# Example 3.6: Inertia forces in a four-bar linkage (graphical method)
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The four-bar linkage of {doc}`L03_14_example5` now moves with no external load, with the crank rotating at a constant $\omega_2 = 15$ rad/s counterclockwise. The crank and the coupler have negligible mass, and the rocker is a uniform bar of mass $m_4 = 3$ kg, with its center of mass $G_4$ at the midpoint of $O_4B$. At position $\theta_2 = 45^\circ$, compute graphically:
1. The inertia loads of the rocker and its equivalent inertia force.
2. The joint reactions and the required input torque $M_2$.
3. The total input torque if the force $\vec{P}$ of example 3.5 also acts.
Data: $\overline{O_2A} = 0.20$ m, $\overline{AB} = 0.50$ m, $\overline{O_4B} = 0.35$ m, $\overline{O_2O_4} = 0.45$ m. Neglect weight and friction.
:::{dropdown} 1. Kinematics and inertia loads
**Kinematics.** From the velocity and acceleration analysis (lesson 2, with polygons or analytically) at this position:
$$ \omega_3 = -2.649~\text{rad/s}, \quad \omega_4 = 5.397~\text{rad/s}, \quad \alpha_3 = 104.2~\text{rad/s}^2, \quad \alpha_4 = 154.7~\text{rad/s}^2 $$
(positive means counterclockwise). The acceleration of the center of mass of the rocker, with $\vec{r}_{O_4G_4} = \frac{1}{2}\vec{r}_{O_4B} = (0.0811, 0.1551)$ m, is:
$$ \vec{a}_{G_4} = \vec{\alpha}_4 \times \vec{r}_{O_4G_4} - \omega_4^2\, \vec{r}_{O_4G_4} = (-26.35, 8.02)~\text{m/s}^2 = 27.55~\text{m/s}^2 \angle 163.1^\circ $$
**Inertia loads.** The moment of inertia of the bar about its center is $I_{G_4} = \frac{1}{12} m_4 L_4^2 = \frac{1}{12} \cdot 3 \cdot 0.35^2 = 0.03063$ kg·m². Then:
$$
\begin{aligned}
\vec{J}_4 &= -m_4\, \vec{a}_{G_4} = (79.06, -24.05)~\text{N} = 82.64~\text{N} \angle {-16.9^\circ} \\
M_{G_4} &= -I_{G_4}\, \alpha_4 = -0.03063 \cdot 154.7 = -4.738~\text{N·m} \quad (\text{clockwise})
\end{aligned}
$$
**Equivalent inertia force.** We replace $\vec{J}_4$ and $M_{G_4}$ by a single force $\vec{J}'_4 = \vec{J}_4$, shifted parallel to itself a distance:
$$ h = \frac{|M_{G_4}|}{|\vec{J}_4|} = \frac{4.738}{82.64} = 0.0573~\text{m} = 57.3~\text{mm} $$
towards the side where its moment about $G_4$ is **clockwise**, like $M_{G_4}$. Since $\vec{J}_4$ points to the right and slightly downwards, it must be shifted **upwards**: its line of action goes through point $K = (0.5477, 0.2100)$ m.
:::
:::{dropdown} 2. Reactions and input torque
The problem is now the same as in example 3.5, with $\vec{J}'_4$ in place of $\vec{P}$:
**Coupler.** No mass and no loads: two-force member, $\vec{F}_{34}$ along $AB$.
**Rocker.** Three forces: $\vec{J}'_4$ (known), $\vec{F}_{34}$ (along $AB$) and $\vec{F}_{14}$ (through $O_4$). The line of $\vec{J}'_4$ and line $AB$ meet at $Q = (0.4313, 0.2454)$ m, and $\vec{F}_{14}$ acts along $O_4Q$. The force triangle gives:
$$ \vec{F}_{34} = 79.86~\text{N} \angle {-160.3^\circ}, \qquad \vec{F}_{14} = 51.16~\text{N} \angle 94.4^\circ $$
Now $\vec{F}_{34}$ points from $B$ towards $A$: the coupler pulls the rocker (**tension**).
