11판/12. 평형과 탄성

12-13 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 6. 29. 15:29
{h=3.00[cm]r=6.00[cm]m=0.415[kg]g=9.80665[m/s2] \begin{cases} h&=3.00\ut{cm}\\ r&=6.00\ut{cm}\\ m&=0.415\ut{kg}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} {x=r2y2y=rh \begin{cases} x&=\sqrt{r^2-y^2}\\ y&=r-h\\ \end{cases} Στ=yF+xmg=0F=xymg=h(2rh)rhmg=833200g=1627903934000000[N]7.049030661598812[N]7.05[N] \begin{aligned} \Sigma \tau&=-yF+xmg=0\\ F&={x\over y} mg\\ &=\frac{ \sqrt{h (2r-h)}}{r-h}mg\\ &={83\sqrt3\over200}g\\ &=\frac{16279039 \sqrt{3}}{4000000}\ut{N}\\ &\approx 7.049030661598812\ut{N}\\ &\approx 7.05\ut{N}\\ \end{aligned}