2
$\begingroup$

I have a multibody system with Coulomb friction:

enter image description here

I'm able to write horizontal linear momentum conservation equations:

  1. $m_1 \ddot{x}_1 = F_1 - F_{f12}$
  2. $m_2 \ddot{x}_2 = F_2 + F_{f12} - F_{f23}$
  3. $m_3 \ddot{x}_3 = F_3 + F_{f23} - F_{f34}$
  4. $m_4 \ddot{x}_4 = F_4 + F_{f34}$

And I'm also able to calculate $F_{f12}$ and $F_{f34}$ as below:

  1. $F_{f12}=\left\{\begin{matrix} if \, \dot{x_1}=\dot{x_2} \, and \, \left| F_1-m_1 \ddot{x}_1 \right|<\mu_s F_{n12} \, then & F_1-m_1 \ddot{x}_1 \\ else & \mu_k F_{n12}sgn\left(\dot{x_1}-\dot{x_2}\right) \end{matrix}\right.$
  2. $F_{f34}=\left\{\begin{matrix} if \, \dot{x_3}=\dot{x_4} \, and \, \left| F_4+m_4 \ddot{x}_4 \right|<\mu_s F_{n34} \, then & -F_4-m_4 \ddot{x}_4 \\ else & \mu_k F_{n12}sgn\left(\dot{x_4}-\dot{x_3}\right) \end{matrix}\right.$

Here the static friction opposes any external force up to the maximum static friction.

I have two issues:

  1. when the conditions for static friction $\dot{x_i}=\dot{x_j} \, and \, \left| F_i-m_i \ddot{x}_i \right|<\mu_s F_{nij}$ are valid the equation becomes the exact same equation as the linear momenta, and I end up with less equations than my unknowns.

  2. I'm not able to write $F_{f23}$ because the opposing forces here include other static frictions and we end up with two different equations!

For example if we write $F_{f23}$ based on mass 3:

$F_{f23}=\left\{\begin{matrix} if \, \dot{x_2}=\dot{x_3} \, and \, \left| F_{f34} -F_3 -m_3 \ddot{x}_3\right|<\mu_s F_{n23} \, then & F_{f34} -F_3 -m_3 \ddot{x}_3 \\ else & \mu_k F_{n23}sgn\left(\dot{x_2}-\dot{x_3}\right) \end{matrix}\right.$

But if we write $F_{f23}$ based on mass 2 we get:

$F_{f23}=\left\{\begin{matrix} if \, \dot{x_2}=\dot{x_3} \, and \, \left| F_2 + F_{f12}-m_2 \ddot{x}_2\right|<\mu_s F_{n23} \, then & F_2 + F_{f12} -m_2 \ddot{x}_2 \\ else & \mu_k F_{n23}sgn\left(\dot{x_2}-\dot{x_3}\right) \end{matrix}\right.$

And the equality of $F_2 + F_{f12}=F_{f34}-F_3$ is not necessarily valid.

I would appreciate if you could help me know

  1. if my math is correct so far?
  2. how to write $F_{f23}$?
$\endgroup$
14
  • $\begingroup$ The true equations for the upper and the lower side must naturally be true at the same time. So if they disagree, then they aren't correct and it must be another one of the possible cases. $\endgroup$
    – Steeven
    Jul 23 '17 at 21:22
  • $\begingroup$ @Steeven do not follow, would you please elaborate? $\endgroup$
    – Foad
    Jul 23 '17 at 21:23
  • $\begingroup$ I don't see any immediate errors in the math, so assuming everything holds, you now have yourself a nice equality to check up with to know which situation you have. If the forces happen to fit, so that the final equality $$F_2+F_{f12}=F_{f34}-F_3$$is true, then you know that the first cases of both $F_{f23}$'s are true. If they do not, then the first cases can't both be true (only one of them or maybe none of them). $\endgroup$
    – Steeven
    Jul 23 '17 at 21:26
  • 1
    $\begingroup$ I don't think you set up the form of $F_{f12}$ or $F_{f34}$ correctly. The way they're currently set up, neither $m_1$ nor $m_4$ can accelerate at all unless they're sliding. But if $m_2$ is accelerating, then $m_1$ can accelerate at the same rate without having $F_{f12}=F_1$. $\endgroup$ Jul 23 '17 at 21:30
  • 1
    $\begingroup$ I Think I have solved the issue. would you guys please take a look at my answer and see if it is correct? $\endgroup$
    – Foad
    Jul 23 '17 at 22:27
0
$\begingroup$

New:

Since posting this question I was able to rewrite the equations in a more elegant way. Actually regardless of being a multibody problem friction between each individual bodies is the same and other frictional forces are just forces! Static friction tries to keep two bodies as one as far as it can. it means for a two body problem:

enter image description here

  1. $m_1 \ddot{x}_1 = F_1 - F_{f12}$
  2. $m_2 \ddot{x}_2 = F_2 + F_{f12}$
  3. $\ddot{x}_1=\ddot{x}_2$
  4. $\implies \, F_{s12}=\frac{m_2 F_1-m_1F_2}{m_1+m_2}$

considering that the maximum static friction is $F_{s_{max}12}=\mu_s F_{n12}$

$\implies F_{f12}=\left\{\begin{matrix} if \, \dot{x_1}=\dot{x_2} \, and \, \left| F_{s12} \right|<F_{s_{max}12} \,\, then & F_{s12} \\ else & \mu_k F_{n12}sgn\left(\dot{x_1}-\dot{x_2}\right) \end{matrix}\right.$

I have also implemented this in Modelica language and SIMULINK. you may see them here and here

Old:

I think I have figured the issue out. Thanks to the questions asked in the comments I had to rewrite some of the equations and then the main misunderstanding was revealed to me. Actually when the conditions for static frictions hold the two parts act as one. we can write the 3 frictions as below:

  1. If $\dot{x_1}=\dot{x_2}$ and $\left| F_1-m_1 \ddot{x}_1 \right|<\mu_s F_{n12}$ and $\left| m_2 \ddot{x}_2-F_2+F_{f23} \right|<\mu_s F_{n12}$ then $\ddot{x}_1=\ddot{x}_2$ , else $F_{f12}=\mu_k F_{n12}sgn\left(\dot{x_1}-\dot{x_2}\right)$

  2. If $\dot{x_3}=\dot{x_4}$ and $\left| F_4-m_4 \ddot{x}_4 \right|<\mu_s F_{n34}$ and $\left| F_3 + F_{f23}-m_3 \ddot{x}_3 \right|<\mu_s F_{n34}$ then $\ddot{x}_3=\ddot{x}_4$ , else $F_{f34}=\mu_k F_{n34}sgn\left(\dot{x_4}-\dot{x_3}\right)$

  3. If $\dot{x_2}=\dot{x_3}$ and $ \left| F_2 + F_{f12}-m_2 \ddot{x}_2\right|<\mu_s F_{n23} $ and $ \left| F_{f34} -F_3 -m_3 \ddot{x}_3\right|<\mu_s F_{n23} $ then $\ddot{x}_3=\ddot{x}_2$ , else $F_{f23}=\mu_k F_{n23}sgn\left(\dot{x_2}-\dot{x_3}\right)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.