B Confusion with orientation of coordinate axis in inclined plane

[rudransh verma]

When we take the x-axis parallel to incline surface its clear that the horizontal component of weight is causing the block to come down but when we take the standard orientation its not so clear to me. Is horizontal component of $F_N$ causing the block to come down?

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[BvU]

The acceleration is along the incline, perpendicular to $F_N$. So: no.
The force that causes the block to come down is the component of $mg$ that is along the incline.
The other component of $mg$ is offset (compensated) by $F_N$.

$\$

 

[gmax137]

You need to write out the magnitudes of the components. What do you get for F_Ny and F_Nx? You should find expressions containing mg, and sines and cosines of the incline angle.

This gets messy but you need to do it, in order to understand why incline problems are always solved using the parallel and perpendicular axes (as shown on the left). The answer has to come out the same either way, but one way is much easier.

 

[Lord Jestocost]

https://docs.google.com/drawings/d/10UhNcINgO14PWv8UikwHvNpWxQzwIJvLGuwB2L1pS2o/edit?usp=sharing
When we take the x-axis parallel to incline surface its clear that the horizontal component of weight is causing the block to come down but when we take the standard orientation its not so clear to me. Is horizontal component of $F_N$ causing the block to come down?

See "Mass on incline with friction" on http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html

 

[Herman Trivilino]

https://docs.google.com/drawings/d/10UhNcINgO14PWv8UikwHvNpWxQzwIJvLGuwB2L1pS2o/edit?usp=sharing
When we take the x-axis parallel to incline surface its clear that the horizontal component of weight is causing the block to come down but when we take the standard orientation its not so clear to me. Is horizontal component of $F_N$ causing the block to come down?

Well, first of all, any horizontal direction is always perpendicular to $m \vec{g}$, by definition. For example, a plumb bob points in the direction of $m \vec{g}$. A carpenter's level is oriented in a horizontal direction when the bubble is in the middle.

We typically choose the x-axis to be parallel to the surface of the ramp. We do this so that the block's acceleration will have a zero component in the y-direction.

If instead you choose the x-axis to be horizontal (what you are calling the "standard orientation") then the acceleration will have generally nonzero x- and y-components. When you follow the scheme outlined by @gmax137 in Post #3 you will have values for $a_x$ and $a_y$, the perpendicular components of $\vec{a}$.

 

[rudransh verma]

View attachment 299037You need to write out the magnitudes of the components. What do you get for F_Ny and F_Nx? You should find expressions containing mg, and sines and cosines of the incline angle.

This gets messy but you need to do it, in order to understand why incline problems are always solved using the parallel and perpendicular axes (as shown on the left). The answer has to come out the same either way, but one way is much easier.

The eqns are $F_N\cos \theta-mg=-ma\sin \theta$
$F_N\sin \theta=ma\cos \theta$.
I understand that it’s easier when one of the components of a is zero.
But my main confusion is with $F_N$. As the previous post suggest it’s not responsible for the acceleration of the block since they are perpendicular to each other.
But if we take the horizontal and vertical axis and resolve components then we can see one component of acceleration is in direction of one component of normal and other component of acceleration is parallel to mg. So are both the forces normal and mg together causing the acceleration down the incline?

 

[BvU]

So are both the forces normal and mg together causing the acceleration down the incline?

No.

 

[rudransh verma]

No.

Why? Isn’t the resultant of $F_N \sin \theta$ and $mg$ causing the acceleration down the incline?

 

[BvU]

No. $F_N \sin\theta$ is not in the direction of the incline.
The component of $F_N \sin\theta$ that is along the incline is $F_N\sin\theta\cos\theta$.
But then so is the component of $F_N\cos\theta$ that is along the incline: $F_N\cos\theta\sin\theta$ and that is pointing in the opposite direction, nicely canceling. In short:

$F_N$ has NO component along the incline. Period.

$\$

 

[rudransh verma]

No. $F_N \sin\theta$ is not in the direction of the incline.
The component of $F_N \sin\theta$ that is along the incline is $F_N\sin\theta\cos\theta$.
But then so is the component of $F_N\cos\theta$ that is along the incline: $F_N\cos\theta\sin\theta$ and that is pointing in the opposite direction, nicely canceling. In short:

$F_N$ has NO component along the incline. Period.

