I want to show that the typical fibre of the associated bundle $(P_F,\pi_F,M)$ is $F$.
Set-up: Consider a principal $G$-bundle $(P,\pi,M)$ with $\pi:P\rightarrow M$. Let $F$ be a smooth manifold equipped with a left $G$-action and construct the associated bundle $(P_F,\pi_F,M)$ by defining $$P_F:=(P\times F)/\sim_G$$ where $(p',f')\sim_G(p,f):\iff \exists g\in G: p'=p\triangleleft g : f'=g^{-1}\triangleright f$ and
$$\begin{alignat*}{2} \pi_{F}:P_F&\longrightarrow M \\ [(p,f)]&\longmapsto \pi_F([(p,f)]):=\pi(p) \end{alignat*}$$
My attempt: I've tried to simply use the definition of a bundle's fibre to see what comes out the other end:
$$ \text{Fibre}_x:=\text{preim}_{\pi_F}(\{x\}) \\ =\{[(p,f)]\in P_F \mid \pi_F([(p,f)])=x\} \\ =\{\{(p',f')\in P\times F \mid \exists g\in G: p'=p\triangleleft g : f'=g^{-1}\triangleright f\} \mid \pi(p)=x\} $$ but now I'm stuck. I was hoping (perhaps with a lot of wishful thinking) that I could get rid of the elements of $P$ somehow and be left with something like "$\text{Fibre}_x=\{f\}=F$", but this doesn't look at all likely. A possible (?) next step could be that $p'=p\triangleleft g \iff \pi(p)=\pi(p')=x$ and so I could maybe get rid of one of these conditions...? Though I'm not sure how much that would help. Any push in the right direction would be much appreciated.