Condition to not let the block descend

[decentfellow]

Homework Statement

In the system shown in Fig. $2E.5 (a)$. block $m_2$ is being prevented from descending by pulling $m_1$ to the right with force $F$. Assuming all the surfaces to be frictionless, Find $F$

Homework Equations

F.B.D of $m_1$

From the F.B.D of $m_1$ we get the following equations:-
$$F=m_{1}a_{m_1}\tag{1}$$
$$N_1+m_1g-N_2=0 \implies N_1+m_1g=N_2\tag{2}$$

F.B.D of $m_2$

From the F.B.D of $m_2$, we get the following equation:-
$$m_2g-T=m_2a\tag{3}$$

F.B.D of $m_3$

From the F.B.D of $m_3$ , we get the following set of equations:-
$$m_3g-N_1=0 \implies m_3g=N_1\tag{4}$$
$$T=m_3a\tag{5}$$

The Attempt at a Solution

[/B]
From eq. $(3)$ and $(5)$, we get
$$a=\frac{m_2}{m_2+m_3}g$$

Now, for $m_2$ to not descend, the condition required is $$\vec{a}_{m_3/m_2}=0\implies \vec{a}_{m_3}-\vec{a}_{m_2}=0\implies \vec{a}_{m_3}=\vec{a}_{m_2}$$
From $(1)$, we have
$$\vec{a}_{m_1}=\frac{\vec{F}}{m_1}$$
$$\therefore F=\frac{m_1m_2}{m_2+m_3}g$$

The book gives the answer as $$F=(m_1+m_2+m_3)\frac{m_2}{m_3}g$$.

 

[gneill]

Hi decentfellow, Welcome to PF!

If all surfaces are frictionless as stated in the problem, how can $m_2$ pull $m_1$ in a horizontal direction? What supports the pulley?

By the way, so insert inline LaTeX, use '##' as a delimiter rather than '$'.

 

[decentfellow]

Hi decentfellow, Welcome to PF!

If all surfaces are frictionless as stated in the problem, how can $m_2$ pull $m_1$ in a horizontal direction? What supports the pulley?

By the way, so insert inline LaTeX, use '##' as a delimiter rather than '$'.

Okay, now that I see, the book had not attached a support for the pulley to the mass $m_1$, so do consider that. I think that attaching a support to the mass $m_1$ ,resolves the issue as to how does the $m_3$ gets pulled due to the weight of mass $m_2$. Sorry for poorly stating my question. Also in my solution consider that the $N_2$ in the F.B.D of $m_1$ at the top of the mass is $N_1$.

 

[gneill]

I think I see where some confusion might arise from the way the problem is stated. It's not m₂ that pulls on the block m₁. Consider that some external agency is providing a force F on block m₁ and is accelerating the system to the right:

What can you say about any forces that might exist between m₁ and m₂ if the system is being accelerated to the right?

 

[decentfellow]

I think I see where some confusion might arise from the way the problem is stated. It's not m₂ that pulls on the block m₁. Consider that some external agency is providing a force F on block m₁ and is accelerating the system to the right:

View attachment 105170

What can you say about any forces that might exist between m₁ and m₂ if the system is being accelerated to the right?

If what you say is that the mass $m_2$ does not pull $m_3$, then there is no need of any external force to be put on the block $m_1$, so as to not let $m_2$ slip. I think the weight of $m_2$ does produce some tension in the string, which connects it to $m_3$, hence producing an acceleration in $m_3$ along the $x-$ direction.

 

[gneill]

If what you say is that the mass $m_2$ does not pull $m_3$, then there is no need of any external force to be put on the block $m_1$, so as to not let $m_2$ slip. I think the weight of $m_2$ does produce some tension in the string, which connects it to $m_3$, hence producing an acceleration in $m_3$ along the $x-$ direction.

No, $m_2$ definitely exerts a force on $m_3$ via the tension in the string, but the reason that this does not cause block $m_2$ to descend is that something else counters that tension force.

Suppose for a moment that block $m_1$ was fixed in place. Then I think you'd agree that $m_2$ would descend and pull $m_3$ to the right. What the question proposes is that block $m_1$ is free to move and that some external force F is applied to it in such a way that the result is block $m_2$ does not descend. The idea is to find that force F which accomplishes this "static" situation for $m_2$ and $m_3$, and the whole system is accelerated by F without $m_2$ descending.

 

[decentfellow]

No, $m_2$ definitely exerts a force on $m_3$ via the tension in the string, but the reason that this does not cause block $m_2$ to descend is that something else counters that tension force.

Suppose for a moment that block $m_1$ was fixed in place. Then I think you'd agree that $m_2$ would descend and pull $m_3$ to the right. What the question proposes is that block $m_1$ is free to move and that some external force F is applied to it in such a way that the result is block $m_2$ does not descend. The idea is to find that force F which accomplishes this "static" situation for $m_2$ and $m_3$, and the whole system is accelerated by F without $m_2$ descending.

