Differential equations of a (simplified) loading bridge

Question

I am having trouble to calculate the differential equations of a simplified loading bridge.

The system is build as shown in the picture below (just a sketch):

If I use the Newton approach, I am getting the following equations by neglecting friction, air resistance and changes in the length of rope:

$$ m_k \ddot{x}_{k} = F_{A} + F_{S} \sin(\varphi) \\ m_G \ddot{x}_{G} = -F_{S} \sin(\varphi) \\ m_G \ddot{z}_{G} = m_{G} g - F_{S} \cos(\varphi) $$

When I look at the kinematic relationships from the gripper (the circle with the weight $m_G$) I get the following equations.

$$ x_{G} = x_{k} + l \sin(\varphi) \\ z_{G} = l \cos(\varphi)\\ \varphi = \omega t = \dot{\varphi} t $$

I know the weights $m_k$ and $m_G$ and the length $l$ but the values are not important right now.

The goal is to have two differential equations at the end. One equation shall show the relationship between the driving force $F_A$ and the path of the trolley $x_k$ (with derivations) The other equation shall show the relationship between driving force $F_A$ and angle of the rope $\varphi_G$.

After that I want to make the transfer functions (Laplace transformation etc.) but that is not the problem.

The problem is that I can not seem to find those equations. My best approach so far looks like this:

$$ m_{k} \ddot{x}_{k} = F_{A} + F_{S} \sin(\varphi) $$

So that means if

$$ m_G \ddot{x}_{G} = -F_{S} \sin(\varphi) \\ F_{S} \sin(\varphi) = -m_{G} \ddot{x}_{G} \\ $$

I can say:

$$ m_{k} \ddot{x}_{k} = F_{A} - m_{G} \ddot{x}_{G} \\ $$

and if I derive $x_{G}$ like this:

$$ x_{G} = x_{k} + l \sin(\varphi) \\ \dot{x}_{G} = \dot{x}_{k} + l \dot{\varphi} \cos(\varphi) \\ \ddot{x}_{G} = \ddot{x}_{k} + l \left[ \ddot{\varphi} \cos(\varphi) - \dot{\varphi}^{2} \sin(\varphi) \right] $$

I am actually getting stuck here because I can not find a way to eliminate $\varphi$ from the equations. The addition theorems are not helping me at all (or I'm using them correctly).

Does anyone have an idea of how I should continue at this point? I hope I don't need a complete solution. I am actually more interested at doing this myself and hope to get a push towards the right direction.

Answer 1

Kinematics and dynamics

Those are the steps to solve problems of this nature.

1. Analize the kinematics of the system.

$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $_{o}\vec{r}_{RP}$

$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $R(\varphi) _{B}\vec{r}_{RP}$

$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $\big(x_{k}î + 0j + 0k \big)$ + $\big(\sin(\varphi)l î + 0j + \cos(\varphi)lk\big)$

$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $\big[\big(x_{k}+\sin(\varphi)l\big)î + 0j + \big(\cos(\varphi)l\big)k\big]$

note: $R(\varphi)$ is a rotation matrix and $x_{G} = x_{k}+\sin(\varphi)l$.

Taking the time derivatives:

$\hspace{5.em}$ $\dot{x_{G}}$ = $\dot{x_{k}}+\cos(\varphi)\dot{\varphi}l$

$\hspace{5.em}$ $\ddot{x_{G}}$ = $\ddot{x_{k}} + l\cos(\varphi)\ddot{\varphi} - l\sin(\varphi)\dot{\varphi}^{2}$

2. Use Newton's equation:

$\hspace{5.em}$ $m_{k}\ddot{x_{k}} = F_{A} - m_{G}\ddot{x_{G}}$

Substitute $x_{G}$:

$\hspace{5.em}$ $m_{k}\ddot{x_{k}} = F_{A} - m_{G}\big(\ddot{x_{k}} + l\cos(\varphi)\ddot{\varphi} - l\sin(\varphi)\dot{\varphi}^{2}\big)$

$\hspace{5.em}$ $\big(m_{k}+m_{G}\big)\ddot{x_{k}} + m_{G}\big(l\cos(\varphi)\ddot{\varphi}\big) - m_{G}\big(l\sin(\varphi)\dot{\varphi}^{2}\big)= F_{A}$

For the z axis:

$\hspace{5.em}$ $F_{Z}$ = $m_{G}g-l\big(\cos(\varphi)\dot{\varphi}^{2}+\sin(\varphi)\ddot{\varphi}\big)$

3. Use Newton's second law for rotation:

$\hspace{5.em}$ $I\ddot{\varphi}$ = $F_{Z}l\sin(\varphi)-\big(m_{G}\ddot{x_{G}}\big)l\cos(\varphi)$

$F_{Z}l\sin(\varphi) = m_{G}gl\sin(\varphi)-l^{2}\big(\cos(\varphi)\sin(\varphi)\dot{\varphi}^{2}+\sin(\varphi)^{2}\ddot{\varphi}\big)$

$\big(m_{G}\ddot{x_{G}}\big)l\cos(\varphi) = m_{G}\big(l^{2}\cos(\varphi)^{2}\ddot{\varphi}\big) - m_{G}\big(l^{2}\cos(\varphi)\sin(\varphi)\dot{\varphi}^{2}\big)+m_{G}\ddot{x_{K}}l\cos(\varphi)$

Using trigonometry identities:

$\hspace{5.em}$ $\big(I+m_{G}l^{2}\big)\ddot{\varphi}$ = $m_{G}gl\sin(\varphi)-m_{k}l\cos(\varphi)\ddot{x_{k}}$

4. Done! Now you can rest... $\ddot\smile$

Answer 2

My guess is that you probably need another differential equation for the angular movement, that will involve the inertia, such as:

$$m_G l^2 \ddot{\varphi} = m_G g l \sin(\varphi)$$

which yields:

$$ \ddot{\varphi} = \frac{g}{l} \sin(\varphi)$$

You can then maybe use the small angles approximation:

$$ \sin(\varphi) \simeq \varphi$$

Check out the inverted pendulum example.