11판/12. 평형과 탄성

12-1 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 6. 17. 21:23
{mAg=32[N]mBg=45[N]ϕ=35° \begin{cases} m_Ag&=32\ut{N}\\ m_Bg&=45\ut{N}\\ \phi&=35\degree\\ \end{cases} {ΣFA=0ΣFB=0 \begin{cases} \Sigma \vec F_A&=0\\ \Sigma \vec F_B&=0\\ \end{cases} {ΣFAx=0ΣFBx=0ΣFAy=0ΣFBy=0 \begin{cases} \Sigma F_{Ax}&=0\\ \Sigma F_{Bx}&=0\\ \Sigma F_{Ay}&=0\\ \Sigma F_{By}&=0\\ \end{cases} {0=T2T1sinϕ0=T2+T3sinθ0=T1cosϕmAg0=T3cosθmBg \begin{cases} 0&=T_2-T_1\sin\phi\\ 0&=-T_2+T_3\sin\theta\\ 0&=T_1\cos\phi-m_Ag\\ 0&=T_3\cos\theta-m_Bg\\ \end{cases} (a)\ab{a} T1=mAgcosϕ=32cos35°39.064786840366594[N]39[N] \begin{aligned} T_1&= {m_A g \over \cos\phi}\\ &={32 \over \cos35\degree}\\ &\approx 39.064786840366594\ut{N}\\ &\approx 39\ut{N}\\ \end{aligned} (b)\ab{b} T2=mAgtanϕ=32tan35°22.40664122271071[N]22[N] \begin{aligned} T_2&= {m_A g \tan\phi}\\ &={32 \tan 35\degree}\\ &\approx 22.40664122271071\ut{N}\\ &\approx 22\ut{N}\\ \end{aligned} (c)\ab{c} T3=T22+(mBg)250.269847531927915[N]50[N] \begin{aligned} T_3&= \sqrt{{T_2}^2+\br{m_Bg}^2}\\ &\approx 50.269847531927915\ut{N}\\ &\approx 50\ut{N}\\ \end{aligned} (d)\ab{d} θ=tan1(T2mBg)0.4619865204671019[rad]0.46[rad] \begin{aligned} \theta &=\tan ^{-1}\br{{T_2\over m_Bg}}\\ &\approx 0.4619865204671019\ut{rad}\\ &\approx 0.46\ut{rad}\\ \end{aligned}