# Masses on a frictionless table with the bottom mass connected to a wall by a spring

Vigorous
Homework Statement:
A block of mass m sits atop a mass M which rests on a frictionless table. The mass
M is connected to a spring of force constant k attached to the wall. (i) How far can
mass M be pulled so that upon release, the upper mass m does not slip off? The
coefficient of friction between the two masses is µ. (ii) Repeat if µ' is the coefficient of friction between M and the table.
Relevant Equations:
F=ma F=-kx
If M is displaced by an amount + x from equilibrium.What happens to the two masses at the point of release for displacements of x and less?
Will they remain static because mass m provides whatever it takes to stop mass M from moving
till some x where m slips and M oscillates
or
Will they decelerate (mass m stays on top of mass M) as the force of static friction decreases with the spring force until equilibrium point is reached but what would be the source of force providing the push to mass m towards the equilibrium point. Moreover, at the point of release the velocity is 0 and the force of static friction ranges from 0 till some maximum to retain the initial velocity so I don't think the second scenario takes place

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Homework Helper
Hello @Vigorous, !

You will need more equations, since not all quantities in the problem appear in one of the equations. In particular: how does ##\mu## (possibly indirectly) influence ##x_{max}## ?

Homework Helper
If my question does not help, make free body diagrams for ##M## and ##m## and post

Vigorous
Hello @BvU , thank you!. That is what I am trying to understand, In my attempt I added two scenarios of what could happen to the two masses at the point of release but I am not sure if my reason is correct?
For mass M, there is a force of gravity Mg acting down the y-axis. The frictionless table is exerting an opposite and equal force N. The block is neither flying off the table nor table is exerting an opposite and equal force N. The spring force is acting on mass M to the left and the force of static friction opposing motion to the right. For mass m ,there is the force of gravity mg acting down, mass M is exerting an opposite and equal force. Mass m is neither ploughed into mass M neither flying off mass M. I think there are no known forces acting in the horizontal direction

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For mass M, there is a force of gravity Mg acting down the y-axis. The frictionless table is exerting an opposite and equal force N.
No and no. Mass M is also acted upon by mass m that sits on top of it. Furthermore the table has to support both masses M and m. Therefore the force that it exerts on M can't be just Mg.

As @BvU already suggested, please draw separate free body diagrams for the two masses and post them. We can only imagine what they might look like if you describe them with words in which case if you are wrong, we will not necessarily be able to point out your mistake(s).

Vigorous
No and no. Mass M is also acted upon by mass m that sits on top of it. Furthermore the table has to support both masses M and m. Therefore the force that it exerts on M can't be just Mg.

As @BvU already suggested, please draw separate free body diagrams for the two masses and post them. We can only imagine what they might look like if you describe them with words in which case if you are wrong, we will not necessarily be able to point out your mistake(s).

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Homework Helper
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For the time being let's consider part (a) where there is no friction between block M and the table but there is friction between the two blocks. The problem is asking you to consider what happens immediately after releasing the blocks. Draw the FBDs at that moment. Be sure to show and label all vertical and horizontal forces. Also remember Newton's third law. Action-reaction counterparts must point in opposite direction and have similar labels. Example : Fby B on A is the reaction counterpart of Fby A on B. It really pays to be systematic.

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Will they remain static because mass m provides whatever it takes to stop mass M from moving
No. Remember that the static frictional force is not ##N\mu_s##. That is the maximum value only; in general it will be less.
If the blocks are not accelerating then there is no horizontal force on m, so m exerts no horizontal force on M.

Vigorous
For the time being let's consider part (a) where there is no friction between block M and the table but there is friction between the two blocks. The problem is asking you to consider what happens immediately after releasing the blocks. Draw the FBDs at that moment. Be sure to show and label all vertical and horizontal forces. Also remember Newton's third law. Action-reaction counterparts must point in opposite direction and have similar labels. Example : Fby B on A is the reaction counterpart of Fby A on B. It really pays to be systematic.

I can't identify the horizontal forces acting on each of the blocks because I am not sure of which of the two scenarios happens at the point of release.

Vigorous
No. Remember that the static frictional force is not ##N\mu_s##. That is the maximum value only; in general it will be less.
If the blocks are not accelerating then there is no horizontal force on m, so m exerts no horizontal force on M.

