User:EGM4313.s12.team4.jones/R1.3

=Problem 3=

Problem Statement
For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2) p.1-4.

Solution
FBD: Force balance equation of spring-dashpot-mass system:

$$\sum F_x=f(t)-f_I=ma$$ where $$a={y}''\,\!$$

Rearranging:
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$$m{y}''+f_I=f(t)\,\!$$ Internal force:
 * $$\displaystyle (Eq. 2) $$
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$$f_I=f_k=f_c\,\!$$
 * $$\displaystyle (Eq. 3) $$
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Force relationships:
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$$f_k=ky_k\,\!$$
 * $$\displaystyle (Eq. 4) $$
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$$f_c={y_c}''\,\!$$ where:
 * $$\displaystyle (Eq. 5) $$
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c=dashpot constant

k=spring constant

Displacement:
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$$y=y_k+y_c\,\!$$ -total displacement is displacement from the spring + displacement from the dashpot
 * $$\displaystyle (Eq. 1) $$
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Taking the double integral of $$y$$ you get:
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$${y}={y_k}+{y_c}''\,\!$$
 * $$\displaystyle (Eq. 6) $$
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Get $$y_c$$ in terms of $$y_k$$: $$f_k=f_c\,\!$$

$$ky_k=c{y_c}'\,\!$$

$${y_c}''=\frac{k}{c}y_k\,\!$$

Plug back into $${y}''\,\!$$: $${y}={y_k}+{({y_c}')}'\,\!$$

$${({y_c}')}'=\frac{k}{c}{y_k}'\,\!$$

$$\rightarrow {y}={y_k}+\frac{k}{c}{y_k}'\,\!$$

Now get $$f_I$$ in terms of $$y_k$$: $$f_I=f_k\,\!$$

$$f_k=ky_k\,\!$$

$$\rightarrow f_I=ky_k\,\!$$

When plugging in all variables into original equation you get:

$$m({y_k}''+\frac{k}{c}{y_k}')+ky_k=f(t)\,\!$$