Irodov- 1.123- Simple conceptual question with two masses connected by a spring

[palaphys]

Homework Statement

Two bars of masses m1 and m2 connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to μ. What minimum constant force has to be applied in the horizontal direction to the bar of mass m1 in order to shift the other bar?

Relevant Equations

WET, F=kx

The diagram above describes the situation. Say the spring constant is k. The main goal will be to apply the condition that when the spring force exactly balances the limiting friction, m2 will begin to slide. i.e $kx=μm_2g$
I came up with two approaches for this one.
i) Assume that m1 is in equilibrium, so balance the spring force and frictional force with the external force.
ii)Apply WET between the two points, initial position and the final position, i.e where the velocity of m1 becomes zero.

method i) yields a solution for F which is less than ii), but the result from ii) is given as the right position. Not able to comprehend why.

Help is appreciated. Thanks.

 

[kuruman]

The problem does not specify which of the two masses $m_1$ or $m_2$ is larger. Let's label them as $m_{\text{smaller}}$ and $m_{\text{larger}}.$

Find two answers, one if you apply the force on the larger mass and one if you apply the force on the smaller mass. Compare your answers.

 

[PeroK]

Homework Statement: Two bars of masses m1 and m2 connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to μ. What minimum constant force has to be applied in the horizontal direction to the bar of mass m1 in order to shift the other bar?
Relevant Equations: WET, F=kx

View attachment 364380
The diagram above describes the situation. Say the spring constant is k. The main goal will be to apply the condition that when the spring force exactly balances the limiting friction, m2 will begin to slide. i.e $kx=μm_2g$
I came up with two approaches for this one.
i) Assume that m1 is in equilibrium, so balance the spring force and frictional force with the external force.
ii)Apply WET between the two points, initial position and the final position, i.e where the velocity of m1 becomes zero.

method i) yields a solution for F which is less than ii), but the result from ii) is given as the right position. Not able to comprehend why.

Help is appreciated. Thanks.

What I do with these problems is try to think them through first. My instinct is forget formulas for now and try to figure out what will physically happen. In this case, what happens when $F$ is large enough to move $m_1$?

Hint: acceleration and velocity are not the same thing!

 

[haruspex]

i) Assume that m1 is in equilibrium, so balance the spring force and frictional force with the external force.

You can certainly reason that it is not speeding up, but there's no obvious reason it is not slowing down.

method i) yields a solution for F which is less than ii),

Really? I would have guessed it was more.

 

[Gavran]

The total force on the bar m₁ is not zero, so the external force F is not in balance with the spring force and the frictional force.

 

[palaphys]

What I do with these problems is try to think them through first. My instinct is forget formulas for now and try to figure out what will physically happen. In this case, what happens when $F$ is large enough to move $m_1$?

Hint: acceleration and velocity are not the same thing!

When F is large enough to move m1, m1 will start with some acceleration, such that F>kx+friction where x is the elongation in the spring. At one instant, when the mass has maximum velocity, F=kx+ friction , after which the mass m1 will begin to decelerate, i.e its velocity will decrease.

This is what I think.. how will this help to solve the problem?

 

[palaphys]

Really? I would have guessed it was more.

Well, force balance leads to the answer: $F= μg(m_1 +m_2)$using work done (energy balance equation) leads to $F= μg(m_2 + 2m_1)/2$, which is given as the right answer.
So yes, you are right. Force balance is giving a solution which is more than the required force.

 

[Steve4Physics]

When F is large enough to move m1, m1 will start with some acceleration, such that F>kx+friction where x is the elongation in the spring. At one instant, when the mass has maximum velocity, F=kx+ friction , after which the mass m1 will begin to decelerate, i.e its velocity will decrease.

This is what I think.. how will this help to solve the problem?

When $m_1$ first comes to rest, its acceleration will be maximum (not zero). So the net force on it will be maximum.

Have you covered simple harmonic motion yet? Think of a comparable system - a mass hanging from a vertical spring and oscillating vertically. When is the spring tension maximum? What is the mass’s velocity at this moment? If you sketch the mass’s velocity time-graph (sinusoidal) you will see that the mass’s acceleration (i.e. $\frac {dv}{dt}$, the gradient) is greatest when the velocity is zero.

 

[palaphys]

When $m_1$ first comes to rest, its acceleration will be maximum (not zero). So the net force on it will be maximum.

Have you covered simple harmonic motion yet? Think of a comparable system - a mass hanging from a vertical spring and oscillating vertically. When is the spring tension maximum? What is the mass’s velocity at this moment? If you sketch the mass’s velocity time-graph (sinusoidal) you will see that the mass’s acceleration (i.e. $\frac {dv}{dt}$, the gradient) is greatest when the velocity is zero.

When the acceleration is maximum, the velocity will be zero, and vice verca. I'm aware of this. But how is it going to help in this case?

 

[haruspex]

When the acceleration is maximum, the velocity will be zero, and vice verca. I'm aware of this. But how is it going to help in this case?

To move the other block, just, you need the point where the spring is at max tension. Since the other forces on m1 are constant (as long as it is moving) that means max leftward acceleration of m1, hence zero velocity. So your energy method is correct whereas the assumption of equilibrium is wrong.

