# Do hanging masses contribute to the mass of the entire system?

Consider the pedagogical machine on a frictionless surface from the book by Kleppner and Kolenkow:

When we are talking about the acceleration of the whole system, do we use $$F=(M_{\text{net}}\text{ of system})a$$ or $$a=\frac{F}{M_{\text{net}}}?$$

The $$M_{\text{net}}$$ of the system should be $$M_1+M_2$$. We don't include the third mass, because it is not touching the ground and there is not a normal force to cancel it (I am not talking about the normal force provided by the wall of $$M_1$$). There is not a contribution of $$M_3$$ on the whole system, but when I do the calculation and find how much force $$F$$ should be applied on the system to make $$M_3$$ stay in equilibrium with $$M_2$$, I get $$F=\frac{(M_1+M_2)M_3g}{M_2}$$ but it is wrong and the correct answer comes when we take $$M_3$$ into account of the $$a$$ of the system.

Is there a fault in this reasoning? It's like holding a string with mass $$m$$ above a block of $$M$$ and calculating acceleration to be $$F/(m+M)$$.

• Voting to reopen. Clearly a question about understanding the physics principles involved and not a homework or engineering question or a question requiring opinion based philosophical discussions.
– KDP
Commented Apr 21 at 16:43
• @KDP Agreed. I have also voted to reopen. Commented Apr 21 at 17:04
• I agree. Edited the question to make it clearer. Commented Apr 21 at 17:48
• To apply a single $F = ma$ equation to the assembly of blocks, treating it as a single system, you need to be sure that when you apply the force all 3 will move together (maintaining their relative positions). If you think that is the case then you would use $F = (M_1 + M_2 + M_3)a$. If you think that $M_3$ will not be affected by a force you apply to the large block, then you don't have a single system. You have (at least) two systems, interacting, so a single $F = ma$ equation will not be enough to say what happens when you apply a force to the large block.
– Ben
Commented Apr 22 at 2:53
• Note that the question in the book is "what force is needed to keep M3 from rising or falling?". This implies that M2 does not move relative to M1, which in turn implies (if M3 cannot move horizontally) that horizontal speed and acceleration is equal for all three masses, which is why we can treat it as one mass. If the forces do not cancel out, the case is more difficult (and, arguably, more interesting). Commented Apr 22 at 12:32

If $$M_3$$ is accelerating to the right with the same acceleration as the rest of the mechanism then we must include its mass in the mass of the whole system because there must be some horizontal force acting on it to accelerate it.

If $$M_3$$ were not accelerating to the right then we could ignore its mass. However, this can only be a temporary state of affairs, because $$M_3$$ will eventually come into contact with the side wall of its channel, which will exert a normal force on it, giving $$M_3$$ the same horizontal acceleration as the rest of the mechanism, which takes us back to the first case above.

• So we only take that masses the force is disturbing.
– alam
Commented Apr 21 at 9:30
• If you know that all the masses are accelerating with the same horizontal acceleration then you can treat them as one mass $M_1+M_2+M_3$ with some unknown horizontal acceleration $a$. If not (if M_2 is sliding on M_1 for example) then you will have to draw a free body diagram for each object showing the forces acting on it, and use what you know bout the geometry of the mechanism to relate their accelerations. Commented Apr 21 at 9:43
• The issue is not M3 whose horizontal velocity (and hence, acceleration) is always the same as that of M1 (if we assume a perfect fit in its vertical slot with no play). The issue is M2 which can move horizontally with respect to the M1/M2 system. Because M2 may move relatively to M1/M3, The center of mass of the complete M1/M2/M3 system will accelerate and consequently move faster or slower than M1/M3, depending on whether the force F is larger or smaller than the gravitational force on M3. If M3 -> 0, M1/M3simply slide out under M2. Commented Apr 22 at 12:28
• Anas "So we only take that masses the force is disturbing" not really, it's simpler than that. See my answer Commented Apr 23 at 2:14
• This is a great answer but even if the hanging mass never hit the side wall, it would still be pulled along. I'm just mentioning that for the sake of completeness. Commented Apr 23 at 15:34

Consider what happens if we move the whole system to the right by one metre. M3 will also have moved 0ne metre to the right so M3 is part of the total mass that has to be accelerated by force F. This is the case whether M3 is being held at a constant horizontal height or if it is in free fall.

how much force F should be applied on the system to make M3 stay in equilibrium with M2?

