lab frame?
You also need the kinematic relationship between the accelerations and the relationship between the two tensions.

 

[Adesh]

On the LHS you have all the forces acting on mass 2, no? Why does that not yield the acceleration in the lab frame?

With respect to lab frame, we have
$$
T_1 /2 +T_2 -m_2g = m_2 a’$$
(this time $a’$ is the acceleration of mass 2 with respect to lab frame).

You also need the kinematic relationship between the accelerations and the relationship between the two tensions.

Can you please give some hint for finding that relation?

 

[haruspex]

With respect to lab frame, we have
$$T_1 /2 +T_2 -m_2g = m_2 a’$$
(this time $a’$ is the acceleration of mass 2 with respect to lab frame).

This is a common error. Mass 2 is not attached to cable 1, so "knows" nothing about it. It is subject to two forces: the tension in cable 2 and gravity. This is one reason students are encouraged to draw free body diagrams for each separate rigid body.

For the relationship between the tensions, draw the free body diagram for pulley 2. Remember, it has no mass, so what does its ΣF=ma equation tell you?

For the kinematic relationship, haven't we been through this before?
Assign variables to the various straight lengths of cable. Write the equations representing how these relate to the total lengths of the cables of which they are part, and for how they relate to the vertical height of mass 2 and distance down the slope of mass 1.
You can then differentiate twice to obtain equations relating accelerations.

 

[Adesh]

This is a common error. Mass 2 is not attached to cable 1, so "knows" nothing about it. It is subject to two forces: the tension in cable 2 and gravity. This is one reason students are encouraged to draw free body diagrams for each separate rigid body.

For the relationship between the tensions, draw the free body diagram for pulley 2. Remember, it has no mass, so what does its ΣF=ma equation tell you?

For the kinematic relationship, haven't we been through this before?
Assign variables to the various straight lengths of cable. Write the equations representing how these relate to the total lengths of the cables of which they are part, and for how they relate to the vertical height of mass 2 and distance down the slope of mass 1.
You can then differentiate twice to obtain equations relating accelerations.

For pulley 2 we have
$$
T_1 - 2 T_2 = m a$$
$$
T_1 = 2T_2$$.
The acceleration of the cables connected to pulley will be equal (due to that 2nd derivative method). So, finally we have from lab’s reference frame
$$
T_1 - m_1g \sin \alpha= m_1 a $$
$$T_2 - m_2 g= m_2 a’$$
Using the relationship between the two tensions, we have
$$
2T_2 - m_1g \sin \alpha = m_1 a$$
$$T_2 -m_2 g = m_2 a’$$
Am I correct this far? What should I do next?

 

[haruspex]

What should I do next?

Obtain the kinematic relationship for a, a' as outlined in post #10.

 

[Adesh]

Obtain the kinematic relationship for a, a' as outlined in post #10.

Sir, I don't know but I'm unable to make equations of length of strings this time, nevertheless let me try

.

$l_1 + l_2 +l_3 +l_4 = c$. Differentiating this expression twice w.r.t time,
$$
\frac{d^2}{dt^2} l_1 + 0 + \frac{d^2}{dt^2} l_3 + 0 = 0$$
If I take pulley 1 to be origin then $\frac{d^2}{dt^2} l_3$ can't be called the acceleration of mass 2, because at any time the position of mass 2 is $l_2 (t) + l_3(t)$ and if we use this fact we would have
$$
\frac{d^2}{dt^2} l_1 + \frac{d^2}{dt^2} (l_2 +l_3) + 0 = 0$$
$$
a = -a'$$.

And now, using this equation we have
$$
T_2 = \frac{-ma' + m_1g \sin \alpha}{2}$$
$$
T_2 = m_2 a' + m_2 g$$
$$
-m_1 a' + m_1 g \sin \alpha = 2m_2 a' + 2m_2 g $$
$$
a' = \frac{(2n-\sin\alpha)g}{2n+1}$$
(In the last equation I used the given condition that $m_2= m_1 n$)

 

[haruspex]

$l_1 + l_2 +l_3 +l_4 = c$.

