# Example 3.5: Graphical static analysis of a four-bar linkage

A horizontal force $\vec{P}$ of $100$ N, pointing to the left, acts at point $C$ of the rocker of the four-bar linkage in the figure. Compute graphically the reactions at the four joints and the torque $M_2$ to be applied to the crank to keep the mechanism in equilibrium at this position. The masses of the links and friction are neglected.

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 (horizontal), $\overline{O_4C} = 0.25$ m, $\theta_2 = 45^\circ$.

<img src="figs/L03_fourbar_static.png" alt="Four-bar linkage with a load" width="480px">

:::{dropdown} Solution
**Position.** From the position analysis (for instance, drawing to scale with ruler and compass): $A = (0.1414, 0.1414)$ m, $B = (0.6121, 0.3102)$ m, and $C$, $0.25$ m from $O_4$ along $O_4B$, at $(0.5658, 0.2216)$ m. The coupler is at $19.7^\circ$ from the horizontal.

**Link 3 (coupler).** It only receives forces at its two joints: it is a **two-force member**, so $\vec{F}_{34}$ (the force of the coupler on the rocker) acts along $AB$.

**Link 4 (rocker).** It is a **three-force member**: $\vec{P}$, known; $\vec{F}_{34}$, along $AB$; and $\vec{F}_{14}$, through $O_4$ with unknown line of action. The three must meet at a point:

1. Extend the line of action of $\vec{P}$ (horizontal through $C$) and line $AB$. They meet at $Q = (0.3649, 0.2216)$ m.
2. $\vec{F}_{14}$ must go through $O_4$ and $Q$: its line of action is now known ($-69.0^\circ$).
3. Draw $\vec{P}$ to scale and, from its ends, lines parallel to $AB$ and to $O_4Q$. The closed triangle $\vec{P} + \vec{F}_{34} + \vec{F}_{14} = \vec{0}$ gives the magnitudes and senses:

$$ \vec{F}_{34} = 93.4~\text{N} \angle 19.7^\circ, \qquad \vec{F}_{14} = 33.8~\text{N} \angle {-69.0^\circ} $$

<img src="figs/L03_fourbar_static_sol.png" alt="Graphical solution" width="760px">

$\vec{F}_{34}$ points from $A$ towards $B$: the coupler pushes the rocker, it works in **compression**.

**Link 2 (crank).** The coupler exerts $\vec{F}_{32} = -\vec{F}_{34} = 93.4~\text{N}\angle{-160.3^\circ}$ on the crank, at $A$. For force equilibrium, $\vec{F}_{12} = -\vec{F}_{32} = 93.4~\text{N}\angle 19.7^\circ$. Both forces are parallel and opposite, and form a couple of value $|\vec{F}_{32}|\, d_2$, where $d_2 = 0.0854$ m is the distance from $O_2$ to line $AB$ (measured on the drawing). The torque $M_2$ must balance it:

$$ |M_2| = 93.4 \cdot 0.0854 = 7.97~\text{N·m} \quad (\text{clockwise}) $$

The sense follows from noting that $\vec{F}_{32}$, applied at $A$ and pointing down and to the left, tends to turn the crank counterclockwise about $O_2$; the input torque must be clockwise.

**Ground.** It receives $\vec{F}_{21} = -\vec{F}_{12}$ at $O_2$ and $\vec{F}_{41} = -\vec{F}_{14}$ at $O_4$. Their resultant equals $\vec{P}$ (the external load is transmitted to the ground), and the moment they produce, together with $M_2$, is the moment the anchors must withstand.

**Analytical check.** By the principle of virtual power (lesson 4), $M_2\,\omega_2 + \vec{P}\cdot\vec{v}_C = 0$ for any compatible velocity. With $\omega_2 = 1$ rad/s, the velocity analysis gives $\omega_4 = 0.360$ rad/s and $\vec{v}_C = \vec{\omega}_4\times\vec{r}_{O_4C} = (-0.0797, 0.0416)$ m/s, so $M_2 = -\vec{P}\cdot\vec{v}_C/\omega_2 = -7.97$ N·m (clockwise). ✓
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