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accelerating inclined plane


nir99
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hello, the question is: there is block m on top of an inclined plane M 

the ratio of the masses M/m is given by k. 

angle alpha is given.

find the acceleration of the masses.

i tried to solve this with fictitious force (F(G)) that exerts on mass m to the left with the magnitude of m*a(M)

and my question is if the a(m) that i got is relative to the ground or to the plane, and if it is to the plane then how do i find the acceleration of mass m relative to ground?

another question is how can i solve this without using a fictitious force.

thanks for the help

20211203_200401.jpg

Edited by nir99
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I see you have mg and N; I’m not sure why you would include a fictitious force. mg acts down (-y), and N has a y component and an x component.

What can you say about the normal force that m exerts on M, and that M exerts on m?

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i learned that by including fictitious force i can solve the problem of mass m as i would if it wouldn't be on accelerating plane.

in my free body diagram the N that m exert on M has 2 components on x axis as i wrote Nsin(alpha)=Ma(M) and y component that N(with floor)=Mg+Ncos(alpha)

and N that M exert on m if i dont add the fictitious force is equal to mgcos(alpha)

Edited by nir99
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2 hours ago, nir99 said:

i learned that by including fictitious force i can solve the problem of mass m as i would if it wouldn't be on accelerating plane.

in my free body diagram the N that m exert on M has 2 components on x axis as i wrote Nsin(alpha)=Ma(M) and y component that N(with floor)=Mg+Ncos(alpha)

and N that M exert on m if i dont add the fictitious force is equal to mgcos(alpha)

What does the third law tell you about these normal forces?

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On 12/3/2021 at 6:05 PM, nir99 said:

hello, the question is: there is block m on top of an inclined plane M 

the ratio of the masses M/m is given by k. 

angle alpha is given.

find the acceleration of the masses.

i tried to solve this with fictitious force (F(G)) that exerts on mass m to the left with the magnitude of m*a(M)

and my question is if the a(m) that i got is relative to the ground or to the plane, and if it is to the plane then how do i find the acceleration of mass m relative to ground?

another question is how can i solve this without using a fictitious force.

thanks for the help

20211203_200401.jpg

You say you want to find the accelleration of the masses , don't you mean the acceleration of the ''free mass'' ? 

Also in which direction do you want to apply the acceleration ?

Do you want the ''free mass'' to lose grip and recede -ve or do you want the mass to assend +ve ? 

F=ma2 in this situation remember and also consider Newtons third law if you want to assend . 

Also remember the Newtons of down force is the same as if it weren't on an incline . 

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On 12/3/2021 at 6:05 PM, nir99 said:

hello, the question is: there is block m on top of an inclined plane M 

the ratio of the masses M/m is given by k. 

Seems to me that

Your diagram does not show the small block at the top of the wedge , as stated in your words. 

Why not ?

You have not mentioned friction anywhere or conversely smooth contacts.

Are we ignoring friction ?

What do you understand as the significance of the ratio M/m = k  ?

Can you describe in words what movements you think could take place ?

You should have done this last step before starting force analyses.

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sorry for not being clear , i'm not good at english.

there is no friction between the plane and the block. the floor and the plane

the movements i think that happen is : block m slides on top of inclinced plane M , inclined plane M has acceleration a(M) reletive to the floor only in x axis.

block m has x and y component . his y component that reletive to the ground equal to the one that reletive to the inclined plane because plane M doesnt have y component.

i chose the left direction to be the positive x d of m and downward to be the positive y axis

 mass M right direction is the positive x.

Edited by nir99
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11 hours ago, Pbob said:

You say you want to find the accelleration of the masses , don't you mean the acceleration of the ''free mass'' ? 

Also in which direction do you want to apply the acceleration ?

Do you want the ''free mass'' to lose grip and recede -ve or do you want the mass to assend +ve ? 

F=ma2 in this situation remember and also consider Newtons third law if you want to assend . 

Also remember the Newtons of down force is the same as if it weren't on an incline . 

Correction: F=ma

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12 hours ago, Pbob said:

You say you want to find the accelleration of the masses , don't you mean the acceleration of the ''free mass'' ? 

