# Obtaining mathematical model for the kinetic system

1. May 20, 2012

### ssulun

1. The problem statement, all variables and given/known data

See attachment.

f(t) is the input force and b1 and b2 are kinetic friction constants. There is no static friction.

2. Relevant equations

$$ƩF = m \ddot{x}$$
$$F_s=kx$$
$$F_f=b\dot{x}$$

Ff is the force from friction and Fs is the force from spring.

3. The attempt at a solution

$$m_1\ddot{x_1}=-b_1\dot{x_1}-k_1x_1-k_1x_2$$
$$m_2\ddot{x_2}=-b_2\dot{x_2}-k_1x_1-k_1x_2-k_2x_2$$
$$m_3\ddot{x_3}=f$$

I have two questions:

1) Should I include the friction between m1 and m3 and m2 and m3 in the equation for x3 and why?

2) When I imagine this system, I think that f(t) should definitely affect x1 and x2, but in my equations it doesn't. Where am I doing wrong?

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Last edited: May 20, 2012
2. May 20, 2012

### azizlwl

Let me try, hope somebody will correct me.
Taking springs as massless.

m1a=k1x1+b1m1g
m2a=k2x2+b2m2g-k1x1
(m1+m2+m3)a=f(t)

3. May 21, 2012

### ssulun

b1 and b2 are kinetic friction constants, and k1 spring is squeezed from both x1 and x2 so I can change those parts, but your work gave me new ideas, thank you.

4. May 21, 2012

### ssulun

I have modified it as:

$$m_1\ddot{x_1}=-b_1\dot{x_1}-k_1x_1-k_1x_2$$
$$m_2\ddot{x_2}=-b_2\dot{x_2}-k_1x_1-k_1x_2-k_2x_2$$
$$(m_1+m_2+m_3)\ddot{x_3}=f$$

But still, x1 and x2 don't depend on f and that bothers me.

5. May 21, 2012

### ssulun

I should have written the effect of k1 in the equation 1 as k1(x1+x2), not k1x1.

Also, the k2 spring will pull m3 with k2x2 (to balance the forces on the k2 spring). So should I write the equation for x3 as:
$$m_3\ddot{x_3}=f-k_2x_2$$
or should I write an overall system with (m1+m2+m3) and include the unbalanced forces from b1 and b2?

I am very confused.