There are two threads. Each is of fixed length.

Differentiating this expression twice w.r.t time,
$$
\frac{d^2}{dt^2} l_1 + 0 + \frac{d^2}{dt^2} l_3 + 0 = 0$$

l₂ and l₄ vary in length.

 

[Adesh]

There are two threads. Each is of fixed length.

l₂ and l₄ vary in length.

If we consider origin as the pulley 1 and focus on first cable, we have
$$
l_1 + l_2 = c$$
$$\frac{d^2}{dt^2} l_1 + \frac{d^2}{dt^2} l_2 = 0$$
$$a_{mass~1} = -a_{pulley 1}$$

Now, coming to frame of reference of pulley 2, we have
$$
l_3 + l_4 = c$$
$$
a_{mass ~2}= -\frac{d^2}{dt^2} l_4 $$

I don’t know what to call $\frac{d^2}{dt^2} l_4$.

 

[haruspex]

If we consider origin as the pulley 1 and focus on first cable, we have
$$
l_1 + l_2 = c$$
$$\frac{d^2}{dt^2} l_1 + \frac{d^2}{dt^2} l_2 = 0$$
$$a_{mass~1} = -a_{pulley 1}$$

Now, coming to frame of reference of pulley 2, we have
$$
l_3 + l_4 = c$$
$$
a_{mass ~2}= -\frac{d^2}{dt^2} l_4 $$

I don’t know what to call $\frac{d^2}{dt^2} l_4$.

There is also a simple relationship between l₂ and l₄.

Careful with signs. Bear in mind that e.g. $\frac{d^2}{dt^2} l_1$ is the 'acceleration' of the length l₁. Whether that is the same as the acceleration of mass 1 depends on which way you are defining as positive for mass 1's acceleration. Increasing l₁ corresponds to mass 1 descending the slope.

 

[Adesh]

There is also a simple relationship between l2 and l4.

Sorry! But I cannot see it. Can you please tell me what’s the relation between them?

 

[Lnewqban]

Sorry! But I cannot see it. Can you please tell me what’s the relation between them?

Since the distance between the anchoring point of L4 to the wall and the center of pulley 1 is constant, so is the addition of L4 and L2.

It is a simple set up to reproduce, using a fine string and two paperclips, they work just like pulleys.
If you could experiment with it, you could easily see those relations.

 

[Adesh]

Since the distance between the anchoring point of L4 to the wall and the center of pulley 1 is constant, so is the addition of L4 and L2.

Oh okay! that means
$$\frac{d^2}{dt^2} l_4 = - \frac{d^2}{dt^2} l_2$$
Right?

 

[Lnewqban]

I am not sure (too much math for me).
As its end is anchored down, L4 is not moving or accelerating, it is just becoming shorter or longer.

 

[Adesh]

All right, for cable 1 we have
$$
\begin{aligned}
&\frac{d^2}{dt^2} l_1 (t) = - \frac{d^2}{dt^2} l_2 (t) \\
&a = \frac{d^2}{dt^2} l_2
\end{aligned}
$$
For cable 2 we have
$$
\begin{aligned}
&\frac{d^2}{dt^2} l_3 (t) = - \frac{d^2}{dt^2} l_4 (t) \\
& a' = - \frac{d^2}{dt^2} l_4 (t)
\end{aligned}
$$
We must remember that $a'$ here is the acceleration of mass 2 w.r.t pulley 2.

From post #19 we know that $\frac{d^2}{dt^2} l_4 = - \frac{d^2}{dt^2} l_2$, now using this equation to obtain $a =a'$

Dynamical equations are :
$$
\begin{aligned}
T_1 = 2T_2 \\
2T_2 - m_1 g \sin \alpha = m_1 a \\
T_2 - m_2g = m_2 (a+ a')
\end{aligned}
$$
Solving this system of equations by using $a=a'$ will give us
$$
a = \frac{(2n - \sin \alpha)}{4n+1} g$$
Therefore, the net acceleration of mass 2 is $a+a' = 2a= 2\frac{(2n - \sin \alpha)}{4n+1}$. Yes, we are done.