 

[Steve4Physics]

When the acceleration is maximum, the velocity will be zero, and vice verca. I'm aware of this. But how is it going to help in this case?

It explains why your approch in Post #7 is wrong. In Post #7 you wrote :

Well, force balance leads to the answer: $F= μg(m_1 +m_2)$

There is no force balance*. Forces aren't balanced because $m_1$ is accelerating.

*Edit. Let me qualify that - the forces on $m_2$ are balanced at the moment it is about to start moving.

 

[PeroK]

When F is large enough to move m1, m1 will start with some acceleration, such that F>kx+friction where x is the elongation in the spring. At one instant, when the mass has maximum velocity, F=kx+ friction , after which the mass m1 will begin to decelerate, i.e its velocity will decrease.

This is what I think.. how will this help to solve the problem?

If $m_2$ does not move, then that gives a maximum possible extension of the spring, based on Hooke's law. That gives a maximum value for $F$, based on moving $m_1$ this same maximum distance.

 

[palaphys]

To move the other block, just, you need the point where the spring is at max tension.

why is this? why not the eqb. position?

 

[palaphys]

There is no force balance*. Forces aren't balanced because $m_1$ is accelerating.

but when velocity is maximum, there is a momentary equilibrium, isn't it? this means forces at that point of time will be balanced. is it not?

 

[palaphys]

If $m_2$ does not move, then that gives a maximum possible extension of the spring, based on Hooke's law. That gives a maximum value for $F$, based on moving $m_1$ this same maximum distance.

so are you implying, that the value of F I got using force balance, IS a valid value for some a force to get the block moving? but it's not the maximum value. Did I get you right?

 

[haruspex]

why is this? why not the eqb. position?

Because to move the other block the tension has to exceed $\mu m_2g$. If the peak tension exceeds it the block will move and if it does not the block will not move. Therefore the peak tension is what matters.

but when velocity is maximum, there is a momentary equilibrium, isn't it? this means forces at that point of time will be balanced. is it not?

Yes, but what makes you think maximum velocity is relevant to the question?
Maximum force, maximum acceleration and zero velocity is the relevant combination; the zero net force, zero acceleration and maximum velocity combination is not.

 

[Steve4Physics]

but when velocity is maximum, there is a momentary equilibrium, isn't it? this means forces at that point of time will be balanced. is it not?

Yes, but the equilibrium position is not relevant.

It’s essential to realise that to find the minimum value of $F$, we must consider the case where $m_1$ gets as far right as possible (giving maximum spring extension). We need the spring tension to just reach the minimum value needed to make $m_2$ slip.

Or to say it slightly differently, $m_1$ will pass through its equilibrium position at max’ speed. It will then continue to move right but slowing down - stretching the spring more till it momentarily stops. We want the smallest value of $F$ which results in the maximum spring tension equalling the limiting frictional force on $m_2$.

 

[wrobel]

It is interesting to think about $k\to\infty$

 

[Sarinle]

A wonderful way to build intuition for this problem is to first recall the classic case of two masses connected by an ideal rope. An ideal rope is taut and stores no elastic potential energy; its sole role is to enforce a distance constraint between the two blocks:

$$\lvert x_2 - x_1 \rvert = L.$$

Differentiating gives

$$\dot{x}_2 - \dot{x}_1 = 0 ;;\Rightarrow;; v_1 = v_2 ;;\Rightarrow;; a_1 = a_2.$$

In that scenario, both blocks must share the same acceleration.

For the spring case, the situation is no longer as simple. The relative distance between the blocks is free to change, which means their accelerations can differ as the spring stretches or compresses.

The natural starting point is Newton’s second law. For this system, one should first write down the force equations for each block separately, i.e. the equations of motion for $m_1$ and $m_2$, which yield formulas for $\ddot{x}_1$ and $\ddot{x}_2$.

The difficult part about this question is remembering that initially the acceleration of object 2 is zero while object 1 has some acceleration. Equation 2 is for the condition; equation 1 is for linking to the applied force. Perhaps the cleanest way to begin is simply by writing down the two subsystem equations explicitly.

 

[TSny]

It is interesting to think about $k\to\infty$

Yes. For any value of $k$, you get the same expression for the minimum force necessary to cause movement of $m_2$. [Edit: This expression is independent of $k$.] Therefore, taking the limit as $k$ goes to infinity still gives the same expression.

Yet, if we replace the spring with an inextensible rope, we get a different expression for the minimum force to move $m_2$.

It's interesting to think about how to reconcile the difference.

 

[haruspex]

Yes. For any value of $k$, you get the same expression for the minimum force necessary to cause movement of $m_2$. [Edit: This expression is independent of $k$.] Therefore, taking the limit as $k$ goes to infinity still gives the same expression.

Yet, if we replace the spring with an inextensible rope, we get a different expression for the minimum force to move $m_2$.

It's interesting to think about how to reconcile the difference.

In reality, not only are all ropes somewhat elastic, so are all blocks and all ropes have mass. Starting with such a completely general model, the idealisations can be at different rates, and the limit result can depend on the relationship between the rates.