From the figure that you provided, I get that the masses $m_1$ and $m_2$ are in contact hence due to a force acting on $m_1$, there will be a normal reaction acting between them, so the F.B.D of $m_1$ and $m_2$ should be changed as follows:-
1)F.B.D of $m_1$

2)F.B.D of $m_2$

From the above FBDs, we get the following set of equations:-
$$N_1+N_2=m_1g\tag{1}$$
$$F-N_3=m_1a_{m_1}\tag{2}$$
$$m_2g-T=m_2a\tag{3}$$
$$N_3=m_2a_{m_1}\tag{4}$$

From the above set of equations, we get
$$a_{m_1}=\dfrac{F}{m_1+m_2}$$

Also, as we have already found $a_{m_3}=\dfrac{m_2}{m_3}g$.

Now for no relative motion between $m_3$ and $m_1$, $a_{m_1m_2}=0$

So, $$a_{m_1}=a_{m_3}\implies \dfrac{F}{m_1+m_2}=\dfrac{m_2}{m_3}g\implies F=(m_1+m_2)\dfrac{m_2}{m_3}g$$

Still, the book's answer is different from mine so I think I am still doing something wrong, if that is so please point it out.

 

[gneill]

From the above set of equations, we get
$$a_{m_1}=\dfrac{F}{m_1+m_2}$$

If the pulley is attached to $m_1$ then the horizontal and vertical forces applied to it via the tension in the ropes will be transferred to $m_1$ (we assume a massless pulley here). So $m_3$ should have a contribution to the overall acceleration of the system via the tension in the string.

Looking at it another way (perhaps a bit more intuitively), as none of the masses are allowed to move with respect to each other under the specified conditions, the entire system of blocks moves as a whole and so all of them are being accelerated by force F. So what's the total mass being accelerated by F?

 

[decentfellow]

If the pulley is attached to $m_1$ then the horizontal and vertical forces applied to it via the tension in the ropes will be transferred to $m_1$ (we assume a massless pulley here). So $m_3$ should have a contribution to the overall acceleration of the system via the tension in the string.

Looking at it another way (perhaps a bit more intuitively), as none of the masses are allowed to move with respect to each other under the specified conditions, the entire system of blocks moves as a whole and so all of them are being accelerated by force F. So what's the total mass being accelerated by F?

Finally understood it, thanks for all the time, I will write the final solution below.

So from the first part of your suggestion the modified F.B.D of $m_1$ will be:-

So, we get $$F-T-N_3=m_1a_{m_1}$$
Now, as $$N_3=m_2a_{m_1}$$
$$\therefore F=T+(m_1+m_2)a_{m_1}$$
As, on applying the force all the masses are going to be accelerated with te same acceleration and $m_2$ would not be descending so we have $$T=m_3a_{m_1}$$
And,$$T=m_2g$$
$$\therefore a_{m_1}=\frac{m_2}{m_3}g$$
So, $$F=T+(m_1+m_2)a_{m_1} \implies F=(m_1+m_2+m_3)\frac{m_2}{m_3}g$$

 

[gneill]

Bravo! Looks good.

 

[decentfellow]

Bravo! Looks good.

One, thing I noticed was that when the mass $m_1$ doesn't move, i.e. there is no force acting on it, then the tension in the string is the weight of the reduced mass of the system which includes $m_3g, m_2g$ and $T$. I will be showing my work below.

From equations $(3)$ and $(5)$ in the first post, we get
$$T=m_3a$$ And, $$m_2g-T=m_2a$$
So, we get $$T=\dfrac{m_2m_3}{m_2+m_3}g$$
As we know, the reduced mass of $m_2$ and $m_3$ is given by $\mu=\dfrac{m_2m_3}{m_2+m_3}$
So, $T=\mu g$

Why is it that the tension comes out to be the reduced mass of $m_2$ and $m_3$.

 

[SammyS]

One, thing I noticed was that when the mass $m_1$ doesn't move, i.e. there is no force acting on it, then the tension in the string is the weight of the reduced mass of the system which includes $m_3g, m_2g$ and $T$. I will be showing my work below.

From equations $(3)$ and $(5)$ in the first post, we get
$$T=m_3a$$ And, $$m_2g-T=m_2a$$

No. T only gives the vertical component of force on m₂.

Therefore, $\ m_2g-T=0\ .$

The vertical component of the acceleration of m₂ is zero.

 

[decentfellow]

No. T only gives the vertical component of force on m₂.

Therefore, $\ m_2g-T=0\ .$

The vertical component of the acceleration of m₂ is zero.

Yes, that would have been the case if i had been looking for the solution and would have considered that $m_2$ should not descend, but as i stated in my last post "One, thing I noticed was that when the mass $m_1$ doesn't move, i.e. there is no force acting on it", which clearly says that I am not referring to the condition that the question wants, but another case where the block $m_1$ is stationary.

 

[SammyS]

Yes, that would have been the case if i had been looking for the solution and would have considered that $m_2$ should not descend, but as i stated in my last post "One, thing I noticed was that when the mass $m_1$ doesn't move, i.e. there is no force acting on it", which clearly says that I am not referring to the condition that the question wants, but another case where the block $m_1$ is stationary.

Yes. Of course that's right.

I missed the change in conditions.