Both masses will accelerate

12 hours ago, Pbob said:

Also in which direction do you want to apply the acceleration ?

There is no applied acceleration

12 hours ago, Pbob said:

Do you want the ''free mass'' to lose grip and recede -ve or do you want the mass to assend +ve ? 

F=ma2 in this situation remember and also consider Newtons third law if you want to assend . 

This is not part of the problem.

12 hours ago, Pbob said:

Also remember the Newtons of down force is the same as if it weren't on an incline . 

No, it's not.

My advice to the OP is to ignore your post entirely. (if you want to discuss your misconceptions it should be done in a different thread)

3 hours ago, nir99 said:

i chose the left direction to be the positive x d of m and downward to be the positive y axis

 mass M right direction is the positive x.

It's usually better to choose one coordinate system for the entire problem. In this case the motion of the two blocks are related, and you've introduced an easy way to make a sign error when you relate the two sets of equations. Extra care must be taken to keep track of the information

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2 hours ago, nir99 said:

sorry for not being clear , i'm not good at english.

there is no friction between the plane and the block. the floor and the plane

the movements i think that happen is : block m slides on top of inclinced plane M , inclined plane M has acceleration a(M) reletive to the floor only in x axis.

block m has x and y component . his y component that reletive to the ground equal to the one that reletive to the inclined plane because plane M doesnt have y component.

i chose the left direction to be the positive x d of m and downward to be the positive y axis

 mass M right direction is the positive x.

 

No problem about the clarity I understood the question well enough and that's what discussion in Homework help is for.

Yes you are correct that the wedge M slides along the table or floor in the horizontally opposite direction to the mass m.
It is important to understand that this happens while the mass m is sliding down the slope.
So the motion of the wedge M is simple, but the motion of mass m is complicated because it's total acceleration is the resultant of three accelerations.

Can you see what these are ?

 

 

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3 minutes ago, studiot said:

 

No problem about the clarity I understood the question well enough and that's what discussion in Homework help is for.

Yes you are correct that the wedge M slides along the table or floor in the horizontally opposite direction to the mass m.
It is important to understand that this happens while the mass m is sliding down the slope.
So the motion of the wedge M is simple, but the motion of mass m is complicated because it's total acceleration is the resultant of three accelerations.

Can you see what these are ?

 

 

i cant think of 3 but 2. it there was no movement of the wedge  then it would slide at g*sin(alpha) in angle (aplha) but it has the acceleration of the wedge affect on it too.

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3 minutes ago, nir99 said:

i cant think of 3 but 2.

The wedge M is moving sideways to the right only. It has zero vertical movement.

mass m is moving downwards and to the horizonally to left relative to the floor, but at the same time it is sitting on wedge M so is also moving to the right the same as wedge M.

So mass m has a net sideways movement relative to the floor of two motions.

 

Does this help?

Hint think about what swansont said about coordinate systems.

 

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3 minutes ago, studiot said:

The wedge M is moving sideways to the right only. It has zero vertical movement.

mass m is moving downwards and to the horizonally to left relative to the floor, but at the same time it is sitting on wedge M so is also moving to the right the same as wedge M.

So mass m has a net sideways movement relative to the floor of two motions.

 

Does this help?

Hint think about what swansont said about coordinate systems.

 

so because mass M is moving to the right and mass m is on top of it, mass m has a net acceleration to the right in the x axis. 2 components ax(relative to M) to left and a(M) to right and it's resultant in x axis is: a(relative to ground)= -ax(relative to M)+a(M)?

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i can't understand to what you aim, the 2 objects move in x axis relative to the floor and mass m moves also in y axis.

the 2 objects, N*sin(aplha) is the resultant force that cause the acceleration in x axis but opposite direction so i get from there that -m(ax)=Ma(M)

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26 minutes ago, nir99 said:

i can't understand to what you aim, the 2 objects move in x axis relative to the floor and mass m moves also in y axis.

It may be that you haven't studied the concept of center of mass yet. But that's not necessary in order to solve the problem

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