12-4 할리데이 11판 솔루션 일반물리학

(모든 블록의 y축 두께는 문제의 결론에 영향을 주지 않으므로 무시합니다.) $$\ab{a}$$ $$ \put \begin{cases} T : \text{Top}\\ L : \text{Middle Left}\\ R : \text{Middle Right}\\ B : \text{Bottom}\\ l : \text{length} \end{cases} $$ $$ \begin{cases} \Sigma F_T&=0=N_T-mg\\ \Sigma F_{L}&=0=N_{L}-N_T-mg\\ \Sigma F_{R}&=0=N_{R}-mg\\ \Sigma F_B&=0=N_B-N_L-N_R-mg\\ \end{cases} $$ $$ \begin{cases} N_T&=mg\\ N_{L}&=2mg\\ N_{R}&=mg\\ N_B&=4mg\\ \end{cases} $$ $$\title{Stable Case}$$ $$ -{l\over 2}\le \com(\text{Block},x)\le{l\over 2}\\ \Updownarrow\\ \text{Block}\rarr\text{Stable}$$ $$\title{About T Block}$$ $$ \com(T,x)=0,$$ $$ T\rarr\text{Stable},\\ \Updownarrow\\ -{l\over 2}\le 0\le{l\over 2}\\ $$ $$\title{About L+T Block}$$ $$ \com(L+T,x)={-{l\over2}+a_1},$$ $$ L+T\rarr\text{Stable},\\ \Updownarrow\\ -{l\over 2}\le -l+a_1\le{l\over 2}\\ $$ $$\therefore {l\over 2}\le a_1\le {3\over2}l\taag1$$ $$\title{About R Block}$$ $$ \com(R,x)={a_1},$$ $$ R\rarr\text{Stable},\\ \Updownarrow\\ -{l\over 2}\le a_1\le{l\over 2}\\ $$ $$\therefore -{l\over 2}\le a_1\le{l\over 2}\taag2$$ $$(1),(2)\rarr a_1={l\over 2}$$ $$\title{About All Block}$$ $$ \begin{aligned} \com(\text{All},x) &={\Sigma xm\over\Sigma m}\\ &={0\cdot 2m+l\cdot m+{l\over2}\cdot m\over4 m}\\ &={3\over8}l\\ \end{aligned} $$ $$ \begin{aligned} \com(\text{All},x)&\le l-a_2\\ {3\over8}l&\le l-a_2\\ a_2&\le{5\over8}l \end{aligned} $$ $$ \begin{aligned} h_a &=a_1+a_2\\ &={l\over 2}+{5\over8}l\\ &={9\over8}l\\ \end{aligned} $$ $$\ab{b}$$ $$ \put \begin{cases} T : \text{Top}\\ L : \text{Middle Left}\\ R : \text{Middle Right}\\ B : \text{Bottom}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_T&=0=N_{TL}+N_{TR}-mg\\ \Sigma F_{L}&=0=N_{L}-N_{TL}-mg\\ \Sigma F_{R}&=0=N_{R}-N_{TR}-mg\\ \Sigma F_B&=0=N_B-N_L-N_R-mg\\ \Sigma \tau_L&=0=-{l\over2}\cdot mg+b_1 \cdot N_L-l\cdot N_{TL}\\ \Sigma \tau_R&=0={l\over2}\cdot mg-b_1 \cdot N_R+l\cdot N_{TR}\\ \end{cases} $$ $$ \begin{cases} N_{TL}&={1\over2}mg\\ N_{TR}&={1\over2}mg\\ N_{L}&={3\over2}mg\\ N_R&={3\over2}mg\\ N_B&=4mg\\ b_1&={2\over3}l \end{cases} $$ $$\title{About All Block}$$ $$ \begin{aligned} \com(\text{All},x)={l\over2}&\le l-b_2\\ b_2&\le{l\over2} \end{aligned} $$ $$ \begin{aligned} h_b&=b_1+b_2\\ &={2\over3}l+{l\over2}\\ &={7\over6}l\\ \end{aligned} $$