$\$

Post in thread 'To move a block up to the top of the wedge'
https://www.physicsforums.com/threa...-to-the-top-of-the-wedge.1013106/post-6609210
See in this post using the horizontal as x-axis both the weight and normal force components are playing a role in slowing the block down. So why not here the normal force and the weight together causing the block to slide down?

 

[gmax137]

You need to write out the magnitudes of the components. What do you get for F_Ny and F_Nx?

The eqns are FNcos⁡θ−mg=−masin⁡θ
FNsin⁡θ=macos⁡θ.

No. What is the magnitude of F_N? You have done this before in other problems.

$\left\|F_N \right\|=mg \cos\theta$

Now, write expressions for the F_Nx and F_Ny components of F_N. This is just trigonometry.

 

[BvU]

Post in thread 'To move a block up to the top of the wedge'

That thread is not about a block on an incline.
Or are you saying that the current thread is not about a block on an incline ?

$\$

 

[rudransh verma]

Now, write expressions for the F_Nx and F_Ny components of F_N.

That is what I wrote.

The eqns are FNcos⁡θ−mg=−masin⁡θ
FNsin⁡θ=macos⁡θ.

 

[rudransh verma]

That thread is not about a block on an incline.

In that thread the block collides with the wedge/incline sets it in motion and then the block moves up the wedge with the wedge in motion and in the end both have same velocity.

the current thread is not about a block on an incline ?

In this thread I am talking about the block sliding down the stationary incline.
If in that thread components of normal and weight are causing the block to slow down then why not in this thread helps to accelerate it down the incline?

 

[nasu]

You forget that the acceleration is related to the net force. In this example the net force is equal to the tangential component of the weight. This is so no matter how you choose the axes.
In other problems the net force may have a different direction, even though there is an inclined plane involved.
In that problem the incline was moving, here it is stationary. It would be silly to expect to see no difference, assuming that what you say about that one is right.

You should have started with the stationary incline and after you understand this simpler case move to more complicated problems with moving inclines.

 

[gmax137]

That is what I wrote.

Yes I know what you wrote, it is not correct. Try again.

 

[rudransh verma]

Yes I know what you wrote, it is not correct. Try again.

If you are asking the components of F_N then it’s $F_N \cos \theta$(vertical) and $F_N \sin \theta$(horizontal).
I actually wrote the eqns involving those components. You can see I know what the components are.

 

[rudransh verma]

You forget that the acceleration is related to the net force. In this example the net force is equal to the tangential component of the weight. This is so no matter how you choose the axes.

Yes , I did not. How will you show that block will come down using horizontal and vertical axis?

 

[nasu]

Yes , I did not. How will you show that block will come down using horizontal and vertical axis?

Do you understand how you do it for axes parallel and perpendicular to the plane?

 

[rudransh verma]

Do you understand how you do it for axes parallel and perpendicular to the plane?

Yes! $mg\sin \theta$ will cause the block to come down.

 

[nasu]

Even though you claim the contrary, you still forget about the net force. How do you prove that the net force is equal to the tangential component of the weight? For any choice of axes. Ifyoudon't prove this, you did not prove that it moves the way you say it does.

 

[rudransh verma]

How do you prove that the net force is equal to the tangential component of the weight?

$F_N-mg\cos \theta=0$
Because there is no vertical movement.
$mg\sin \theta=ma$
Are you convinced?

 

[nasu]

There is vertical movement. There is no movement perpendicular to the plane. But did you prove this or you took it as an assumption? How do you know that there is no motion perpendicular to the plane? Newton's laws alone do not allow you to prove this. The equation for Newton's law along the normal direction is a result of the asumption that the normal acceleration is zero. It is not proving it.

 

[rudransh verma]

How do you know that there is no motion perpendicular to the plane? Newton's laws alone do not allow you to prove this. The equation for Newton's law along the normal direction is a result of the asumption that the normal acceleration is zero.

Well we can clearly see that the block in the end stays on the incline and also during the acceleration. It does not sink in or leave the incline at any time. It’s visible in front of us. No need to prove it.

