# Example 3.1: Robotic gripper (quasi-static analysis)

The figure shows the mechanism of a [robotic gripper](https://www.youtube.com/watch?v=shUN5VXV8mE), with dimensions in cm. A gripping force $F_g = 25$ N is required at each finger. Neglecting gravity, what must the actuator force $F_a$ be? To design the support structure of the gripper, compute also the reaction at the pin $O_2$.

<img src="./figs/ejemplo3_01_media/image_0.jpg" alt="Robotic gripper" width="520px">

The upper finger (link 2) pivots about $O_2$ and is at $15^\circ$ from the horizontal; bar $AB$ (link 3) is at $45^\circ$; the actuator moves block $B$ (link 4) horizontally. The mechanism moves slowly, so the analysis is **quasi-static**: inertia loads are negligible.

:::{dropdown} Graphical solution (equilibrium of each link)
By symmetry, it is enough to study the upper finger. Bar $AB$ carries no loads and has no mass: it is a **two-force member**, and the force $F_{AB}$ it transmits acts along the bar, at $45^\circ$.

**Link 2.** Taking moments about $O_2$ (counterclockwise positive), the reaction at $O_2$ does not appear:

<img src="./figs/ejemplo3_01_media/image_1.jpeg" alt="Upper finger" width="400px">

$$ F_{AB} \cos 45^\circ \cdot 1.5 \sin 15^\circ + F_{AB} \sin 45^\circ \cdot 1.5 \cos 15^\circ - 25 \cdot 4 \cos 15^\circ = 0 $$

which gives $F_{AB} = 74.4$ N.

**Link 4.** Block $B$ receives the forces of both bars (the one of the upper finger and the one of the lower finger, symmetric) and the actuator force:

<img src="./figs/ejemplo3_01_media/image_2.jpeg" alt="Actuator block" width="220px">

The vertical components cancel out by symmetry, and the horizontal ones add up:

$$ F_a = 2\, F_{AB} \cos 45^\circ = 105.15~\text{N} $$

**Reaction at $O_2$.** It is obtained by closing the force polygon of link 2, $\vec{F}_g + \vec{F}_{AB} + \vec{F}_{12} = \vec{0}$:

<img src="./figs/ejemplo3_01_media/image_3.jpeg" alt="Force polygon" width="320px">

$$ \vec{F}_{12} = 52.6\,\hat{i} - 77.6\,\hat{j}~\text{N} $$
:::

:::{dropdown} Analytical solution (system of equations)
The systematic procedure is to write the force and moment equilibrium equations of all bodies and solve them simultaneously. We keep studying the upper half; its share of the actuator force is $F_a/2$.

Position vectors (in cm), with $\hat{i}$ to the right and $\hat{j}$ upwards:

$$
\vec{r}_{O_2A} = 1.5\,(\cos 15^\circ, \sin 15^\circ), \quad
\vec{r}_{O_2C} = -4\,(\cos 15^\circ, \sin 15^\circ), \quad
\vec{r}_{AB} = 2.5\,(\cos 45^\circ, -\sin 45^\circ)
$$

**Body 2.** From the free-body diagram:

<img src="./figs/ejemplo3_01_media/image_4.jpeg" alt="FBD body 2" width="320px">

$$
\begin{aligned}
\vec{F}_{32} + \vec{F}_{12} + \vec{F}_g &= \vec{0} \\
\vec{r}_{O_2A} \times \vec{F}_{32} + \vec{r}_{O_2C} \times \vec{F}_g &= \vec{0}
\end{aligned}
$$

with $\vec{F}_g = 25\,\hat{j}$ N. All unknown components are written as positive; the sign of the result gives their sense.

**Body 3.**

<img src="./figs/ejemplo3_01_media/image_5.jpeg" alt="FBD body 3" width="220px">

$$
\begin{aligned}
\vec{F}_{43} - \vec{F}_{32} &= \vec{0} \\
\vec{r}_{AB} \times \vec{F}_{43} &= \vec{0}
\end{aligned}
$$

We use $\vec{F}_{23} = -\vec{F}_{32}$ to avoid duplicating unknowns.

**Body 4** (half of the block, actuator force $-F\,\hat{i}$, horizontal guide):

<img src="./figs/ejemplo3_01_media/image_6.jpeg" alt="FBD body 4" width="220px">

$$ F_{14y}\,\hat{j} - \vec{F}_{43} - F\,\hat{i} = \vec{0} $$

**Solution.** These are 8 scalar equations with 8 unknowns ($F_{12x}$, $F_{12y}$, $F_{32x}$, $F_{32y}$, $F_{43x}$, $F_{43y}$, $F_{14y}$, $F$). In MATLAB:

```matlab
syms F12x F12y F32x F32y F43x F43y F14y F real
rO2A = 1.5*[cosd(15) sind(15) 0];
rO2C = -4*[cosd(15) sind(15) 0];
rAB  = 2.5*[cosd(45) -sind(45) 0];
Fg = [0 25 0];
F12 = [F12x F12y 0]; F32 = [F32x F32y 0]; F43 = [F43x F43y 0];
eq1_2 = F32 + F12 + Fg == 0;
eq3   = cross(rO2A, F32) + cross(rO2C, Fg) == 0;
eq4_5 = F43 - F32 == 0;
eq6   = cross(rAB, F43) == 0;
eq7_8 = [0 F14y 0] - F43 - [F 0 0] == 0;
sol = solve([eq1_2(1:2), eq3(3), eq4_5(1:2), eq6(3), eq7_8(1:2)], ...
    [F12x, F12y, F32x, F32y, F43x, F43y, F14y, F]);
structfun(@double, sol)
```

Result: $\vec{F}_{32} = \vec{F}_{43} = -52.58\,\hat{i} + 52.58\,\hat{j}$ N (the bar is in compression with $74.4$ N), $\vec{F}_{12} = 52.58\,\hat{i} - 77.58\,\hat{j}$ N, $F = 52.58$ N, and therefore $F_a = 2F = 105.15$ N, as in the graphical solution. The reaction $F_{14y} = 52.58$ N of the upper half cancels out with that of the lower half, so the actuator guide carries no load.
:::