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The material points is massed by springs, found the second law of Newton for every point.

I already do this for one point

https://i.stack.imgur.com/DFS1D.jpg

And get

$ \begin{cases} m\ddot{x}= -cx - u\cos{a} \\ m\ddot{y}=-cy - u \sin{a} \end{cases}$

Where $c$ - elasticity (same for all springs), $a$ - angle between force and OX ($m\ddot{x}$) and OY ($m\ddot{y}$)

But how to do this for

https://i.stack.imgur.com/Bf4fn.jpg

I try do this, and get (but this is wrong I think)

$ \begin{cases} m\ddot{x_{1}}= -cx + 2u\cos{a} \\ m\ddot{y_{1}}=-cy + 2u \sin{a} \\ m\ddot{x_{2}}= -cx + 2u\cos{a} \\ m\ddot{y_{2}}=-cy + 2u \sin{a} \\ m\ddot{x_{3}}= -cx + 2u\cos{a} \\ m\ddot{y_{3}}=-cy + 2u \sin{a} \\ m\ddot{x_{4}}= -cx + 2u\cos{a} \\ m\ddot{y_{4}}=-cy + 2u \sin{a} \end{cases}$

Stepan Vanzuriak
  • 15

1 Answers1

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For the first we have $u(t)=(x(t),y(t))$ thus

\begin{cases} m\ddot{x}= -cx \\ m\ddot{y}=-cy \end{cases}

For the second case we have $4\times2$ degrees of freedom $u_i(t)=(x_i(t),y_i(t))$ and we need to write down the equation of motion accordigly taking into account the relative displacement for the spring connecting two masses and also for the forces $F_i$.

For example for the mass $m_1$, assuming the spring constant "c" for the springs and an angle of 45° for the inclined spring, we have

$$\begin{cases} m_1\ddot{x_1}= -c(x_1-x_3)-c\frac{\sqrt2}2x_1+\frac{\sqrt2}2F_1 \\ m_1\ddot{y_1}= -c(y_1-y_4)-c\frac{\sqrt2}2y_1+\frac{\sqrt2}2F_1\end{cases}$$

for the mass $m_2$ we have

$$\begin{cases} m_2\ddot{x_2}= -c(x_2-x_4)-cx_2+\frac{\sqrt2}2F_2 \\ m_2\ddot{y_2}= -c(y_2-y_3)-cy_2-\frac{\sqrt2}2F_2\end{cases}$$

user
  • 154,566
  • Show example for $x_{2}$ please – Stepan Vanzuriak May 15 '18 at 14:16
  • @StepanVanzuriak I've added also the equations for mass $m_2$ and corrected some typo. Try to understand the idea behind and how to obtain the equation. You need to consider for the masses 4 displacement in x direction (right direction) and 4 displacement in y direction (up direction). Then write down the Newton's equation accordingly. – user May 15 '18 at 14:23
  • In $m_1\ddot{y_1}= -c(y_1-y_3)-c\frac{\sqrt2}2y_1+\frac{\sqrt2}2F_1$, is there must be $y_3$ or $y_4$? – Stepan Vanzuriak May 15 '18 at 14:32
  • @StepanVanzuriak Yes it is a typo of course, in vertical direction $m_1$ is affected only by $y_4$. I fix. – user May 15 '18 at 14:48