Understanding Forces: Explaining maa and mbb in a Simple Example"

[pluspolaritons]

Hi,

Please take a look at this figure:
Screenshot by Lightshot

There is a force that is applied so that block B will not fall under the influence of gravity.

I have set up the problem as:
F= (m_a+m_b)a

μR=m_bg

where μ is the coefficient of friction and R is the reaction force.

Now my question is I can't figure out why m_ba is the reaction force.

What exactly is m_aa and m_bb? Shouldn't m_aa be the force of block a on b? Then what about m_bb? What is this force? The force direction should be pointing to the right side (since the applied force is to the right)? So it is as if m_bb is acting on thin air?

I can't wrap my head around what is m_aa and m_bb, please help.

Thanks.

 

[Andrew Mason]

Hi,

Please take a look at this figure:
Screenshot by Lightshot

There is a force that is applied so that block B will not fall under the influence of gravity.

I have set up the problem as:
F= (m_a+m_b)a

μR=m_bg

where μ is the coefficient of friction and R is the reaction force.

Now my question is I can't figure out why m_ba is the reaction force.

The 3rd law pair to the force of A on B is the equal and opposite force of B on A. But you do not really need to use the third law. This is a second law problem.

The force of A on B, F_A→B = m_ba, which is normal to the surfaces between A and B. The maximum static friction force between B and A is the force normal to the surface between B and A multiplied by the co-efficient of static friction (μ_s).

F_fmax = μ_sF_A→B = μ_sm_ba

In order for B not to fall, how must this maximum static friction force be in relation to m_bg ?

AM

 

[pluspolaritons]

Thanks Andrew. I know the solution to this problem, but I need to know what is m_aa and m_ba. Why is the force of A on B equals to m_ba instead of m_aa? Then what is m_aa?

 

[Andrew Mason]

Thanks Andrew. I know the solution to this problem, but I need to know what is m_aa and m_ba. Why is the force of A on B equals to m_ba instead of m_aa? Then what is m_aa?

Draw a freebody diagram and apply the second law.

F = m_a+ba = m_aa + m_ba

so: m_aa = F - m_ba

Examining the forces on A: there is F pushing A forward and the force from B on A (= -m_ba) which is in the opposite direction to F. Together they must sum to the net force on A which is necessarily m_aa (second law).

For B, there is only the force of A on B. This must be equal to m_ba (second law).

AM

 

[pluspolaritons]

Draw a freebody diagram and apply the second law.

F = m_a+ba = m_aa + m_ba

so: m_aa = F - m_ba

Examining the forces on A: there is F pushing A forward and the force from B on A (= -m_ba) which is in the opposite direction to F. Together they must sum to the net force on A which is necessarily m_aa (second law).

For B, there is only the force of A on B. This must be equal to m_ba (second law).

AM

Thanks, I think it's clearer now.

So, F only acts on A but not B right? This is because F is in direct contact with A but not B, correct? But since B is in direct contact with A, it will feel the force of A on B.

 

[Andrew Mason]

Thanks, I think it's clearer now.

So, F only acts on A but not B right? This is because F is in direct contact with A but not B, correct? But since B is in direct contact with A, it will feel the force of A on B.

You can divide the bodies any way you like. The forces that apply to all parts of A and B all originate with F: eg. without F, A could apply no force to B.

If you want to make a distinction between A and B, then you can say that F acts on A and A acts on B.

You could divide A into two parts and say that the left half of A applies a force to the right half of A so that the force applied by the left half of A causes the right half to accelerate.

Or you could make no distinction between bodies or parts of bodies treating A+B as a single mass and simply say that F applies a force to a mass A+B.

AM