## Example 3.1

The figure shows a [robotic gripper](https://www.youtube.com/watch?v=shUN5VXV8mE) mechanism, with dimensions in cm. If a gripping force of $F_g =25N$ is required, and gravity is not considered, what must the actuator force $F_a$ be? In order to properly design the gripper's support structure, we are also asked to compute the reactions at point $O_2$.

<img src="./figs/ejemplo3_01_media/image_0.jpg" alt="ejemplo301" width="520px">

### Quasi-static analysis

This exercise can be solved by applying the laws of statics. Due to symmetry, the exercise can be solved by focusing on the upper finger. Considering that force $F_a$ is transmitted as a tension in bar AB, at 45º, the analysis of body AC gives, taking moments about point $O_2$ and considering the counterclockwise direction as positive:
$ F_{AB} \cos 45\cdot 1.5\sin 15+F_{AB} \sin 45\cdot 1.5\cos 15-25\cdot 4\cdot \cos 15=0 $ 

<img src="./figs/ejemplo3_01_media/image_1.jpeg" alt="ejemplo301b" width="400px">

from which:

 $ F_{AB} =74.4\,N $ 

Once tension $F_{AB}$ is known, using the action-reaction principle we can compute force $F_a$ as,

<img src="./figs/ejemplo3_01_media/image_2.jpeg" alt="ejemplo301c" width="220px">

 $ F_a =2\cdot F_{AB} \cos 45 $ 

from which:

 $ F_a =105.15\,N $ 

Finally, we can compute the reaction at $O_2$ by graphically solving the static equilibrium equation of body 2:

<img src="./figs/ejemplo3_01_media/image_3.jpeg" alt="ejemplo301d" width="320px">

which gives us ${\vec{F} }_{12} =52.60\hat{i} -77.60\hat{j} \,N$ 

### Systematic analysis

There is another, systematic procedure that consists of writing all the force and moment equilibrium equations for every body and solving them simultaneously. To do this, we analyze the free body diagrams of the different elements.

#### Body 2

We define the equations from the free body diagram:

<img src="./figs/ejemplo3_01_media/image_4.jpeg" alt="ejemplo301e" width="320px">

 $ \begin{array}{l} {\vec{F} }_{32} +{\vec{F} }_{12} +{\vec{F} }_g =0\newline {\vec{r} }_{O2A} \times {\vec{F} }_{32} +{\vec{r} }_{O2C} \times {\vec{F} }_g =0 \end{array} $ 

Note that under this approach we treat all components as positive. On one hand, the known position- and force-vector components will give us the positive sense of the moments. On the other hand, the results we obtain when solving the system of equations will determine the sign, and therefore the direction, of the unknown forces.


#### Body 3

We define the equations from the free body diagram:

<img src="./figs/ejemplo3_01_media/image_5.jpeg" alt="ejemplo301f" width="220px">

 $ \begin{array}{l} {\vec{F} }_{43} -{\vec{F} }_{32} =0\newline {\vec{r} }_{AB} \times {\vec{F} }_{43} =0 \end{array} $ 

Note that we use negative signs for those forces that, by the action-reaction principle, are equal and opposite to others we have already defined, in order to reduce the number of unknowns.

#### Body 4

We define the equations from the free body diagram:

<img src="./figs/ejemplo3_01_media/image_6.jpeg" alt="ejemplo301g" width="220px">

 $ {\vec{F} }_{14} -{\vec{F} }_{43} +{\vec{F} }_a =0 $ 

#### Solution:

To solve the system made up of the 8 equations we must use MATLAB's *solve* function, as in the previous exercises, keeping in mind that we must group the equations into a vector. The 8 unknowns are entered separated by commas, just as in the previous exercises.

```matlab
% Solve the complete system:
solucion=solve([eq1_2, eq3, eq4_5, eq6, eq7_8],...
    F12x, F12y, F32x, F32y, F43x, F43y, F14y, F);```