Newton's Second Law of Motion -- Three masses, an inclined plane and a pulley

[mustafamistik]

Homework Statement

It's not Homework.

Relevant Equations

F=m*a

I did part a and part b but stuck in part c. Could you help me? (Part a and b attached.)

 

[haruspex]

Homework Statement:: It's not Homework.
Relevant Equations:: F=m*a

I did part a and part b but stuck in part c. Could you help me?

Please post your answers to parts a and b.

 

[mustafamistik]

Please post your answers to parts a and b.

I edited. You can help even if you tell me a solution way. Cause I don't even know what to do .

 

[haruspex]

I edited. You can help even if you tell me a solution way. Cause I don't even know what to do .

You are missing a force on m₁.

 

[mustafamistik]

You are missing a force on m₁.

Is it Friction force ?

 

[haruspex]

Is it Friction force ?

Yes.

 

[mustafamistik]

Yes.

Thanks. Could you give a hint for part c?

 

[PeroK]

Thanks. Could you give a hint for part c?

Think about the maximum acceleration that is possible with the given coefficient of friction.

 

[mustafamistik]

Think about the maximum acceleration that is possible with the given coefficient of friction.

friction-m*g*sin(16)=m*a
Is this equation true ?

 

[PeroK]

friction-m*g*sin(16)=m*a
Is this equation true ?

It's hard to make sense of that out of context. Please describe what you are trying to say with that equation.

 

[mustafamistik]

It's hard to make sense of that out of context. Please describe what you are trying to say with that equation.

This equation to find maximum acceleration. (m2*g*cos(16)* μ) -(m2*g*sin(16))=m2*a (F=ma)

 

[PeroK]

This equation to find maximum acceleration. (m2*g*cos(16)* μ) -(m2*g*sin(16))=m2*a (F=ma)

Okay, and the mass cancels out, of course.

What is the relationship between the large mass, $M$, and the acceleration of $m_2$?

 

[mustafamistik]

Okay, and the mass cancels out, of course.

What is the relationship between the large mass, $M$, and the acceleration of $m_2$?

I think,
$T=M*g$
$M*g - (m_1+m_2)*sin(16)=(m_1+m_2)*a$
The acceleration above is the acceleration of the $m_1 and m_2$ since, acceleration of $m_1 and m_2$ must be equal.
But maximum acceleration is,
(m_2*cos(16)* μ )-(m_2*sin(16) )=m_2 * a_max
a_max=cos(16)* μ -sin(16)
Am i right?

 

[haruspex]

$T=M*g$

How are you getting that from Newton's Laws? What are you wrongly assuming?

 

[mustafamistik]

How are you getting that from Newton's Laws? What are you wrongly assuming?

Is it wrong? I don't understand.

 

[haruspex]

Is it wrong? I don't understand.

Answer my question: which of Newton's Laws did you use to write the equation?

 

[mustafamistik]

Answer my question: which of Newton's Laws did you use to write the equation?

I used Newton's Second Law.

 

[haruspex]

I used Newton's Second Law.

Which says..?

 

[mustafamistik]

Which says..?

F=ma

 

[haruspex]

F=ma

And what is a for the mass M?

 

[mustafamistik]

And what is a for the mass M?

g? I get it its not. We should consider all the system. Am i right ?

 

[haruspex]

g? I get it its not

That would be free fall.
What is the relationship between M's acceleration and the acceleration of the masses on the slope?

 

[mustafamistik]

That would be free fall.
What is the relationship between M's acceleration and the acceleration of the masses on the slope?

They are same.

 

[haruspex]

They are same.

Right. And what is the net force on M?

 

[mustafamistik]

Right. And what is the net force on M?

(M*g)-((m_1+m_2)*g*sin(16)) ?

 

[haruspex]

(M*g)-((m_1+m_2)*g*sin(16)) ?

Only consider the forces that act directly on M. What are they?