**Crank.** $\vec{F}_{32} = -\vec{F}_{34} = 79.86~\text{N}\angle 19.7^\circ$ at $A$, with the same arm $d_2 = 0.0854$ m as before about $O_2$. Its moment is clockwise, so the input torque is counterclockwise:
$$ M_2 = 79.86 \cdot 0.0854 = 6.82~\text{N·m} \quad (\text{counterclockwise}) $$
The motor must supply this torque to accelerate the rocker at this position.
**Power check.** $M_2\,\omega_2 + \vec{J}_4\cdot\vec{v}_{G_4} + M_{G_4}\,\omega_4 = 0$, with $\vec{v}_{G_4} = \vec{\omega}_4 \times \vec{r}_{O_4G_4} = (-0.837, 0.438)$ m/s:
$$ M_2 = -\frac{79.06\cdot(-0.837) + (-24.05)\cdot 0.438 + (-4.738)\cdot 5.397}{15} = 6.82~\text{N·m} \;✓ $$
:::
:::{dropdown} 3. Superposition with the external load
If the force $\vec{P}$ of example 3.5 also acts, the problem is still linear (there is no friction), and by the **superposition principle** the reactions and the input torque are the sum of those of both cases:
$$ M_2 = \underbrace{-7.97}_{\text{static}} + \underbrace{6.82}_{\text{inertia}} = -1.15~\text{N·m} \quad (\text{clockwise}) $$
$$ \vec{F}_{34} = 93.4\angle 19.7^\circ + 79.86\angle{-160.3^\circ} = 13.5~\text{N}\angle 19.7^\circ $$
At this position, $\vec{P}$ acts in favor of the motion (on its own, it would force the motor to brake with $7.97$ N·m), but the rocker is accelerating ($\omega_4$ and $\alpha_4$ have the same sign) and absorbs almost all that power: the motor only has to brake with $1.15$ N·m. The reactions at $O_4$ are added vectorially in the same way: $\vec{F}_{14} = (12.10, -31.52) + (-3.89, 51.01) = (8.21, 19.49)$ N.
This is the **complete kinetostatic analysis**: static (loads) + dynamic (inertia).
:::
## Interactive construction
Go through the construction step by step and drag the crank to see how the inertia force, its offset $h$, the reactions and the input torque change over a revolution:
```{raw} html
```
:::{dropdown} 1. Kinematics and inertia loads
**Kinematics.** From the velocity and acceleration analysis (lesson 2, with polygons or analytically) at this position:
$$ \omega_3 = -2.649~\text{rad/s}, \quad \omega_4 = 5.397~\text{rad/s}, \quad \alpha_3 = 104.2~\text{rad/s}^2, \quad \alpha_4 = 154.7~\text{rad/s}^2 $$
(positive means counterclockwise). The acceleration of the center of mass of the rocker, with $\vec{r}_{O_4G_4} = \frac{1}{2}\vec{r}_{O_4B} = (0.0811, 0.1551)$ m, is:
$$ \vec{a}_{G_4} = \vec{\alpha}_4 \times \vec{r}_{O_4G_4} - \omega_4^2\, \vec{r}_{O_4G_4} = (-26.35, 8.02)~\text{m/s}^2 = 27.55~\text{m/s}^2 \angle 163.1^\circ $$
**Inertia loads.** The moment of inertia of the bar about its center is $I_{G_4} = \frac{1}{12} m_4 L_4^2 = \frac{1}{12} \cdot 3 \cdot 0.35^2 = 0.03063$ kg·m². Then:
$$
\begin{aligned}
\vec{J}_4 &= -m_4\, \vec{a}_{G_4} = (79.06, -24.05)~\text{N} = 82.64~\text{N} \angle {-16.9^\circ} \\
M_{G_4} &= -I_{G_4}\, \alpha_4 = -0.03063 \cdot 154.7 = -4.738~\text{N·m} \quad (\text{clockwise})
\end{aligned}
$$
**Equivalent inertia force.** We replace $\vec{J}_4$ and $M_{G_4}$ by a single force $\vec{J}'_4 = \vec{J}_4$, shifted parallel to itself a distance:
$$ h = \frac{|M_{G_4}|}{|\vec{J}_4|} = \frac{4.738}{82.64} = 0.0573~\text{m} = 57.3~\text{mm} $$
towards the side where its moment about $G_4$ is **clockwise**, like $M_{G_4}$. Since $\vec{J}_4$ points to the right and slightly downwards, it must be shifted **upwards**: its line of action goes through point $K = (0.5477, 0.2100)$ m.
