Mechanics Question -- The direction of the pseudo force

[Suyogya]

Homework Statement

mechanics problem

Relevant Equations

laws of motion

I'm often got confused about why the direction of pseudo force is taken opposite to the direction of motion of the body. As specific, I have this question. In this, I didn't understood about why the direction of pseudo force is taken in right, in the diagram of solution provided to this question. In my knowledge it should be in the direction of motion of the bead (i.e. in left side).

 

[etotheipi]

Suppose in an inertial frame, a force $\vec{F}$ acts on a body resulting in an acceleration $\vec{a}$ as measured in that frame. Evidently, $\vec{F} = m\vec{a}$.

Now, let's consider a non-inertial frame whose acceleration is $\vec{a}_2$ relative to our inertial frame. The acceleration of the body relative to this non-inertial frame is thus $\vec{a} - \vec{a}_2$. Since this frame is non-inertial, we'll need to add some fictitious forces in order to make Newton II work. So the second law becomes

$\vec{F} + \vec{F}_{fict} = m(\vec{a} - \vec{a}_2) = m\vec{a} - m\vec{a}_2$

We can cancel out the $\vec{F} = m\vec{a}$ on both sides, and that leaves us with

$\vec{F}_{fict} = -m\vec{a}_2$.

As in, the fictitious force is directed in the opposite direction to the acceleration (not necessarily motion) of the non-inertial frame. Hopefully that clears up a few things - can you see how it applies to your problem?

 

[PeroK]

Homework Statement:: mechanics problem
Relevant Equations:: laws of motion

I'm often got confused about why the direction of pseudo force is taken opposite to the direction of motion of the body. As specific, I have this question. In this, I didn't understood about why the direction of pseudo force is taken in right, in the diagram of solution provided to this question. In my knowledge it should be in the direction of motion of the bead (i.e. in left side).

If you are in a car and you accelerate and you look out the window at an object (a house, say). The house appears to accelerate in the opposite direction to your acceleration. If you study the house in your frame of reference, you must add a pseudo-force opposite to the direction of your (real) acceleration.

From your reference frame you must apply this pseudo-force to everything. Including yourself. In order to have yourself at rest (in your reference frame) you need to add a pseudo-force equal and opposite to the real force you are subject to.

 

[Dr.D]

If you simply Newton's Second Law as F = m*a, there is no need to ever deal with a pseudo force. The very name "pseudo force" admits that this is not a force, so there is no reason at all to enter it into the force summation (left side of Newton's Law as written above).

I've only been doing mechanics for 60 years, but I've never yet needed a pseudo force, d'Alembert force, or any other imaginary force. May be the day will come, but I doubt it.

 

[Doc Al]

If you simply Newton's Second Law as F = m*a, there is no need to ever deal with a pseudo force.

That's true if you restrict yourself to only inertial frames.

The very name "pseudo force" admits that this is not a force, so there is no reason at all to enter it into the force summation (left side of Newton's Law as written above).

If you want to use Newton's laws in a non-inertial frame (something that can be extremely useful!) you'll need that pseudo force.

 

[hutchphd]

Can you not call the "force" of gravity itself a pseudo-force?

 

[Dr.D]

I must be working the wrong problems because I've never yet needed to use a non-inertial frame in a practical problem.

 

[hutchphd]

You've never thrown in a "centrifugal force" on a rotating object? Wow I am amazed..

 

[Cutter Ketch]

You are in a car. The car accelerates. You would like to think of the interior of the car as an inertial reference frame but some mysterious unseen force is pushing you in a direction relative the the interior of the car. Are you being pushed out the windshield or are you being pushed backwards into your seat back?

 

[Dr.D]

@hutchphd Then I suppose you must be amazed. I was very fortunate as a college freshman to have a physics prof who made the distinction between real and imaginary forces very clear.

@Cutter Ketch : You said, "You would like to think of the interior of the car as an inertial reference frame..." Why would anyone want to do that when you know the car is accelerating? That would just be dumb and asking for trouble.

 

[Cutter Ketch]

@Cutter Ketch : You said, "You would like to think of the interior of the car as an inertial reference frame..." Why would anyone want to do that when you know the car is accelerating? That would just be dumb and asking for trouble.

Because, as @Doc Al said

If you want to use Newton's laws in a non-inertial frame (something that can be extremely useful!) you'll need that pseudo force

You can rail all you like, but you aren’t going to get around the fact that this is what they are teaching the OP, and he needs to understand it. It’s fine if you don’t like using this tool but others find it useful. Useful enough that it is taught in physics class and physics students have to learn it.

 

[Dr.D]

You can rail all you like, but you aren’t going to get around the fact that this is what they are teaching the OP, and he needs to understand it. It’s fine if you don’t like using this tool but others find it useful. Useful enough that it is taught in physics class and physics students have to learn it.

I suspect that you are correct about this being taught in physics classes today and students being required to learn it. More is the pity. It is most unfortunate that students are being deliberately confused, but I'm really not surprised. It is what happens when the blind lead the blind.

What examples can you cite where this is necessary rather than confusing? When is there not an inertial coordinate system available if you want to use one?

I had no idea I was railing. How is railing defined exactly?

 

[Dr.D]

Let me tell you why D'Alembert's Principle (which is what this is all about) is a bad idea.

Newtons' 2nd law says Sum of Forces = M*a
Students and anyone else who want to apply this are encouraged to draw a free body diagram (FBD) that shows all of the FORCES. Forces are phenomena that either push or pull, so all of the actual forces belong on the FBD. But the imaginary forces (centrifugal forces, for example) does not.