Now that it has been clarified that the system is frictionless, we can calculate the above.

In order to achieve equilibrium we need the horizontal reaction force acting backward on M2 to be equal to the force acting downward on M3, so:

$$F_2 = F_3 \ \rightarrow \ M_2 \ a_2 = M_3 \ g \ \rightarrow \ a_2 = \frac{M_3}{M_2} g$$

In order to keep M2 stationary with respect to M1 it is necessary to match this acceleration with the horizontal force acting on the total mass of the system, so:

$$F = M_{total} \ a_2 = (M_1+M_2+M_3) \frac{M_3 }{M_2} \ g$$.

• The surface is frictionless
– alam
Commented Apr 21 at 6:58
• So what your are essentially conyeing is that since the effect of force F can be seen on all 3 blocks then they constitute the whole system?
– alam
Commented Apr 21 at 7:02
• Yes, as far as the total mass ($M_t$) in the equation ($F=M_t a$) is concerned for the motion in the horizontal direction is concerned, otherwise M3 will have moved to the right without any energy input. However, as I mentioned in my answer, M3 does alter the friction between the large block and the lower surface.
– KDP
Commented Apr 21 at 7:05
• Oh I think I got it since F=Mnet(a) so if Mnet was m1+m2 then the force can only effect their motion but since because of this force M3 is also getting disturbed then this means M3 contributes in total mass am I right ?
– alam
Commented Apr 21 at 7:10
• This answer is, unfortunately, not correct. I made the same mistake, initially, and it is easy to make. The problem is that you essentially assume all you need is to match the acceleration of $M_2$ with $M_1$'s (and $M_3$'s with it). That this can not be correct can be seen with the border case of a very small $M_2$: $(M_2+M_3)$'s acceleration due to the weight of $M_3$ approaches g ($M_3$ is essentially not braked by a very light $M_2$). (ctd. ...) Commented Apr 23 at 8:31

Without friction

$$M_1\,a_1=F$$

With friction between $$~M_1~$$ and $$~M_2$$

From the FBD, you obtain:

$$M_2\,a_2=T-F_\mu\\ M_2\,a_3=-T-M_3\,g\\ a_2=a_3\\ M_1\,a1=F+F_\mu$$

From here with $$~F_\mu=-\mu\,M_2\,g~$$ you can obtain $$~a_1~,a_2~,a_3,T~$$.

From the requirement that $$~a_1=a_2~$$, you can obtain the friction coefficient $$~\mu~$$.

$$\mu=\frac{M_1}{M_2}\,\frac{M_3} {M_1+M_2+M3}$$

With this friction coefficient, the acceleration

$$M_1\, a_1=F-\frac{M_1\,M_3\,g}{M_1+M_2+M3}$$

• Yes thats correct
– alam
Commented Apr 21 at 11:56
• $M_1 a_1=F$ is certainly incorrect in all cases where $M_3 \neq 0$. Commented Apr 22 at 12:34
• I assumed , no friction and no collision of the mass M3 and the wall. Thus the only force that accelerate the mass M1 is F
– Eli
Commented Apr 22 at 15:33
• @Eli Presumably "no collision of the mass M3 and the wall" is not a valid assumption. Commented Apr 22 at 18:51

Let's first consider the problem the way you phrase it. Then, from first principles, the answer to "When we are talking about the acceleration of the whole system, do I need to consider all three masses" is an unequivocal "yes": The machine is a system with only one external horizontal force F acting on it. We don't need to know anything about its inner workings but can say with confidence that the external force accelerates the entire system (i.e., the center of mass of $$M_a = M_1 + M_2 + M_3$$) corresponding to Newton's laws:

$$F = m_a*a$$.