 

[mustafamistik]

Only consider the forces that act directly on M. What are they?

Gravitational force and T

 

[haruspex]

Gravitational force and T

Right, so write out the ΣF=ma equation for the mass, putting the sum of those forces on the left and its mass times acceleration on the right. Careful with signs.

 

[mustafamistik]

Right, so write out the ΣF=ma equation for the mass, putting the sum of those forces on the left and its mass times acceleration on the right. Careful with signs.

$(M*g)+(m_1+m_2)*g*sin(16) =M*a$

 

[haruspex]

$(M*g)+(m_1+m_2)*g*sin(16) =M*a$

As you correctly stated in post #28, the forces that act on M are gravity and the tension. Those are the only forces that should appear in the equation. Yes, it may turn out that the tension is equal to $(m_1+m_2)*g*\sin(16)$, but take it one step at a time.
And think about the way each force acts on M.

 

[mustafamistik]

As you correctly stated in post #28, the forces that act on M are gravity and the tension. Those are the only forces that should appear in the equation. Yes, it may turn out that the tension is equal to $(m_1+m_2)*g*\sin(16)$, but take it one step at a time.
And think about the way each force acts on M.

T= $(m_1+m_2)*g*\sin(16)$
Gravity = M*g
Tension and Gravity are opposite forces. So we should consider the Gravity negative?

 

[PeroK]

T= $(m_1+m_2)*g*\sin(16)$
Gravity = M*g
Tension and Gravity are opposite forces. So we should consider the Gravity negative?

It doesn't matter which direction you take as positive as long as you are consistent. The important point is that tension and gravity are acting in opposite directions

 

[mustafamistik]

It doesn't matter which direction you take as positive as long as you are consistent. The important point is that tension and gravity are acting in opposite directions

I get it thanks sir. But i don't know how to calculate max M still.

 

[PeroK]

I get it thanks sir. But i don't know how to calculate max M still.

Let's take a step back. If there is no slipping between $m_1$ and $m_2$, then we have a simple pulley system with a mass $M$ pulling a mass $m_1 + m_2$ up a slope.

Can you calculate the acceleration of the system?

 

[mustafamistik]

Using $F=m*a$;
$M*g-(m_1+m_2)*sin(16)*g/(M+m_1+m_2)=a$

 

[PeroK]

Using $F=m*a$;
$M*g-(m_1+m_2)*sin(16)*g/(M+m_1+m_2)=a$

You might need to tidy that up a bit. And check you aren't missing some brackets.

What does that tell you about the acceleration of $m_1$?

 

[mustafamistik]

You might need to tidy that up a bit. And check you aren't missing some brackets.

What does that tell you about the acceleration of $m_1$?

You are right.
$(M*g-(m_1+m_2)*sin(16)*g)/(M+m_1+m_2)=a$
=> $(g*(M-(m_1+m_2)*sin(16)))$
a is acceleration of the all system.

 

[PeroK]

You are right.
$(M*g-(m_1+m_2)*sin(16)*g)/(M+m_1+m_2)=a$
=> $(g*(M-(m_1+m_2)*sin(16)))$
a is acceleration of the all system.

And there's an upper limit on $a$, correct? Depending on $\mu$?

 

[mustafamistik]

And there's an upper limit on $a$, correct? Depending on $\mu$?

This is the point where I stuck.

 

[PeroK]

This is the point where I stuck.

I don't know what you mean. You did that already:

This equation to find maximum acceleration. (m2*g*cos(16)* μ) -(m2*g*sin(16))=m2*a (F=ma)

Maybe it was so long ago!

 

[mustafamistik]

I don't know what you mean. You did that already:
Maybe it was so long ago!

Is that equation true ?

 

[PeroK]

Is that equation true ?

I think so. You just have to put everything together now.

 

[mustafamistik]

I think so. You just have to put everything together now.

Thank you and @haruspex for helping solve and understand this question. Also thanks for your interest.