Free-body diagram of mechanical system

[PainterGuy]

Hi,

Could you please help me with the queries below?

Question 1:
Do you think that the free-body diagram for M1 correct? It looks fine except that, I think, the arrow for $M_{1}\frac{d^{2}}{dt^{2}}x_{1}$ should point downward.

Question 2:
Similarly, I think that the free-body diagram for M2 is okay but the arrow for $M_{2}\frac{d^{2}}{dt^{2}}x_{2}$ should point downward instead. Could you please confirm it?

 

[A.T.]

I think, the arrow for $M_{1}\frac{d^{2}}{dt^{2}}x_{1}$ should point downward.

The arrow directions just indicate the convention for the positive direction. It doesn't matter how you draw them.

 

[PainterGuy]

Thank you!

The arrow directions just indicate the convention for the positive direction. It doesn't matter how you draw them.

"positive direction" for what?

Also you said that it doesn't matter but, in my opinion, it does matter how the arrows are drawn. Please see the diagram for M1 where I've drawn the F=ma arrow downward, and as a result the equation also changes.

I have another related question and came across it when I was trying to understand the involved equations.

-kx - b*v_x = m*a_x

To me the equation above means that at any time, the force accelerate an object in x direction is equal and opposite to the spring force and damping force; in other words, the net force is zero. But then why would the object move at all if some forces are pulling the object backward and the other force pushing it forward? It should remain stationary. Where am I going wrong? Thank you!

 

[A.T.]

"positive direction" for what?

The arrows indicate the sign convention to be used for the forces. If you get a positive value for the force, it means that the force points in the direction that the arrow indicates. If you get a negative value, the force points opposite to the arrow.

I have another related question and came across it when I was trying to understand the involved equations.

-kx - b*v_x = m*a_x

To me the equation above means that at any time, the force accelerate an object in x direction is equal and opposite to the spring force and damping force; in other words, the net force is zero.

The net force is the right hand side. Sometimes you introduce a fictitious force to cancel the net force, because you are not interested in the acceleration, just the internal stresses (quasi static analysis).