We can see that M2 may move horizontally relative to M1; therefore, the speed of the entire system (i.e., its center of mass) may be different from that of its base M1, because of the contribution of M2 moving at a different speed.

Let's investigate a few interesting cases (I'll use "Mx = 0" as a short form for "negligibly small").

### M3 = 0

In this case, any force applied to M1 accelerates only M1 which slides off from under M2. M1 accelerates faster than $$a_a = F/(M_1+M_2+0)$$ would suggest, but only until M3 hits the pulley.

### F = 0

Because there is no external force, the center of mass is not accelerated; M3 will fall and M2 will accelerate right, while M1 and M3 will accelerate left. The horizontal speeds will be inversely proportional to the masses of $$(M_1+M_3)$$ and $$M_2$$.

### F is the force needed to keep M1 stationary

This is an interesting case because energy and momentum are decoupled. The force accelerating M2 acts also, via the pulley, equally but opposite on M1; because our precondition is that we exactly counteract it, M1 does not move. Instead, only M2 is accelerated horizontally (and M3 vertically). This implies that the entire system $$(M_1+M_2+M_3)$$ gains horizontal momentum here, exactly according to the force we apply multiplied by the time: $$p = F*t$$. That the point the force is applied to does not move M1 is irrelevant for the momentum gain1. It is not, however, irrelevant for the energy gain: Pulling against a stationary object does not do any work, i.e., does not transfer energy. The kinetic energy of M2 is part of the potential energy M3 lost on its way down. But, inversely, the vertical acceleration of M3 cannot change the horizontal momentum of M2 — as we said, the change in momentum comes from the external force we exerted. Energy and momentum changes are entirely decoupled.

By the way, the force acting on M2 and oppositely on M1, i.e. the tension in the rope, is not, as I previously thought, M3's weight $$M_3*g$$. The reason is that M3's weight is accelerating both masses. For a very heavy M3 and a light M2, it is obvious that the rope will not carry all of M3's weight. We have

$$F=m*a$$ with $$F=M_3*g$$ and $$m=M_2+M_3$$: $$M_3*g = (M_2+M_3)*a$$ Solving for a: $$\frac{M_3*g}{M_2+M_3} = a$$

Which makes sense: For negligible M2 or huge M3, the acceleration approaches g, for huge M2 0 or negligible M3, it approaches 0.

Now we can compute the force $$F2_{accel}$$ needed to accelerate M2 with this acceleration, which is the tension in the rope and the opposite force on M1 (via the pulley). Again we simply use $$F=m*a$$, with the acceleration we just computed but with m this time only M2:

$$F2_{accel} = M_2*\frac{M_3*g}{M_2+M_3}$$

Interestingly, if we slightly rewrite this it becomes clear that even though the problem appears highly asymmetric, this is an entirely symmetrical formula:

$$F2_{accel} = \frac{M_2*g*M_3}{M_2+M_3}$$

Exchanging masses M2 and M3 does not change the tension in the rope! For equal masses, this is clear; for highly different masses $$M_{tiny}$$ and $$M_{huge}$$, the force cannot exceed $$M_{tiny}*g$$, either because that is the weight (as $$M_3$$) or because $$M_{tiny}$$ as $$M_2$$ will be accelerated with almost g by the falling $$M_{huge}$$ as $$M_3$$.

### F2accel = M3*g

This is a bit tricky and needs exposition.

Let's start with a resting M1. In this case, the weight of M3 accelerates both M2 and M3. The force needed to accelerate M2 — the tension in the rope — is dependent on the acceleration and on M2. For a very light M2, the combined mass approaches M3 and the overall acceleration consequently g; very little force is needed to accelerate M2 with almost g. Correspondingly, there is very little tension on the rope. By contrast, for a very heavy M2, the overall mass that needs to be accelerated by the constant weight of M3 is large, and the resulting acceleration will be small. The tension in the rope will approach the weight of M3, $$M_3*g$$. If the tension in the rope is exactly $$M_3*g$$, M3's weight will be entirely held by the rope, and consequently M3 will not accelerate vertically, and M2 will not accelerate horizontally relative to M1. (It will, of course, accelerate in absolute terms according to the force acting on it, as will the other parts of the machine.)

