Physics I Homework 2

EX 2

Given	Figure
$m=0.25kg\ R=0.7m\ T_{max}=30N\ g=9.8m/s^2\ 1)\vec{P_{T_{max}}}-?\ 2)u_{max}-?$

Answer: At the lowest point.

$\begin{cases} u=const\ R=const\ m=const \end{cases}=> \begin{cases} ma_c=const\ mg=const \end{cases}$

$\phi$ is the angle between string and string at the lowest point

$T-mg\cos(\phi)=ma_c\ T=c_1+c_2\cos(\phi)$

we can reach $T_{max}$ if $\cos(\phi)=0<=>\phi=2\pi k, k\in R$

$\phi=2\pi k, k\in R$ (lowest point)

Given	Figure
$m=50kg\ \mu=0.25\ \theta=30\degree\ T_2$ is horizontal$\ g=9.8m/s^2\ M-?$

$\begin{cases} mg=N\ F_{fr}=\mu N => F_{fr} = \mu mg\ T_3=F_{fr}\ Mg=T_1\ T_2 \sin(30 \degree)=T_1\ T_2 \cos(30 \degree)=T_3 \end{cases}$

$\begin{cases} T_2=\frac{Mg}{\sin(30\degree)}\ T_2=\frac{\mu mg}{\cos(30\degree)} \end{cases}$

$M\not{g}\cos(30\degree)=\mu m\not{g}\sin(30\degree)$

$M=\frac{\mu m\sin(30\degree)}{\cos(30\degree)}$

$M=\frac{0.25\times 50kg\times \not{\frac{1}{2}}}{\sqrt{3}/\not{2}}=\frac{12.5\sqrt{3}}{3}kg=\frac{25\sqrt{3}}{6}kg\approx 7.2kg$

$Answer: 7.2kg$