Relative Acceleration - Particle and Wedge

[Edward_C]

I would be very grateful for help with deriving the following equation...

"A smooth fixed plane is inclined at 30 degrees to the horizontal. A wedge of mass M and angle 30 degrees is held on the surface so that its upper face is horizontal, and a particle of mass m rests on this face. The system is released from rest. Show that the resultant acceleration of the particle is:

(M + m)g / 4M + m
"
I do not understand how to derive this equation. Right now, I can only envisage the particle adding to the wedge's overall weight, and thereby increasing the normal reaction of the surface and consequently the wedge's acceleration down the slope. It seems to me that the acceleration of the wedge and particle should be the same.

Any solution or assistance would be most gratefully received

 

[berkeman]

I would be very grateful for help with deriving the following equation...

"A smooth fixed plane is inclined at 30 degrees to the horizontal. A wedge of mass M and angle 30 degrees is held on the surface so that its upper face is horizontal, and a particle of mass m rests on this face. The system is released from rest. Show that the resultant acceleration of the particle is:

(M + m)g / 4M + m
"
I do not understand how to derive this equation. Right now, I can only envisage the particle adding to the wedge's overall weight, and thereby increasing the normal reaction of the surface and consequently the wedge's acceleration down the slope. It seems to me that the acceleration of the wedge and particle should be the same.

Any solution or assistance would be most gratefully received

Welcome to the PF.

In the future, please organize your schoolwork post by using the Homework Help Template that you are provided when posting here. I'll paste in a copy of the Template below so you can see what it looks like.

Also, can you please Upload a diagram that shows the situation? Please also show us your free body diagram (FBD) for the problem. That will make it a lot easier for us to help you.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Edward_C]

Thank you very much for your response and my apologies for not adhering to the template. Based on the problem description (which was not accompanied by an image) I came up with the free body diagram attached.

Does this look correct? As you can probably see, the main problem is that I'm uncertain as to the acceleration of the particle relative to the wedge. I can see the wedge being acted on by the surface and particle, but no net forces seems to act on the particle. I clearly don't understand the system and would really appreciate some assistance.

 

[Edward_C]

 

[berkeman]

Thank you very much for your response and my apologies for not adhering to the template. Based on the problem description (which was not accompanied by an image) I came up with the free body diagram attached.

Does this look correct? As you can probably see, the main problem is that I'm uncertain as to the acceleration of the particle relative to the wedge. I can see the wedge being acted on by the surface and particle, but no net forces seems to act on the particle. I clearly don't understand the system and would really appreciate some assistance.

Ah, the diagram helps a lot, thanks.

Are they assuming that all surfaces are frictionless? They say the inclined surface is "smooth", but that doesn't necessarily imply frictionless.

If there is no friction between the wedge and the inclined plane, and there is friction between the wedge and the top box/particle, then you are right that the masses would just add. But if there is no friction between the wedge and the top box/particle, I can see that the situation would be changed. The top box would have no horizontal motion as the wedge slipped down and to the left. Can you see how that would change the acceleration of the wedge and particle?

 

[Edward_C]

Thank you very much for your reply. As all the earlier exercises in the textbook indicated frictionless surfaces, I assume the same applies here and that there's no friction between the wedge and the top particle. Honestly, though, I just don't understand how the system produces an acceleration of '(M + m)g / 4M + m' in the particle.

 

[berkeman]

(M + m)g / 4M + m

Could you use parenthesis to clarify what you mean by this equation? If it's (M + m)g / 4(M + m) I think I know how they got it.

 

[Edward_C]

Could you use parenthesis to clarify what you mean by this equation? If it's (M + m)g / 4(M + m) I think I know how they got it.

Thank you very much indeed for your response. The equation is definitely printed in the book as being:

(M+m)g / (4M + m),

I.e. M only is multiplied by four in the denominator. The system description preceding the equation is faithfully copied in this post. The book is "Introducing Mechanics" by Jefferson (OUP). Nevertheless, I would certainly be very interested to see the derivation of (M + m)g / 4(M + m).

 

[berkeman]

(M+m)g / (4M + m)

Hmm, I will need to think more about it then...

I would certainly be very interested to see the derivation of (M + m)g / 4(M + m).

The reasoning was that since the top mass cannot be moved sideways (because there is no friction) by the motion of the wedge, it just translates straight downward as the wedge accelerates down and left along the inclined plane.

So the acceleration of the wedge is (M+m) * cos(60) = (M+m)/2.

And since the top mass only translates downward, I thought that the acceleration downward was half of the wedge's acceleration down and to the left. But there must be more going on here, so back to the drawing board... Have you made any more progress?

 

[Edward_C]

I haven't solved it yet, but your your reasoning has generated new impetus and I've now gotten closer to the desired result. I'm going to have another try at getting the actual result, but I might have to resume tomorrow evening as I need to be up early tomorrow.

 

[Edward_C]

The best I've been able to come up with for now is the following updated equations of motion (where $R_{m}$ is the normal reaction of particle on wedge):

Horizontal Forces on wedge: $\frac{1}{2}R = \frac{3}{\sqrt{2}}Ma$

Vertical Forces on wedge: $Mg + R_{m} - \frac{3}{\sqrt{2}}R = \frac{1}{2}Ma$

Vertical Forces on Particle: $mg - R_{m} = \frac{1}{2}ma$ (acceleration of wedge downwards = 1/2a)

These gave me $a = \frac{2(M+m)}{19M + m}$, which really sucks. Obviously I'm making a big error but I'll have to look into it again tomorrow.