This is the state we want to achieve. For that, we can accelerate M1 which will exert a horizontal accelerating force on M2 and a corresponding, opposing vertical force on M3 via the pulley; this will increase the tension in the rope. The tension — the accelerating force — must equal the weight of M3 if M3 is supposed to be vertically stationary; all its weight will be on the rope:

$$F2_{accel} = M_3*g$$

With $$F2_{accel}=M_2*a$$:

$$M_2*a = M_3*g$$ Solving for a: $$a = g*M_3/M_2$$

The acceleration a extends to the entire system M1 + M2 + M3, since no part is horizontally moving relative to each other, therefore, with $$F = m*a$$ and $$a = g*M_3/M_2$$:

$$F = (M_1+M_2+M_3)*g*M_3/M_2$$

We see that M3 has a quadratic influence on the needed force: It accelerates M2 more, and its inertia must be overcome when accelerating the overall system.

We also see that when M2 shrinks, the acceleration needed for an inertial force that matches $$M_3*g$$ grows; to achieve higher accelerations, in turn, we need a larger F to accelerate the overall mass. This is true even though $$(M_1+M_2+M_3)$$ shrinks somewhat with a shrinking M2: With a shrinking M2, the fraction $$M_3/M_2$$ approaches infinity while the overall mass needing acceleration settles at the very finite $$M_1+M_3$$.

1 It should be noted that in all these experiments, the momentum changes of the overall system, horizontally through the external force F and vertically by the accelerating M3, are exactly compensated by corresponding opposite changes in Earth's momentum. (Yes, a hammer falling on its own hits the ground earlier than a feather falling on its own because it accelerates the Earth more towards itself while it's falling.)

• That's a very great answer thank you
– alam
Commented Apr 23 at 1:51
• The terrificest answer! Commented Apr 23 at 2:18
• @anasalam I realize I still made a mistake with the $F = M_3*g$ case because that is not the tension in the string, i.e., not the force we need to apply to make $M_1$ stationary. (As I said in the last part, that tension depends on $M_2$! And is always smaller than $M_3*g$.) Commented Apr 23 at 14:11
• @anasalam Corrected. The correct result is also quite interesting! Commented Apr 23 at 14:56

It's easiest to break this up into two parts:

1. $$M_2$$ and $$M_3$$ with $$F = 0$$
2. Add F after solving for F = 0.

With F = 0, if T is the tension in the string and a is the net acceleration then:

$$M_3g - T = M_3a \text{ and } M_2a = T$$

Solving for a gives: $$a = \frac{M_3g}{M_2+M_3}$$

how much force F should be applied on the system to make M3 stay in equilibrium with M2

For this to occur, the relative velocity between $$M_1$$ and $$M_2$$ must be zero. This means the net acceleration between $$M_1$$ and $$M_2$$ is zero. Since $$M_1$$ cannot be accelerated without also accelerating $$M_3$$ ($$M_3$$ is in a slot within $$M_1$$), the total force includes all three masses:

$$F = (M_1+M_2+M_3)a$$ Substituting a gives:

$$F = \frac{(M_1+M_2+M_3)M_3g}{M_2+M_3}$$

• Yes thats what I got after considering all masses
– alam
Commented Apr 22 at 4:51

I think the question posed is, "What must $$F_1$$ be so $$M_3$$ neither rises nor falls?" If that is the question, then all the parts are stationary with respect to each other and $$F_1 = (M_1+M_2+M_3)a_1$$. However, this does not complete the answer yet because we need to know a1. (Note that $$a_1 = a_2$$.) The tension in the string is $$gM_3$$. This force is applied to $$M_2$$ so $$gM_3 = M_2a_1$$. Therefore $$a_1 = gM_3/M_2$$ and the final solution is

$$F_1 = \frac{(M_1+M_2+M_3)gM_3}{M_2}$$

This makes sense because if $$M_3=M_2$$, then $$F_1 = (M_1+M_2+M_3)g$$ and the acceleration of the system is $$g$$, resulting in the acceleration force on $$M_2$$ exactly balancing the gravitational force on $$M_3$$. Just do a free body diagram and don't forget the string tension applies a force $$T$$ to $$M_1$$ in the left direction thru the pulley.