The problem here is that treating M*a as simply another force discourages folks from properly formulating the acceleration. Almost everyone "simply guesses" at the form for a, but if you get a incorrect, the whole thing goes up in smoke. It is important to carefully and properly formulate a, and that means looking carefully at the kinematics of the problem. (Many folks may be surprised to learn that there is more, indeed much more, to kinematics than the so-called SUVAT formulas.)

That's where the rub comes.

 

[hutchphd]

It is most unfortunate that students are being deliberately confused, but I'm really not surprised. It is what happens when the blind lead the blind.

That seems a bit harsh to me and I will respectfully disagree with this assessment.
I was taught the notion of using a pseudo force in certain circumstances about a half century ago. The fellow who taught me this could see well enough to have helped design the first atomic weapon. I find it occasionally useful. I can also see just fine..
I will agree that students can become confused by it. But that alone does not disqualify its occasional utility.

 

[Dr.D]

@hutchphd You are certainly free to disagree. I don't think anyone should deny you that right. Let us look at one aspect of your answer:

The fellow who taught me this could see well enough to have helped design the first atomic weapon.

Allow me to ask how he helped. Was his effort in the mechanics of the device, or was it in the atomic physics side? I know a great many physicists today who are interested in things nuclear, atomic, relativistic, or molecular. I know of exceedingly few who continue to take mechanics seriously.

To continue you comment on Why D'Alembert's Principle is a bad idea," let me make one further observation. There are times in working a complicated mechanics problem when it is useful to account for all the reaction forces. According to Newton's 3rd Law, every (true) force has an equal and opposite reaction. Now if you are searching to see where each force is reacted, you can search the entire cosmos without finding the reaction to an imaginary m*a force. Thus, accepting D'Alembert's Principle means invalidating Newton's 3rd Law. Many of us are not comfortable with that.

 

[Suyogya]

Suppose in an inertial frame, a force $\vec{F}$ acts on a body resulting in an acceleration $\vec{a}$ as measured in that frame. Evidently, $\vec{F} = m\vec{a}$.

Now, let's consider a non-inertial frame whose acceleration is $\vec{a}_2$ relative to our inertial frame. The acceleration of the body relative to this non-inertial frame is thus $\vec{a} - \vec{a}_2$. Since this frame is non-inertial, we'll need to add some fictitious forces in order to make Newton II work. So the second law becomes

$\vec{F} + \vec{F}_{fict} = m(\vec{a} - \vec{a}_2) = m\vec{a} - m\vec{a}_2$

We can cancel out the $\vec{F} = m\vec{a}$ on both sides, and that leaves us with

$\vec{F}_{fict} = -m\vec{a}_2$.

As in, the fictitious force is directed in the opposite direction to the acceleration (not necessarily motion) of the non-inertial frame. Hopefully that clears up a few things - can you see how it applies to your problem?

What I can infer from your explanation is that the pseudo force should be applied on right hand side on the frame because the reference frame is moving on left with acceleration $\vec{a}$ , along with the bead. Hence the net acceleration on the frame is now zero, and the from the frame the acceleration of bead is now appears to be in left.
And further I solved the question like this, but the final answer didn't matches with the book. Is there any error I have done?

 

[hutchphd]

Let us look at one aspect of your answer:

Sorry, but no. If it doesn't work for you, then by all means don't use it

 

[PeroK]

What examples can you cite where this is necessary rather than confusing? When is there not an inertial coordinate system available if you want to use one?

The surface of the Earth itself is a rotating reference frame. An analysis of this throws up centrifugal and Coriolis terms. I'm not sure how you avoid those altogether. They are conceptually similar to the fictitious forces of simple linearly accelerated systems. Using non-inertial frames in simple cases should be good practice for more complicated systems like the Earth as a global system.

The most useful example I would say is adding the fictitious force to gravity to get an effective gravity. Moreover, fundamentally, this is the gravity we experience: the Earth's gravity towards the centre plus a fictitious centrifugal force (or minus a required centripetal force), which apart from the equator is not towards the centre of the Earth.

It's ironic, of course, that gravity itself turns out to be a fictitious force and an object at rest on the Earth's surface does not have an inertial rest frame: it experiences a single upward force and its state of rest in the Earth's reference frame requires the addition of the fictitious gravitational force.

 

[etotheipi]

What I can infer from your explanation is that the pseudo force should be applied on right hand side on the frame because the reference frame is moving on left with acceleration $\vec{a}$ , along with the bead. Hence the net acceleration on the frame is now zero, and the from the frame the acceleration of bead is now appears to be in left.
And further I solved the question like this, but the final answer didn't matches with the book. Is there any error I have done?

When you transform into the accelerating frame, the bead is acted on by all of the previous forces (weight, normal reaction) as well as a fictitious force $ma$ pointing opposite to the direction of acceleration of the frame (shouldn't this be to the right?).

You're then asked to find the acceleration of the bead relative to the rod. The accelerating frame we have just set up is the rest frame of the rod, so it suffices to compute the acceleration of the bead in this frame. It essentially reduces to resolving and finding the acceleration for an inclined plane problem!

 

[PeroK]

What I can infer from your explanation is that the pseudo force should be applied on right hand side on the frame because the reference frame is moving on left with acceleration $\vec{a}$ , along with the bead. Hence the net acceleration on the frame is now zero, and the from the frame the acceleration of bead is now appears to be in left.
And further I solved the question like this, but the final answer didn't matches with the book. Is there any error I have done?

I think you still have not understood the concept. Try a simpler question first.

Let's assume that the bead is not attached to the rod. Let's assume that the bead is resting on the ground independent of the rod.

The rod is accelerated to the left with acceleration $a$.

What is the acceleration of the bead relative to the rod? (Remember, the bead remains "at rest" on the ground.)