Once again, many thanks for your most valuable input.

 

[Rafa_El]

Which subject is it? In vector or in Linear?

 

[Edward_C]

Which subject is it? In vector or in Linear?

Hi Rafa_El, thanks for joining the thread. I'm not sure that I quite understand your question - please could you clarify the "it" you refer to? The question text in the original post is copied straight from the book and fully defines the problem. It needs solving using Newton's laws and straightforward vector/linear algebra.

 

[Rafa_El]

Hi Rafa_El, thanks for joining the thread. I'm not sure that I quite understand your question - please could you clarify the "it" you refer to? The question text in the original post is copied straight from the book and fully defines the problem. It needs solving using Newton's laws and straightforward vector/linear algebra.

Oh I'm sorry I am not really fluent english. I was misunderstand it. I was referring to your question

 

[Edward_C]

Oh I'm sorry I am not really fluent english. I was misunderstand it. I was referring to your question

No problem at all! The subject of the question is Introductory Mechanics, specifically using forces represented as vectors to solve for an acceleration.

 

[Rafa_El]

No problem at all! The subject of the question is Introductory Mechanics, specifically using forces represented as vectors to solve for an acceleration.

I ever face something like this. But usually we use common derivative in my school. Like for horizontal FX = Fx1+Fx2... Fxn then FY= Fy1... for vertical then we use Fr=square root(Fx+Fy)^2 then F=M.a then we substitute it for desired result like accele,mass or even the force

 

[Rafa_El]

I ever face something like this. But usually we use common derivative in my school. Like for horizontal FX = Fx1+Fx2... Fxn then FY= Fy1... for vertical then we use Fr=square root(Fx+Fy)^2 then F=M.a then we substitute it for desired result like accele,mass or even the force

And based on your question i think it ask amount of vertical force which mean against the gravity mean Fy= xN.sin○
correcte if I'm wrong

 

[Edward_C]

Yes, I think the consensus in the thread is that the particle can only move vertically. We just need to show that it's acceleration is:
(M + m)g / 4M + m
I.e. the actual value of the acceleration doesn't matter.

 

[Rafa_El]

Yes, I think the consensus in the thread is that the particle can only move vertically. We just need to show that it's acceleration is:
(M + m)g / 4M + m
I.e. the actual value of the acceleration doesn't matter.

well i need to work on it and hope i get the derivative

 

[Rafa_El]

well i need to work on it and hope i get the derivative

Btw i want to see your textbook about it. Maybe i can conclude something

 

[gneill]

Suppose you were given the vertical acceleration of the wedge surface; Call it $a_v$. What will be the normal force $f_n$ of the particle on the wedge? (write an expression for $f_n$ in terms of $m$ and $a_v$). You should also be able to relate $a_v$ with the downlsope acceleration of the wedge, $a$, via the geometry.

Then it's a matter of doing the usual block on a slope stuff.

 

[TSny]

The best I've been able to come up with for now is the following updated equations of motion (where $R_{m}$ is the normal reaction of particle on wedge):

Horizontal Forces on wedge: $\frac{1}{2}R = \frac{3}{\sqrt{2}}Ma$

Vertical Forces on wedge: $Mg + R_{m} - \frac{3}{\sqrt{2}}R = \frac{1}{2}Ma$

Vertical Forces on Particle: $mg - R_{m} = \frac{1}{2}ma$ (acceleration of wedge downwards = 1/2a)

These equations look (almost) correct. Should each $3/\sqrt{2}$ be $\sqrt{3}/2$?

 

[Edward_C]

Talk about embarrassing! Many thanks indeed TSny for revealing my stupid error, which most likely explains the larger coefficient of M mentioned earlier. I'll update the equations and post an update this evening.

 

[Edward_C]

I would like to thank you all for your generous and patient assistance, and also apologise for the howler above! Anyway, here are the correct equations:

Horizontal Forces on wedge: $\frac{1}{2}R = \frac{\sqrt{3}}{2}Ma$

Vertical Forces on wedge: $Mg + R_{m} - \frac{\sqrt{3}}{2}R = \frac{1}{2}Ma$

Vertical Forces on Particle: $mg - R_{m} = \frac{1}{2}ma$ (acceleration of wedge downwards = 1/2a)

$R_{m} = mg - \frac{1}{2}ma$
$R = \sqrt{3}Ma$
$(M + m)g - \frac{1}{2}ma - \frac{3}{2}Ma = \frac{1}{2}Ma$
...
$\frac{a}{2} = \frac{(M + m)g}{4M+ m}$ - particle's acceleration as required

 

[TSny]

Horizontal Forces on wedge: $\frac{1}{2}R = \frac{\sqrt{3}}{2}Ma$

Vertical Forces on wedge: $Mg + R_{m} - \frac{\sqrt{3}}{2}R = \frac{1}{2}Ma$

Vertical Forces on Particle: $mg - R_{m} = \frac{1}{2}ma$ (acceleration of wedge downwards = 1/2a)

$\frac{a}{2} = \frac{(M + m)g}{4M+ m}$ - particle's acceleration as required

Looks good!

You might check to see if you can work it by replacing the vertical and horizontal force equations on the wedge by just a single wedge equation for forces acting parallel to the plane. (I think this might have been what gneill was hinting at in post #21.)