Yes I know that the force of static friction ranges from zero to some maximum equal to ##N\mu_s##. But say I displaced M x meters from equilibrium and I released. At the point of release, M and m have an initial velocity of zero, If the spring force is greater than the maximum of the varying force of static friction will m slip off M and fall or accelerate with M and the force of static friction changes to the force of kinetic friction but what would of the source of the force acting on m to make it move. If it is less will the two blocks remain static?

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But say I displaced M x meters from equilibrium and I released. At the point of release, M and m have an initial velocity of zero, If the spring force is greater than the maximum of the varying force of static friction will m slip off M and fall or accelerate with M and the force of static friction changes to the force of kinetic friction but what would of the source of the force acting on m to make it move. If it is less will the two blocks remain static?
This is still part (i), with the frictionless table, right? Even if m were glued to M, the two would move.
To figure out whether m will slide on M you need to analyse the system. Draw a FBD for each block, showing all the forces. You can write all the possible forces, as unknowns, without having to predict what will happen.
Assume, at first, that m does not slide, so there is only one acceleration to consider. Find an expression for that acceleration and check whether the forces on m can produce it.

Vigorous
This is still part (i), with the frictionless table, right? Even if m were glued to M, the two would move.
To figure out whether m will slide on M you need to analyse the system. Draw a FBD for each block, showing all the forces. You can write all the possible forces, as unknowns, without having to predict what will happen.
Assume, at first, that m does not slide, so there is only one acceleration to consider. Find an expression for that acceleration and check whether the forces on m can produce it.

The spring force can cause mass M to accelerate but mass m does not feel any horizontal force for it to undergo the same fate as mass M

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I assume that the diagram you appended to post #12 is pursuant to the suggestions by @haruspex in post #11 and that the masses accelerate as one. In what direction is the acceleration of the top mass? Does your free body diagram for the top mass show forces that can conceivably be added together to give a net force in that direction?

Vigorous
I assume that the diagram you appended to post #12 is pursuant to the suggestions by @haruspex in post #11 and that the masses accelerate as one. In what direction is the acceleration of the top mass? Does your free body diagram for the top mass show forces that can conceivably be added together to give a net force in that direction?

I showed no forces acting in the horizontal direction on mass m because I don't know the source of the force acting to cause it to accelerate in the same direction as mass M.

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Not knowing the source is insufficient reason for not drawing the force because you know that there is a horizontal force. Can you make an educated guess about its origin?

Hint: There are only two items in the universe that interact, i.e. exchange forces, with the top block: the Earth and the bottom block. You have correctly drawn the force exerted by the Earth as vertical. So that leaves ...

Vigorous
I am going to guess with reason, the block of mass M which is connected to the spring has a natural tendency to move towards the equilibrium point so the force of static friction will act accordingly to the right . Whereas for mass m natural state of motion is to remain at a displacement x from equilibrium of the spring hence a force of friction will act to the left.

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Yes. The bottom block exerts a force on the top block and the top block exerts a force on the bottom block. These forces have equal magnitudes and opposite directions. The line along which they line is at an angle with respect to the vertical. Each of these forces has a component perpendicular to the interface between the blocks which we call the "normal" component and a component parallel to the interface which we call "friction". Obviously, these components are in opposite directions, a pair along the verticl and a pair along the horizontal.

In view of all this, please redraw your FBDs and post them. If you can write equations expressing Newton's second law based on each FBD. You should have 4 equations, because there are two directions (horizontal and vertical) and two FBDs.

Vigorous
Yes. The bottom block exerts a force on the top block and the top block exerts a force on the bottom block. These forces have equal magnitudes and opposite directions. The line along which they line is at an angle with respect to the vertical. Each of these forces has a component perpendicular to the interface between the blocks which we call the "normal" component and a component parallel to the interface which we call "friction". Obviously, these components are in opposite directions, a pair along the verticl and a pair along the horizontal.

In view of all this, please redraw your FBDs and post them. If you can write equations expressing Newton's second law based on each FBD. You should have 4 equations, because there are two directions (horizontal and vertical) and two FBDs.

