11판/12. 평형과 탄성

12-31 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 10. 9. 17:48
{m1=85.0[kg]m2=30.0[kg]L2=2.00[m]m3=20.0[kg]d=0.500[m]L1=L2+2d \begin{cases} m_1&=85.0\ut{kg}\\ m_2&=30.0\ut{kg}\\ L_2&=2.00\ut{m}\\ m_3&=20.0\ut{kg}\\ d&=0.500\ut{m}\\ L_1&=L_2+2d\\ \end{cases} {ΣF1y=0Στ1=0ΣF2y=0Στ2=0 \begin{cases} \Sigma F_{1y}&=0\\ \Sigma \tau_1&=0\\ \Sigma F_{2y}&=0\\ \Sigma \tau_2&=0\\ \end{cases} {0=T1L+T1RT2LT2Rm1g0=dT2LL12m1g(L1d)T2R+L1T1R0=T2L+T2Rm2gm3g0=dm3gL22m2g+L2T2R \begin{cases} 0&=T_{1L}+T_{1R}-T_{2L}-T_{2R}-m_1g\\ 0&=-dT_{2L}-{L_1\over2}m_1g-\br{L_1-d}T_{2R}+L_1T_{1R} \\ 0&=T_{2L}+T_{2R}-m_2g-m_3g\\ 0&=-dm_3g-{L_2\over2}m_2g+L_2T_{2R} \\ \end{cases} T1R=12(m1g+m2g+4dL1m3g)=3856g=1510224124000[N]629.2600416666667[N]629[N] \begin{aligned} T_{1R}&=\frac{1}{2} \br{m_1 g+m_2g+\frac{4 d }{L_1}m_3 g}\\ &={385\over6}g\\ &={15102241\over24000}\ut{N}\\ &\approx 629.2600416666667\ut{N}\\ &\approx 629\ut{N}\\ \end{aligned}