What Is the Net Force on the Middle Mass?

[InertialRef]

Homework Statement

Three masses (m1 = 4.4 kg, m2 = 13.2 kg and m3 = 8.8) hang from three identical springs in a motionless elevator. The springs all have the same spring constant, which is 268 N/m. The elevator is moving downward with a velocity of v = -3.7 m/s but accelerating downward at an acceleration of a = -1.6 m/s2 (speeding up). What is the magnitude of the net force on the middle mass?

Homework Equations

F = -kx
Net Force = m(a - g)

The Attempt at a Solution

I drew a free body diagram and found out the forces acting upon the middle mass. However, I don't think this is the way to go, since what I did was calculate those forces and then subtract one from the other.

 

[frogjg2003]

Is the elevator moving or not? The question suggests both. As for drawing the force diagram, you have to draw one for each mass. Also keep in mind that they are probably accelerating.

 

[InertialRef]

The elevator is moving downward with a velocity of v = -3.7 m/s but accelerating downward at an acceleration of a = -1.6 m/s2 (speeding up). So yes, it was moving. The question was weirdly worded and threw me off at first too.

Is there a different way to solve it other than drawing a force diagram?

 

[frogjg2003]

The elevator is moving downward with a velocity of v = -3.7 m/s but accelerating downward at an acceleration of a = -1.6 m/s2 (speeding up). So yes, it was moving. The question was weirdly worded and threw me off at first too.

Yes, badly worded. I'm assuming that it's telling you that the springs are motionless with respect to the elevator.

Is there a different way to solve it other than drawing a force diagram?

No. With practice, you don't have to physically draw one, but you still end up doing it in your head. You need some way to keep track of all the forces on each of the objects.

 

[InertialRef]

I'm still confused. Even after drawing the force diagrams, I'm completely unsure as to which forces contribute to the net force of the middle box. Can someone help me figure out which ones actually contribute to the net force?

 

[frogjg2003]

It would help if we could see your force diagrams. If the force diagrams are incorrect, there is no way you can get the question right.

 

[InertialRef]

Okay, so I have attached the force diagram.

The only forces that should be acting on the middle mass are the forces due to gravity and the acceleration of the elevator, which is -1.6 m/s. Since the force of the middle spring depends upon the mass of both the middle and the bottom springs, I'm assuming that you would have to add those together in order to calculate the net force. So what I did was:

Net Force = (total mass) x (acceleration)
= (13.2+8.8) x (-1.6 + (-9.81))
= -251.02 N

This answer was wrong. So I suppose that there is another force at work here (presumably due to the top most mass). However, whenever I calculate the force of the top most mass, either from gravity or from the acceleration of the elevator, I still do not get the right answer. I am at a loss, since I can't figure out which force it is that I am not accounting for.

 

[Simon Bridge]

Erm - no, you don't use the total mass in ƩF=ma - because that is just the sum of the forces on the one mass alone.

You are only asked for the magnitude of the net force on the middle mass.
You are given it's acceleration and it's mass... what is the problem?

 

[frogjg2003]

You are only asked for the magnitude of the net force on the middle mass.
You are given it's acceleration and it's mass... what is the problem?

Simon is right. I was trying to make you do too much work.
F_net=ma and you know a (-1.6m/s^2) and m (13.2kg not 26.4kg), so just plug the numbers into the equation.

On the other hand, you still don't have a good understanding of what forces act on the masses. Gravity does act on the mass, but so do the springs. You have completely neglected the springs acting on the mass. If you were given a more difficult problem, this would have seriously harmed you.
The other thing is that you included the acceleration as a force. Acceleration is not a force, it is the result of a net imbalance of forces. There are two ways to handle it. You can either not include it in the diagram, add up the forces, and have the imbalanced net force be equal to the mass times acceleration. The other way is to be in the accelerating frame and have a fictitious force in the opposite direction. Then you just add up the forces as if they were balanced and there was no net force.

