1
$\begingroup$

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) Kinematic chain 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 ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.