 

[nasu]

Aha, this is the point. What you say here is what is called a "constraint" in mechanics. The constraints are conditions imposed on the system, independently of Newton's laws. In this case we have a body sliding down the plane, right? This is how the constraint is described usually in the text of the problem. This means that the acceleration has to be along the plane and so the net force has to be along the plane. This is a constraint and not something you can prove from Newton's laws. And is independent on the axes, of course.
Now, if you want to see how this work for your vertical-horizontal axes, you need to see how the constraint is expressed for this case. Now you have acceleration components along both x and y axes. But you know that the body slides along the plane so there should be a relationship between the components of the acceleration so that the resultant is along the plane. If you use this condition, together with the equations provided by Newton's laws along the x and y, you will find again that the normal force is equal to mg*cos(α).

 

[gmax137]

If you are asking the components of F_N then it’s FNcos⁡θ(vertical) and FNsin⁡θ(horizontal).

Good. Now we know the magnitude of the normal force is
$\left\| F_N \right\| = mg \cos\theta$
do you agree? Do you see why it is independent of the coordinate axes we use?

So we have for the components of the normal force:
$F_Nx = F_N \sin\theta = mg\cos\theta \sin\theta$
$F_Ny = F_N \cos\theta = mg \cos\theta \cos\theta = mg \cos^2\theta$

In addition to the normal force, there is also gravity:

$F_Gx = 0$
$F_Gy = -mg$

So the components of the total force are
$F_x = F_Nx + F_Gx = mg\cos\theta \sin\theta + 0 = mg\cos\theta \sin\theta$
$F_y = F_Ny + F_Gy = mg \cos^2\theta - mg = mg(\cos^2\theta - 1) = -mg \sin^2\theta$

Do you know how to find the magnitude of the resultant vector? If
$\vec F = \vec F_x + \vec F_y$
Then
$\left\| F \right\| = \sqrt {F_x^2 + F_y^2}$

So
$\left\| F \right\| = \sqrt {(mg\cos\theta \sin\theta)^2 + (-mg \sin^2\theta)^2}$

This looks horrible but it simplifies nicely, can you do that?

 

[rudransh verma]

This looks horrible but it simplifies nicely, can you do that?

$F=mg\sin \theta$.

Now, if you want to see how this work for your vertical-horizontal axes, you need to see how the constraint is expressed for this case. Now you have acceleration components along both x and y axes. But you know that the body slides along the plane so there should be a relationship between the components of the acceleration so that the resultant is along the plane.

Ok.Here the constraint a is expressed as:

$$a=\sqrt {(a\sin \theta)^2+(a\cos \theta)^2}$$
$$a=\sqrt{(-g\cos ^2\theta+g)^2+(g\cos \theta \sin \theta)^2}$$
$a=g\sin \theta$
After putting a in equations I am getting $F_N=mg \cos \theta$
So net force, net acceleration, normal, weight are all constraints here. They will be same no matter what orientation of axis we choose.

And $F_N \sin \theta$ and $mg$ can’t alone produce net force. I just realized after watching the above post I forgot $F_N \cos \theta$. Silly mistake Thanks @gmax137

 

[nasu]

No, the relation you wrote (the first one) is not a constraint. The relationship between the magnitude0 of a vector and his components is a trivial one, true for any vector, in any setup.

In order to have the body sliding down the plane and keeping the contact the vector acceleration needs to make the same angle with the horizontal as the incline does. This means that the ratio between the vertical component of the acceleration and the horizontal one need to be equal to the tangent of the angle. This is the "constarint" for this problem. With this and Newton's laws you can find the normal force and the magnitude of the acceleration as well as the components of the acceleration. The second formula is right but it does not result from the first one which is irelevant to the case as it is valid for any vector.

 

[rudransh verma]

No, the relation you wrote (the first one) is not a constraint. The relationship between the magnitude0 of a vector and his components is a trivial one, true for any vector, in any setup.

Ok! So you mean I forgot about the angle of vector acceleration. That first relation is true for any angle. But to keep the body on the plane we need to have an angle that makes same angle with horizontal and vector and incline. So in a sense the net acceleration vector is the constraint. Right!

And for that problem with moving incline it’s the same three forces slowing the block as here accelerating the block. I got confused. But now I got it.