 

[Simon Bridge]

Yes - I was torn between working through the inherent misconceptions that have come up, which are not involved in the answer, or pointing to the direct path to the answer. It's a tricky call. (I butted in because OP asked me to in private.)

Before students do spring-linked masses, they usually do masses linked by something non-stretchy and use tension forces. So OP will have completed that sort of exercise. Students are sometimes asked to do convoluted things with pulleys at this stage and/or to model a continuous rope as a series of small masses dm separated by a short massless length ds.

As a result, OP should be familiar with isolating each mass in turn by now so the above confusion is of some concern and will be worth spending time on if OP is willing.

The problem above has a lot of red herrings in it - the test is to see if the student understands enough physics to extract the needed information from a lot of data. Usually a problem is formulated to look like there is not enough information so this excess of data approach can cause confusion.

Curiously, IRL, we usually have far too much information and the main task is sorting out the stuff that doesn't matter. (Then we find there's a bit missing.)

 

[InertialRef]

Erm - no, you don't use the total mass in ƩF=ma - because that is just the sum of the forces on the one mass alone.

You are only asked for the magnitude of the net force on the middle mass.
You are given it's acceleration and it's mass... what is the problem?

So would it just be F = (13.2 x -1.6)? That answer is right.

Before students do spring-linked masses, they usually do masses linked by something non-stretchy and use tension forces. So OP will have completed that sort of exercise.

Unfortunately, we weren't. I think if we had done something tension related, as another classmate of mine pointed out, this question would have been easier to solve. I think either my instructors were expecting us to be aware of this already, or simply skipped over it thinking it would be obvious even if we hadn't done something like that before.

The problem above has a lot of red herrings in it - the test is to see if the student understands enough physics to extract the needed information from a lot of data.

That's exactly right. For someone like me who tends to over think every piece of information in a question, this one was particularly difficult because there were a lot of things that were given and because of the assumption I made that all of the information had to somehow be relevant. The tricky thing I'm still learning with physics is when to know if all the information is required, and when some can be disregarded.

Thank you for your help. :) I do struggle with concepts, so it's always nice to learn where I'm going wrong and how I can work to fix the mistakes I make.

 

[frogjg2003]

In this case, a spring and a string are the same. Since they are in equilibrium in the elevator frame, we don't need to know the nature of the forces, just that there are forces. If, on the other hand, they had to calculate the motion if the masses weren't at rest with each other, it would be a different story (and in the advanced physics homework help forum).

 

[Simon Bridge]

So would it just be F = (13.2 x -1.6)? That answer is right.

That is what we are saying.

As a matter of policy I do not normally confirm answers.
You are supposed to realize the correctness of the answer because you understood the physics.

In this case, if a mass m is accelerating at rate a, then it is, by definition, experiencing a net unbalanced force F given by F=ma.

Unfortunately, we weren't. I think if we had done something tension related, as another classmate of mine pointed out, this question would have been easier to solve. I think either my instructors were expecting us to be aware of this already, or simply skipped over it thinking it would be obvious even if we hadn't done something like that before.

Well, if your class has never encountered free body diagrams, that would explain why the problem is so simple. They would not anticipate that you'd go ask the question in a forum like this one where people know about these diagrams and would expect to use them. All they expected you to do is figure out that you didn't need most of the information supplied.

It's one of the problems with knowing a lot ;) When you have a machine-gun and a water-pistol there is a tendency to reach for the machine-gun when the pistol is all you need.

I suspect that the free-body method will be covered soon, then, since it hasn't already.

I'm still learning with physics is when to know if all the information is required, and when some can be disregarded.

It comes with practice ... now you know to watch for it. The main thing is to concentrate on the physics and not the equations.

When I do something like this I sit back and clear my mind and think: what would I need to know if I was doing this from scratch? Then I look through the data for that information - or clues to how I could work it out.