11판/12. 평형과 탄성

12-14 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 6. 29. 16:10
{Fh=13.4[N]Fv=162.4[N] \begin{cases} F_h&=13.4\ut{N}\\ F_v&=162.4\ut{N}\\ \end{cases} Στ=d3Ftsin45°+dFvsin10°+dFhsin80°=0 \begin{aligned} \Sigma \tau&=-{d\over3}F_t\sin45\degree+d F_v \sin10\degree+dF_h\sin80\degree\\ &=0 \end{aligned} Ft=32(Fhcos10°+Fvsin10°)=325(67cos10°+812sin10°)175.63212110909666[N]176[N] \begin{aligned} F_t &=3\sqrt2\br{F_h\cos10\degree+F_v\sin10\degree}\\ &={3\sqrt2\over5}\br{67\cos10\degree+812\sin10\degree}\\ &\approx 175.63212110909666\ut{N}\\ &\approx 176\ut{N}\\ \end{aligned}