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Much better but the horizontal equation for the bottom mass is inconsistent with your convention that "to the left is negative, to the right is positive". Check your signs, fix the equation, and solve for the horizontal acceleration. Then consider what happens to it when you make the initial extension ##x## of the spring larger and larger.

Vigorous
Much better but the horizontal equation for the bottom mass is inconsistent with your convention that "to the left is negative, to the right is positive". Check your signs, fix the equation, and solve for the horizontal acceleration. Then consider what happens to it when you make the initial extension ##x## of the spring larger and larger.

I equate the force of friction for mass m and M and solved for a. a=-kx/(M+m) as the extension to the right increases the acceleration will also increase but to the left.However, the force of friction itself ranges from zero to some maximum so x is constrained by that maximum.

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Very good. For the magnitude of the acceleration you have to remove the negative sign. You can see as x increases, the magnitude of the acceleration increases.

Now can you find the force of static friction between the masses in terms of k, x, m and M?

Vigorous
For mass m the normal force is mg therefore the maximum of the force of static friction mg μ =mkx/(M+m)
x=gμ(M+m)/k. I have a question about the force of static friction, I thought that it exactly balances the force applied on an object and keeps it from moving. However mass m is accelerating looking at from the frame of references (as in my picture)if on the other hand the frame of reference is fixed at mass M mass m would remain static and ax relative=0. But I am not sure if my thought is correct.

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You have the right thoughts here, but let me organize them.
1. When the two masses move as one, their common acceleration is ##a=\dfrac{kx}{M+m}.##
2. The net force on the top mass is the horizontal force of static friction, ##F_{\text{Net,top}}=ma=f_s=\dfrac{m}{M+m}kx.##
3. As ##x## increases, ##f_s## must increase in order for the two masses to move as one.
4. The force of static friction has a maximum value ##f_s^{max}=\mu_sN## beyond which it can increase no longer.
5. At spring elongation ##x=x_{\text{max}}## we have ##f_s=f_s^{max}##. If the spring is elongated past this value, the top mass will not have enough net force to give it the acceleration needed to keep up with the bottom mass and there will be sliding at the interface between masses.

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Further to @kuruman's post #23, it is interesting to consider what would happen next if m slides at first.
The frictional force is now ##mg\mu_k##, where typically ##\mu_k<\mu_s##. So M's acceleration is given by ##Ma_M=kx-mg\mu_k##, and ##ma_m=mg\mu_k##.
(Note: I am taking x as positive to the right, but accelerations as positive to the left. It would be more kosher to keep positive as the same for all.)
Note that m's acceleration is constant while M's is reducing, and may even reverse direction.
Assuming M is sufficiently wide that m does not fall off, this will continue until the two masses happen to have the same velocity again. At this point static friction will be restored.

Vigorous
You have the right thoughts here, but let me organize them.
1. When the two masses move as one, their common acceleration is ##a=\dfrac{kx}{M+m}.##
2. The net force on the top mass is the horizontal force of static friction, ##F_{\text{Net,top}}=ma=f_s=\dfrac{m}{M+m}kx.##
3. As ##x## increases, ##f_s## must increase in order for the two masses to move as one.
4. The force of static friction has a maximum value ##f_s^{max}=\mu_sN## beyond which it can increase no longer.
5. At spring elongation ##x=x_{\text{max}}## we have ##f_s=f_s^{max}##. If the spring is elongated past this value, the top mass will not have enough net force to give it the acceleration needed to keep up with the bottom mass and there will be sliding at the interface between masses.

So if the spring was stretched past this displacement, will the top mass stay at the stretched position and not be pulled back? If I try to write Newtons laws for both masses in this scenario the force of friction would no longer be acting

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If I try to write Newtons laws for both masses in this scenario the force of friction would no longer be acting
That is incorrect. The contact between the masses does not become frictionless all of a sudden. There will still be friction acting on both masses, but of the kinetic variety.

Vigorous
That is incorrect. The contact between the masses does not become frictionless all of a sudden. There will still be friction acting on both masses, but of the kinetic variety.

So What happens when the maximum is exceeded, the top mass and bottom mass start moving with different accelerations, am=μg aM=(μmg-kx)/M=(m*am-kx)/M (previous to that static friction kept mass m glued to mass M causing both blocks to accelerate at the same rate) until at some point the top mass will fall to the ground.