 

[nasu]

Ok! So you mean I forgot about the angle of vector acceleration. That first relation is true for any angle. But to keep the body on the plane we need to have an angle that makes same angle with horizontal and vector and incline. So in a sense the net acceleration vector is the constraint. Right!

And for that problem with moving incline it’s the same three forces slowing the block as here accelerating the block. I got confused. But now I got it.

In that problem the constraint is different. As the incline moves too, the relationship between the components of the acceleration of the block is different in the frame attached to the ground.

 

[rudransh verma]

In that problem the constraint is different. As the incline moves too, the relationship between the components of the acceleration of the block is different in the frame attached to the ground.

First of all to be clear we do measure everything relative to ground in Newtonian mechanics. Right?
Is it $\tan \theta=\frac{a_1\sin \theta}{a_1\cos \theta-a_2}$ where $a_1$ is acceleration along the incline and $a_2$ the horizontal one?

 

[nasu]

First of all to be clear we do measure everything relative to ground in Newtonian mechanics. Right?
Is it $\tan \theta=\frac{a_1\sin \theta}{a_1\cos \theta-a_2}$ where $a_1$ is acceleration along the incline and $a_2$ the horizontal one?

We measure it in respect to any convenient reference frame. The ground is quite common but it is nothing special about it. But in order to apply Newton's laws without "fictitious" forces the reference frame must be inertial. The ground is pretty good aproximation of inertial frame. It does not men that non-inertial frame are not useful in solving problems. Sometimes is quicker to solve in a non-inertial frame.

 

[rudransh verma]

We measure it in respect to any convenient reference frame. The ground is quite common but it is nothing special about it. But in order to apply Newton's laws without "fictitious" forces the reference frame must be inertial. The ground is pretty good aproximation of inertial frame. It does not men that non-inertial frame are not useful in solving problems. Sometimes is quicker to solve in a non-inertial frame.

To make it easier for you the problem is like this. The wedge is initially and rest and the moving block hits it setting it into motion. The block moves up with decreasing velocity and at the top both the wedge and the block have same velocity. The constraint net acceleration's angle I calculated is shown above.Tell me if I am right?

 

[nasu]

To make what easier for me?

 

[rudransh verma]

To make what easier for me?

You did not tell me whether the relationship between the components of the acceleration in moving incline case is right. I thought you did not get what the problem is.
In post #31 is the expression right?

 

[nasu]

If a₂ is non-zero the relationship is not true.
If a₂=0 you just have the definition of the tangent function.
So it is either wrong or useless, depending on the values of a₂. You don't need a diagram to realize this.

 

[rudransh verma]

If a2 is non-zero the relationship is not true.

Then what is the constraint in this case?
“A block of mass m is pushed towards a movable wedge of mass ηm and height h, with a velocity u. All surfaces are smooth. The minimum u for which the block will reach the top of the wedge is”
How is the constraint extracted from the question?

 

[rudransh verma]

Do you see why it is independent of the coordinate axes we use?

Yeah! Because it’s part of the constraint.

 

[gmax137]

Yeah! Because it’s part of the constraint.

What I was thinking is, the normal force is by definition the force perpendicular to the surface. That's what "normal" means in this context. So the normal force is equal and opposite to the component of gravity perpendicular to the surface. This is unrelated to what axes you subsequently choose for the analysis.

The notion that the inclined plane is "hard" such that the mass doesn't sink into the plane is in my opinion a "given" in problems such as these. Calling it a "constraint" is probably correct but to dwell on that idea overmuch misses the point of these problems - namely to correctly identify the forces, create the free body diagram, and practice trigonometry skills.

 

[nasu]

So the normal force is equal and opposite to the component of gravity perpendicular to the surface. This is unrelated to what axes you subsequently choose for the analysis.

But this is not true if the incline is mobile. This is why is better to see where the "rule" comes from rather than applying it blindly as an universal condition. It's not important what you call it but it's important to realize that it something you need besides N's laws to solve the motion.

 

[gmax137]

But this is not true if the incline is mobile.

Ahh, OK, I was not thinking about the mobile incline problem. That one is certainly a level of sophistication up from the current thread, at least as it started.