Minimum mass of a block

[tellmesomething]

Homework Statement

Find out minimum mass of block A so that block B may slide up (img attached)

Relevant Equations

no

Here we know that if block B is going to move up or just be at the verge of moving up
$Mg \sin \theta$ will act downwards and maximum static friction will act downwards $\mu Mg \cos \theta$

Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach
$Mg \sin \theta + \mu Mg \cos \theta$ before when x is not equal to its maximum extension.

Like kx keeps on increasing and block B stays despite $Mg \sin \theta$ being greater than kx because static friction keeps on acting, but say at x= not maximum extension block A reaches a value bigger or just equal to $Mg \sin \theta + \mu Mg \cos \theta$ ?? Wouldn't that invalidate this seemingly generalised approach?

 

[haruspex]

Homework Statement: Find out minimum mass of block A so that block B may slide up (img attached)
Relevant Equations: no

View attachment 365550

What has the spring to do with it? When the system is released from rest, is the spring relaxed, at equilibrium with its suspended mass, or in some other state?

Here we know that if block B is going to move up or just be at the verge of moving up

$Mg \sin \theta$ will act downwards and maximum static friction will act downwards $\mu Mg \cos \theta$

Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers,

The homework statement you provided doesn't ask for the rate. Is this an extra question that occurred to you?

the suggested solution says that when block A is at its maximum extension, then block B will start to move up

I assume you mean when the spring is at maximum extension.

but with a certain set of values couldn't block A reach
$Mg \sin \theta + \mu Mg \cos \theta$ before when x is not equal to its maximum extension.

Yes, but would those values be with A at its minimum mass?

 

[PeroK]

Let's fix the mass of block B ($M$). Start with a small mass of block A $m$. This moves up and down. The maximum force in the sping being insufficient to move block B.

Let's increase the mass of A one small unit at a time. Gradually the extension of the sping increases each time. The maximum force is reached at its maximum extension. Gradually more force is applied to block B each time.

Eventually, we add a tiny amount of mass and the maximum extension increases by a tiny amount and the maximum force on B increases by a tiny amount and that is just enough to move block B.

It's that final mass we a looking for. No mass less than that could move block B. And, every mass greater than that will move block B before it reaches its maximum extension on the spring.

 

[PeroK]

PS if we are being totally precise, doing the mathematical calculation, we will find the maximum mass of A that will not move block B. Given that, when the forces on B are balanced, it will still not move. Technically, we will have found the threshold mass, above which the block definitely moves.

 

[Steve4Physics]

Like kx keeps on increasing and block B stays despite $Mg \sin \theta$ being greater than kx because static friction keeps on acting, but say at x= not maximum extension block A reaches a value bigger or just equal to $Mg \sin \theta + \mu Mg \cos \theta$ ?? Wouldn't that invalidate this seemingly generalised approach?

I’m not entirely sure what the difficulty is, but this might help...

Imagine gradually adding weights to to A, i.e. gradually increasing $m$. While there is no slipping, the spring’s extension, $x$, will gradually increase and the spring tension will equal $mg$.

Eventually. $m$ will have increased to the point when the spring tension equals $Mg \sin \theta + \mu Mg \cos \theta$.

At this point $mg = Mg \sin \theta + \mu Mg \cos \theta$. Then $m$ is the required minimum mass. (Adding more weights to A would cause slipping.)

Note that the values of of $k$ and $x$ are irrelevant. We could replace the spring by string - the minimum value of $m$ would be unaffected.

EDIT.Apologies. I have misinterpreted the question. I thought mass A was in equilibrium, but it is performing simple harmonic motion (SHM). So my post is inapplicable. See also Post #8.

 

[tellmesomething]

What has the spring to do with it? When the system is released from rest, is the spring relaxed, at equilibrium with its suspended mass, or in some other state?

The spring is relaxed when the system is released from rest.

The homework statement you provided doesn't ask for the rate. Is this an extra question that occurred to you?

Yes...

I assume you mean when the spring is at maximum extension.

