Calculating Acceleration on an Inclined Plane with Friction | Two Block System

[Arman777]

Homework Statement

Two blocks,in contact,slide down an inclined plane $AC$ of inclination $30^°$,The coefficient of kinetic friction between the $2.0 kg$ block and the incline is $μ_1=0.20$ and that between the $4.0kg$ block and the incline is $μ_2=0.30$.

Find the magnitude of the acceleration.

Homework Equations

$\vec {F_t }=m\vec a$
$F_f=μN$
$w=mg$

The Attempt at a Solution

Ok I need to draw free body diagram,after that I can solve the question I guess.I stucked in the free body diagram,I don't know its true or not.there's pic in the attached files.

Is there will be any action-reaction force pairs ?

 

[Doc Al]

If you draw individual free body diagrams, don't forget the forces the blocks exert on each other. (Those would be 3rd law pairs.)

But why not treat the two blocks as a single "system"?

 

[Mastermind01]

Is there will be any action-reaction force pairs ?

When will there be action-reaction pairs? What do you think?

 

[Arman777]

If you draw individual free body diagrams, don't forget the forces the blocks exert on each other. (Those would be 3rd law pairs.)

But why not treat the two blocks as a single "system"?

They have differenet friction constant and $N$ are different This 2 body diagrams seem simpler ( maybe not but for me yes )

When will there be action-reaction pairs? What do you think?

If an object exerts a force on something.That something should exert a force on that object

Well,If $2.0 kg$ object exerts a force on $4.0kg$ object. which its I guess, cause there's $mgsin30$ (component of $2.0kg$ block) exerts a force on 4 kg block so there will be action-reaction force pair ? Is it true ? the magnitude of it ? Or the idea ?

 

[Doc Al]

They have different friction constant and N are different I guess.

Sure. So what? (You can include the two friction forces acting on the blocks.)

This 2 body diagrams seem simpler ( maybe not but for me yes )

Using separate diagrams is perfectly fine! But then you must include the unknown contact force that the blocks exert on each other.

 

[Arman777]

Sure. So what? (You can include the two friction forces acting on the blocks.)Using separate diagrams is perfectly fine! But then you must include the unknown contact force that the blocks exert on each other.

I don't know how to think them as a single object..

I want to write down my solution ( I ll add in a minute )

 

[Doc Al]

I don't know how to think them as a single object..

Since they move together, it's easy to treat them as a single system. But you don't have to.

(Think about it: If a single block were sliding down the incline you could -- mentally -- imagine the block divided into two pieces and treat each piece separately. You'll get the same answer as usual, though.)

 

[Arman777]

Parallel to inclined plane is x and and the perpendicular to that axis is y.Down is +x and up is +y

$m_2a=m_2gsin(30^°)-μ_2m_2gcos(30^°)+F_{12}$
$m_1a=m_1gsin(30^°)-μ_1m_2gcos(30^°)-F_{21}$

Since
$F_{12}=-F_{21}$

lets subtract them.

$a(m_2-m_1)=gsin(30^°)(m_2-m_1)+gcos(30^°)(m_1μ_!-μ_2m_2)$

 

[Arman777]

Since they move together, it's easy to treat them as a single system. But you don't have to.

(Think about it: If a single block were sliding down the incline you could -- mentally -- imagine the block divided into two pieces and treat each piece separately. You'll get the same answer as usual, though.)

I guess I understand

 

[Doc Al]

Parallel to inclined plane is x and and the perpendicular to that axis is y.Down is +x and up is +y

$m_2a=m_2gsin(30^°)-μ_2m_2gcos(30^°)+F_{12}$
$m_1a=m_1gsin(30^°)-μ_1m_2gcos(30^°)-F_{21}$

Since
$F_{12}=-F_{21}$

lets subtract them.

$a(m_2-m_1)=gsin(30^°)(m_2-m_1)+gcos(30^°)(m_1μ_!-μ_2m_2)$

Careful with signs.

Let's let $F_{12}$ stand for the magnitude of the contact force. In the first equation, it will appear as $+ F_{12}$; in the second, as $- F_{12}$.

 

[Arman777]

Careful with signs.

Let's let $F_{12}$ stand for the magnitude of the contact force. In the first equation, it will appear as $+ F_{12}$; in the second, as $- F_{12}$.

But $F_{12}$ in the +x direction and $F_{21}$ in the -x direction. that's why there's "-" sign

 

[Mastermind01]

Since they move together, it's easy to treat them as a single system. But you don't have to.

This approach though only works for this question. I think we should check first whether they do move together?

 

[Arman777]

This approach though only works for this question. I think we should check first whether they do move together?

I thought that too but the question asks one acceleration so they must have same acceleration I guess.

 

[Doc Al]

But $F_{12}$ in the +x direction and $F_{21}$ in the -x direction. that's why there's "-" sign

You are counting the sign twice, which is why your final equation is incorrect.

What you meant to do was use $+ F_{12}$ in the first equation and $+ F_{21}$ in the second equation. Then, using $F_{12} = - F_{21}$ would work out fine.

 

[Mastermind01]

I thought that too but the question asks one acceleration so they must have same acceleration I guess.

What you're doing now (along with Doc Al's correction) is correct. For a different question, I think it's better to check first.

 

[Arman777]

I am writing $\vec F_{21}=F_{21} (-i)$ but actually it is $\vec F_{21}=F_{12} (-i)$ my mistake I couldn't notice this.I solved and answer is correct thanks

 

[Arman777]

So is it always $\vec F_{21}=F_{12} (-i)$ like this right but always (of course if $\vec F_{12}=F_{12} (i)$

 

[Doc Al]

I would say it this way: $\vec{F}_{12} = - \vec{F}_{21}$. The directions are opposite but the magnitudes are equal.

And note that when you combine the two equations correctly, you'll get an equation for the two-block system. (The masses will add.)

 

[Arman777]

I would say it this way: $\vec{F}_{12} = - \vec{F}_{21}$. The directions are opposite but the magnitudes are equal.

And note that when you combine the two equations correctly, you'll get an equation for the two-block system. (The masses will add.)

Newtons Third Law yeah...

I noticed two-block system equation when you say right know.Thanks again

Problem solved.I have one more question and I ll open another thread soon If you also help me in that I ll happy