# force transmission in kinematic chain

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 ?