Yes, but would those values be with A at its minimum mass?

Yes i do mean that, sorry for the wording.

Let's fix the mass of block B ($M$). Start with a small mass of block A $m$. This moves up and down. The maximum force in the sping being insufficient to move block B.

Let's increase the mass of A one small unit at a time. Gradually the extension of the sping increases each time. The maximum force is reached at its maximum extension. Gradually more force is applied to block B each time.

Eventually, we add a tiny amount of mass and the maximum extension increases by a tiny amount and the maximum force on B increases by a tiny amount and that is just enough to move block B.

It's that final mass we a looking for. No mass less than that could move block B. And, every mass greater than that will move block B before it reaches its maximum extension on the spring.

After reading post #3 i think i get it, the mass of block A increases gradually, so if we didn't get the amount of force required to move block B even after the spring goes till its maximum extension with this small mass of A then adding another small mass $\delta m$ to A, would mean the block B would move only (if the spring force is sufficient) at the maximum extension of spring attached to A again and all the previous values of spring force have already been checked when we did the experiment first time with initial mass so we can be sure it would move (if sufficient) only at the maximum extension.

This also clears up the fact that if the spring force reaches $Mg \sin \theta + \mu Mg \cos \theta$ at any extension x other than the maximum extension, this would mean that the mass is definitely more than the minimum mass.

 

[tellmesomething]

I’m not entirely sure what the difficulty is, but this might help...

Imagine gradually adding weights to to A, i.e. gradually increasing $m$. While there is no slipping, the spring’s extension, $x$, will gradually increase and the spring tension will equal $mg$.

Eventually. $m$ will have increased to the point when the spring tension equals $Mg \sin \theta + \mu Mg \cos \theta$.

At this point $mg = Mg \sin \theta + \mu Mg \cos \theta$. Then $m$ is the required minimum mass. (Adding more weights to A would cause slipping.)

Note that the values of of $k$ and $x$ are irrelevant. We could replace the spring by string - the minimum value of $m$ would be unaffected.

But since we are increasing the mass gradually, aren't we sure that only at the maximum extension of the spring we will get a new spring force to check if it is sufficient enough to just move block B? wouldn't using kx=mg here give us more than the minimum mass?...

 

[Steve4Physics]

But since we are increasing the mass gradually, aren't we sure that only at the maximum extension of the spring we will get a new spring force to check if it is sufficient enough to just move block B? wouldn't using kx=mg here give us more than the minimum mass?...

At the point when slipping starts, the tension ($kx$) must be equal to both $mg$ and $Mg \sin \theta + \mu Mg \cos \theta$.

EDIT. Apologies. I have misinterpreted the question. I thought mass A was in equilibrium, but it is performing simple harmonic motion (SHM). So my post is inapplicable. See also Post #5.

 

[jbriggs444]

The spring is relaxed when the system is released from rest.

Then should we worry at least briefly about the possibility that the block B begins to slip downward immediately and does not come to rest by the time that the spring reaches maximum extension?

 

[haruspex]

At the point when slipping starts, the tension ($kx$) must be equal to … $mg$

Really?

 

[kuruman]

This problem can be solved in four steps:

1. Find an equation for the threshold critical tension $T_{crit.}$ at which block B will start sliding.
2. Find an equation for the force that a spring-mass system, oscillating vertically with amplitude $A$, exerts at its point of support.
3. Find an equation relating the mass of the hanging block to the amplitude of oscillations $A$ when the block is released from rest with the spring relaxed.
4. Combine the above appropriately to get the answer. Assume that the string is inextensible.

On Edit:
For some reason, for this post only, the LaTeX looks OK in preview but not OK when posted, at least on my screen no matter how many times I refresh.

 

[Steve4Physics]

Really?

In Posts #5 and #8 I was referring only to the case of a non-oscillating mass(A) in equilibrium. I hope this was clear from what I wrote.

I had not realised that the OP was asking about a system with mass A oscillating – that isn't explcitly stated in the question, but I could/should have realised it.

I will edit the posts.