:::
:::{dropdown} 2. Reactions and input torque
The problem is now the same as in example 3.5, with $\vec{J}'_4$ in place of $\vec{P}$:
**Coupler.** No mass and no loads: two-force member, $\vec{F}_{34}$ along $AB$.
**Rocker.** Three forces: $\vec{J}'_4$ (known), $\vec{F}_{34}$ (along $AB$) and $\vec{F}_{14}$ (through $O_4$). The line of $\vec{J}'_4$ and line $AB$ meet at $Q = (0.4313, 0.2454)$ m, and $\vec{F}_{14}$ acts along $O_4Q$. The force triangle gives:
$$ \vec{F}_{34} = 79.86~\text{N} \angle {-160.3^\circ}, \qquad \vec{F}_{14} = 51.16~\text{N} \angle 94.4^\circ $$
Now $\vec{F}_{34}$ points from $B$ towards $A$: the coupler pulls the rocker (**tension**).
**Crank.** $\vec{F}_{32} = -\vec{F}_{34} = 79.86~\text{N}\angle 19.7^\circ$ at $A$, with the same arm $d_2 = 0.0854$ m as before about $O_2$. Its moment is clockwise, so the input torque is counterclockwise:
$$ M_2 = 79.86 \cdot 0.0854 = 6.82~\text{N·m} \quad (\text{counterclockwise}) $$
The motor must supply this torque to accelerate the rocker at this position.
**Power check.** $M_2\,\omega_2 + \vec{J}_4\cdot\vec{v}_{G_4} + M_{G_4}\,\omega_4 = 0$, with $\vec{v}_{G_4} = \vec{\omega}_4 \times \vec{r}_{O_4G_4} = (-0.837, 0.438)$ m/s:
$$ M_2 = -\frac{79.06\cdot(-0.837) + (-24.05)\cdot 0.438 + (-4.738)\cdot 5.397}{15} = 6.82~\text{N·m} \;✓ $$
:::
:::{dropdown} 3. Superposition with the external load
If the force $\vec{P}$ of example 3.5 also acts, the problem is still linear (there is no friction), and by the **superposition principle** the reactions and the input torque are the sum of those of both cases:
$$ M_2 = \underbrace{-7.97}_{\text{static}} + \underbrace{6.82}_{\text{inertia}} = -1.15~\text{N·m} \quad (\text{clockwise}) $$
$$ \vec{F}_{34} = 93.4\angle 19.7^\circ + 79.86\angle{-160.3^\circ} = 13.5~\text{N}\angle 19.7^\circ $$
At this position, $\vec{P}$ acts in favor of the motion (on its own, it would force the motor to brake with $7.97$ N·m), but the rocker is accelerating ($\omega_4$ and $\alpha_4$ have the same sign) and absorbs almost all that power: the motor only has to brake with $1.15$ N·m. The reactions at $O_4$ are added vectorially in the same way: $\vec{F}_{14} = (12.10, -31.52) + (-3.89, 51.01) = (8.21, 19.49)$ N.
This is the **complete kinetostatic analysis**: static (loads) + dynamic (inertia).
:::
## Interactive construction
Go through the construction step by step and drag the crank to see how the inertia force, its offset $h$, the reactions and the input torque change over a revolution:
```{raw} html
```