Công thức vật lý - Vật Lý 24/7

$$\begin{gathered} \overrightarrow{F}={\overrightarrow{F}}_x+{\overrightarrow{F}}_y \\ \Rightarrow \left\{\begin{array}{l}Ox :F_x=F\cos \left(\alpha \right) \\ Oy: F_y=F\sin \left(\alpha \right)\end{array}\right. \end{gathered}$$

${\overrightarrow{v}}_{thuyen/bo}={\overrightarrow{v}}_{thuyen/nuoc}+{\overrightarrow{v}}_{nuoc/bo}$

Trọng lực $\overrightarrow{P}$ xuyên qua $S$

$$\begin{gathered} P_k=\frac{F_k.s.\cos \left(\beta \right)}{t} ; s=v_{0}t+\frac{1}{2}a{t}^2 \\ {F}_k=\frac{m\left(a+g\left(\mu \cos \left(\alpha \right)\pm \sin \left(\alpha \right)\right)\right)}{\cos \left(\beta \right)+\mu \sin \left(\beta \right)} ; N=P\cos \left(\alpha \right)-F_k\sin \left(\beta \right) \end{gathered}$$

$$\begin{gathered} P_k=\frac{A_{F_k}}{t}=F_k.v.\cos \left(\beta \right) ,s=vt \\ {F}_k=\frac{P\left(\mu \cos \left(\alpha \right)\pm \sin \left(\alpha \right)\right)}{\cos \left(\beta \right)+\mu \sin \left(\beta \right)} ; N=P\cos \left(\alpha \right)-F_k\sin \left(\alpha \right) \end{gathered}$$

$$\begin{gathered} {\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=\overrightarrow{0} \\ \Rightarrow \left\{\begin{array}{l}{\overrightarrow{F}}_3⇅{\overrightarrow{F}}_{12}\\ F_3=F_{12}\end{array}\right. \end{gathered}$$

Nằm ngang : $∆l_1+∆l_2=\left|AB-l_{01}-l_{02}\right| ; ∆l_1=\frac{k_2}{k_1}∆l_2$

Thẳng đứng : $∆l_1=∆l_2=\frac{P}{k_1+k_2}$

$h=\frac{4\sigma }{Dgd}$

$F=k.∆l=E.S.\frac{\alpha l_0\left(t_2-t_1\right)}{l_0}=S.\alpha .E.\left(t_2-t_1\right)$

$$\begin{gathered} l_0=\frac{l_2-l_1}{\alpha \left(t_2-t_1\right)} \\ {V}_0=\frac{V_2-V_1}{\beta \left(t_2-t_1\right)} \end{gathered}$$