 

[Adesh]

I want to clear up few things:

1. I'm totally unaware of how to keep the track of signs. For pulley 1, I took the downslope as negative and upslope as positive, and similarly for pulley 2, going down is negative and up is positive. But if we adhere to this system of signs, why we have $\frac{d^2}{dt^2} l_1(t) =- a$ but $\frac{d^2}{dt^2} l_3(t) = a'$? In both cases the increase in length of string would amount to acceleration in negative direction.

2. When we did dynamics of pulley 2, we considered the forces acting on it, and they were $T_1$ and $-2T_2$ but it felt like a strange thing to equate the sum to zero just because the mass of pulley is zero. I know what we did is
$$ T_1 - 2T_2 = m a $$ as mass is zero we got
$$T_1 -2T_2 = 0$$
But this is strange, it is, basically, saying that a massless thing cannot ever move.

3. The last thing: @Lnewqban pointed out that $l_2 + l_4$ will be constant, how can I develop such a hawk eye? Is it just about experience or something more to it?

Thank you so much @haruspex and @Lnewqban sir.

 

[Lnewqban]

No special skills, Adesh
Once you have successfully completed some of these problems, it is not difficult anymore.
Just keep trying, my friend.
Follow the directions and advice of @haruspex, you are in good hands.

Pulleys are one of several types of simple machines.
If you ever must lift a heavy object using a block and tackle contraption, everything will become very clear very soon.

Please, see:
https://en.wikipedia.org/wiki/Simple_machine

https://en.wikipedia.org/wiki/Block_and_tackle

 

[haruspex]

In both cases the increase in length of string would amount to acceleration in negative direction.

My reading of your work is that you consistently took a' as positive down.

saying that a massless thing cannot ever move.

Quite the contrary. It says that the slightest net force would make it move with infinite acceleration, so there is no net force.

how can I develop such a hawk eye?

It helps if you can visualise how the parts of the mechanism will move in relation to each other.

Now, when I solve a problem like this I do not actually go through the steps of writing the sum-of-lengths=constant equations and differentiating them. I just consider how far one thing moves when something else does. In the present case, if the mass on the slope slides up one unit the hanging pulley drops one unit, so the hanging mass drops two units. But I took you through the differentiation path because it is rigorous and less prone to error.
Either way, it is always worth trying to visualise the movement to see whether the kinematic equation makes sense.

Oh, and I agree with your final answer, except that right at the end you dropped 'g'.
There's a useful sanity check you can do. Consider extreme values of n, tending to 0 and tending to infinity. Does you expression give the right answers for those cases?

 

[Adesh]

There's a useful sanity check you can do. Consider extreme values of n, tending to 0 and tending to infinity. Does you expression give the right answers for those cases?

Yes, when $n$ goes to zero, the acceleration of body 2 is double to that of body 1, and when $n$ is very large the acceleration of body 2 is (as expected) $g$.

 

[Lnewqban]

Yes, when $n$ goes to zero, the acceleration of body 2 is double to that of body 1, and when $n$ is very large the acceleration of body 2 is (as expected) $g$.

I am sorry, but I am confused about what mass is multiplied by n.

The OP shows this:
“Find the acceleration of the body 2 in the arrangement shown in the figure, if it mass is ntimes as great as the mass of the body 2 and the angle of inclination plane forms with horizontal is equal to α...”

 

[Adesh]

I am sorry, but I am confused about what mass is multiplied by n.

The OP shows this:
“Find the acceleration of the body 2 in the arrangement shown in the figure, if it mass is ntimes as great as the mass of the body 2 and the angle of inclination plane forms with horizontal is equal to α...”

I’m sorry, it is $m_2=n m_1$. I wrote body 2 instead of body 1 (second time).

 

[Lnewqban]

I’m sorry, it is $m_2=n m_1$. I wrote body 2 instead of body 1 (second time).

Thank you

Yes, when $n$ goes to zero, the acceleration of body 2 is double to that of body 1, ...

Up or down?

 

[Adesh]

Up or down?

Up.