Modelling a second order system

Chickentown

Hi,

I have attached an image of the question/problem I am trying to solve and my attempt at deriving the differential equation that describes the motion of mass m1.

I was wondering if any of you guys could take a look at it any see if it is right or offer some guidance on answering these type of questions.

Thank you

Michael Collins

The system of equations you need comes from

F = m*a i.e. net sum of forces = mass * acceleration

where the net above means you need to take account of direction.

First find the forces, taking right as positive, left as negative. Looking at your free body diagrams, we get

[latex]
\eqalign{
&F_1 = + F_1 \hbox{ (the input force)} \cr
&F_2 = - k x_1 \hbox{ (the sping force)} \cr
&F_3 = - B\dot{x_1} \hbox{ (the damping force from } m_1) \cr
&F_4 = + B\dot{x_2} \hbox{ (the damping force from } m_2)\cr
}
[/latex]

The net sum of forces, is

[latex] \sum F = +F_1 - k x_1 - B\dot{x_1} + B\dot{x_2} [/latex]

which must equal mass times acceleration, but the acceleration of [latex] m_1 [/latex], is just [latex] \ddot{x_1} [/latex], so we get

[latex] F_1 - k x_1 - B\dot{x_1} + B\dot{x_2} = m_1\ddot{x_1}[/latex],

now bring terms in x to one side:

[latex] m_1\ddot{x_1} + B(\dot{x_1} - \dot{x_2}) + k x_1= F_1[/latex].

So the first term is the mass times acceleration, and yes the signs of the terms do matter.

Chickentown

Michael Collins wrote: »

[latex]
\eqalign{
&F_3 = - B\dot{x_1} \hbox{ (the damping force from } m_1) \cr
&F_4 = + B\dot{x_2} \hbox{ (the damping force from } m_2)\cr
}
[/latex]


[latex] m_1\ddot{x_1} + B(\dot{x_1} - \dot{x_2}) + k x_1= F_1[/latex].

Thanks for your help, I understand everything except the signs of the damper, If I apply a force to m1 in the positive direction (left to right) why does it become [latex] B(\dot{x_1} - \dot{x_2})[/latex], to phrase my question better, Why not: [latex] B(\dot{x_2} - \dot{x_1})[/latex]?

Michael Collins

Chickentown wrote: »

Thanks for your help, I understand everything except the signs of the damper, If I apply a force to m1 in the positive direction (left to right) why does it become [latex] B(\dot{x_1} - \dot{x_2})[/latex], to phrase my question better, Why not: [latex] B(\dot{x_2} - \dot{x_1})[/latex]?

The short answer: because we've taken this force to the other side of the equation.

The longer answer:

You apply a force in the positive x direction (this is F1), but the 1st damper acts in the opposite direction to that force (this is what dampers do), so in this case the damping force acts in the negative x direction, so we give it a minus sign:

[latex] -B\dot{x}_1 [/latex]

The other damping force is due to M2 moving, which attempts to pull M1 along with it due to the force M2 exerts on the 2nd damper, in this case the force is to the right, just like you had in your 2nd diagram for M1:

[latex] +B\dot{x}_2. [/latex]

So the net force, to the right, is

[latex] F = -B\dot{x}_1 +B \dot{x}_2 = B(\dot{x}_2 - \dot{x}_1). [/latex]

In the final equation we add up all the forces on the right-hand-side

[latex] \sum F = B(\dot{x}_2 - \dot{x}_1) +F_1 - k x_1 [/latex]

this must equal mass * acceleration [latex] = m_1 \ddot{x_1} [/latex], so

[latex] B(\dot{x}_2 - \dot{x}_1) +F_1 - k x_1 = m_1 \ddot{x_1}.[/latex]

We could leave it like this, but usually we like to get all the unknown terms (terms with x in them) to one side, so when we bring everything on the left over, (except F_1, since there's no x in it), we have

[latex] m_1 \ddot{x_1} + B(\dot{x}_1 - \dot{x}_2) + k x_1 = F_1. [/latex]

Chickentown

Michael Collins wrote: »

The short answer: because we've taken this force to the other side of the equation.

The longer answer:

You apply a force in the positive x direction (this is F1), but the 1st damper acts in the opposite direction to that force (this is what dampers do), so in this case the damping force acts in the negative x direction, so we give it a minus sign:

[latex] -B\dot{x}_1 [/latex]

The other damping force is due to M2 moving, which attempts to pull M1 along with it due to the force M2 exerts on the 2nd damper, in this case the force is to the right, just like you had in your 2nd diagram for M1:

[latex] +B\dot{x}_2. [/latex]

So the net force, to the right, is

[latex] F = -B\dot{x}_1 +B \dot{x}_2 = B(\dot{x}_2 - \dot{x}_1). [/latex]

In the final equation we add up all the forces on the right-hand-side

[latex] \sum F = B(\dot{x}_2 - \dot{x}_1) +F_1 - k x_1 [/latex]

this must equal mass * acceleration [latex] = m_1 \ddot{x_1} [/latex], so

[latex] B(\dot{x}_2 - \dot{x}_1) +F_1 - k x_1 = m_1 \ddot{x_1}.[/latex]

We could leave it like this, but usually we like to get all the unknown terms (terms with x in them) to one side, so when we bring everything on the left over, (except F_1, since there's no x in it), we have

[latex] m_1 \ddot{x_1} + B(\dot{x}_1 - \dot{x}_2) + k x_1 = F_1. [/latex]

Michael you have saved my bacon in the nick of time, thanks so much, that could not have been explained any better.