• I was confused about tension part. The string is not physically attached to m1 so why would it apply force on it. Or are we using are Tension as pseudo force. If that's the case then tension is external force for m1 and internal for both m2 and m3. Is the reasoning correct?
– alam
Commented Apr 23 at 3:40

I don't see why $$M_3$$ should be excluded from the net mass when it's still in interaction with both $$M_2$$ and $$M_3$$ via tension and normal forces. Even when not touching $$M_1$$ $$M_3$$ still affects it via the normal force that the string hanging $$M_3$$ exerts on the pulley of $$M_1$$. None of these forces has anything to do with friction. As long as $$M_1$$ has interaction with any part of the system it should be included. To solve your problem it must be noticed that if $$M_3$$ doesn't move vertically, then the distance between $$M_2$$ and the pulley is the same, meaning that $$M_2$$ and $$M_3$$ has the same acceleration. We have $$F = a_1M_1 + a_2M_2 + a_3M_3 = a_{sys} (M_1+M_2+M_3)$$ Since $$a_1 = a_2 = a_3$$, $$a_1 = a_2 = a_3 = a_{sys}$$. Therefore $$F = M_{sys}a_{sys} = (M_1+M_2+M_3)\frac{F_2}{M_2} = (M_1+M_2+M_3)\frac{gM_3}{M_2}$$

If the mass is

that means it is

# pressing down

on the hook which is part of the object.

You can now immediately see there is absolutely no difference whether it is pressing down on a hook, a ledge, the base, or any other part of the object.

Some of the the other answers and comments have nothing to do with the question: The various calculations offered apply to any of the three parts shown. (Or indeed ... "to anything".)

Your actual question is, does the same apply to all three parts or is the "hanging" one different.

The actual answer is "no different". And you can see that by the "lightbulb moment" that pressing on the hook (or anything else) is no different than pressing on the base, a ledge, etc.

# Here is the "flaw in your reasoning":

"It's like holding a string with mass 𝑚 above a block of 𝑀 ..."

Quite simply it's not. You are not holding it, the block M is holding it, that's all there is to it.

# Note

if you're thinking about wanting to think of the device as not a platonic diagram, but a "real object",

so you are thinking about things like:

• The string on the right you are pulling it by in fact is elastic (there are no inelastic substances)

• "Simultaneous" motion in steel (or any) substance is only transmitted at the speed of sound in that object

• The glue, weld, or whatever between the arm/wheel part is entirely non-platonic, it will bend, vibrate etc

• Pulling at the point up-and-down (where F connects or is joined to a hook or whatever) will in reality massively torque the system, depending drastically on that up-down value, the exact angles involved, the 3D moments of the various systems and so on.

• The vertical bit of rope (or whatever it is) above M3 will behave in an incredibly complex manner (you can imagine one of those ultra-slow motion films of it, demonstrating a vast complexity of curves, bends, vibrations, etc) depending on a huge number of factors, such as the material properties, the exact nature of dozens of things such as the connections at each end, etc.

• Many more similar issues

... if you have in mind any of "that stuff", nobody here would have a clue. This is a physics site. ("The answer is F = ma") You would have to ask on an engineer site and there'd be a lot of talk about the latest simulation techniques to study these sort of systems.

{By way of example, as you obviously know from common experience, when you throw around an object that has moving parts it behaves in erratic ways only generally predictable: For example, roll around a half-full bottle of liquid, slide around an empty box with some heavy small objects in it, or say apply forces to the keys of a piano. All of that is "engineering" and nothing to do with the physics question - the pedagogy of which is simply that pressing down on the hook or any other part is no different, it's all the same object.}

# Note

If you envisage the top mass "falling off the edge" of the object then the object would be different!