In the following kinematic chain, I would like to get equation to go from
$F_{in}$ to $T_{out}$.
(excuse the poor drawing, nodes 3,4,5 are always aligned)
Here's my attempt :
The torque at $1$ is : $C_1 = F_{in} \times a$
So $F_2 = \frac{C_1}{b}$ ( opposite direction of $F_{in}$)
But $F_3 = F_2 \times \cos{(\alpha_1)}$.
Here's where I'm suspecting to go wrong :
$F_5 = F_3 \times \cos{(\beta_1)}$.
$T_{out} = (F_5 \times \cos{(\theta_1)}) \times f$.
TL;DR; Can I chain force transmission property along a kinematic chain ?
