(모든 블록의 y축 두께는 문제의 결론에 영향을 주지 않으므로 무시합니다.) ( a ) \ab{a} ( a ) p u t { T : Top L : Middle Left R : Middle Right B : Bottom l : length \put \begin{cases} T : \text{Top}\\ L : \text{Middle Left}\\ R : \text{Middle Right}\\ B : \text{Bottom}\\ l : \text{length} \end{cases} put ⎩ ⎨ ⎧ T : Top L : Middle Left R : Middle Right B : Bottom l : length { Σ F T = 0 = N T − m g Σ F L = 0 = N L − N T − m g Σ F R = 0 = N R − m g Σ F B = 0 = N B − N L − N R − m g \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} ⎩ ⎨ ⎧ Σ F T Σ F L Σ F R Σ F B = 0 = N T − m g = 0 = N L − N T − m g = 0 = N R − m g = 0 = N B − N L − N R − m g { N T = m g N L = 2 m g N R = m g N B = 4 m g \begin{cases} N_T&=mg\\ N_{L}&=2mg\\ N_{R}&=mg\\ N_B&=4mg\\ \end{cases} ⎩ ⎨ ⎧ N T N L N R N B = m g = 2 m g = m g = 4 m g [Stable Case] \title{Stable Case} [Stable Case] − l 2 ≤ c o m ( Block , x ) ≤ l 2 ⇕ Block → Stable -{l\over 2}\le \com(\text{Block},x)\le{l\over 2}\\ \Updownarrow\\ \text{Block}\rarr\text{Stable} − 2 l ≤ com ( Block , x ) ≤ 2 l ⇕ Block → Stable [About T Block] \title{About T Block} [About T Block] c o m ( T , x ) = 0 , \com(T,x)=0, com ( T , x ) = 0 , T → Stable , ⇕ − l 2 ≤ 0 ≤ l 2 T\rarr\text{Stable},\\ \Updownarrow\\ -{l\over 2}\le 0\le{l\over 2}\\ T → Stable , ⇕ − 2 l ≤ 0 ≤ 2 l [About L+T Block] \title{About L+T Block} [About L+T Block] c o m ( L + T , x ) = − l 2 + a 1 , \com(L+T,x)={-{l\over2}+a_1}, com ( L + T , x ) = − 2 l + a 1 , L + T → Stable , ⇕ − l 2 ≤ − l + a 1 ≤ l 2 L+T\rarr\text{Stable},\\ \Updownarrow\\ -{l\over 2}\le -l+a_1\le{l\over 2}\\ L + T → Stable , ⇕ − 2 l ≤ − l + a 1 ≤ 2 l ∴ l 2 ≤ a 1 ≤ 3 2 l ⋯ (1) \therefore {l\over 2}\le a_1\le {3\over2}l\taag1 ∴ 2 l ≤ a 1 ≤ 2 3 l ⋯ (1) [About R Block] \title{About R Block} [About R Block] c o m ( R , x ) = a 1 , \com(R,x)={a_1}, com ( R , x ) = a 1 , R → Stable , ⇕ − l 2 ≤ a 1 ≤ l 2 R\rarr\text{Stable},\\ \Updownarrow\\ -{l\over 2}\le a_1\le{l\over 2}\\ R → Stable , ⇕ − 2 l ≤ a 1 ≤ 2 l ∴ − l 2 ≤ a 1 ≤ l 2 ⋯ (2) \therefore -{l\over 2}\le a_1\le{l\over 2}\taag2 ∴ − 2 l ≤ a 1 ≤ 2 l ⋯ (2) ( 1 ) , ( 2 ) → a 1 = l 2 (1),(2)\rarr a_1={l\over 2} ( 1 ) , ( 2 ) → a 1 = 2 l [About All Block] \title{About All Block} [About All Block] c o m ( All , x ) = Σ x m Σ m = 0 ⋅ 2 m + l ⋅ m + l 2 ⋅ m 4 m = 3 8 l \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} com ( All , x ) = Σ m Σ x m = 4 m 0 ⋅ 2 m + l ⋅ m + 2 l ⋅ m = 8 3 l c o m ( All , x ) ≤ l − a 2 3 8 l ≤ l − a 2 a 2 ≤ 5 8 l \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} com ( All , x ) 8 3 l a 2 ≤ l − a 2 ≤ l − a 2 ≤ 8 5 l h a = a 1 + a 2 = l 2 + 5 8 l = 9 8 l \begin{aligned} h_a &=a_1+a_2\\ &={l\over 2}+{5\over8}l\\ &={9\over8}l\\ \end{aligned} h a = a 1 + a 2 = 2 l + 8 5 l = 8 9 l ( b ) \ab{b} ( b ) p u t { T : Top L : Middle Left R : Middle Right B : Bottom \put \begin{cases} T : \text{Top}\\ L : \text{Middle Left}\\ R : \text{Middle Right}\\ B : \text{Bottom}\\ \end{cases} put ⎩ ⎨ ⎧ T : Top L : Middle Left R : Middle Right B : Bottom { Σ F T = 0 = N T L + N T R − m g Σ F L = 0 = N L − N T L − m g Σ F R = 0 = N R − N T R − m g Σ F B = 0 = N B − N L − N R − m g Σ τ L = 0 = − l 2 ⋅ m g + b 1 ⋅ N L − l ⋅ N T L Σ τ R = 0 = l 2 ⋅ m g − b 1 ⋅ N R + l ⋅ N T R \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} ⎩ ⎨ ⎧ Σ F T Σ F L Σ F R Σ F B Σ τ L Σ τ R = 0 = N T L + N TR − m g = 0 = N L − N T L − m g = 0 = N R − N TR − m g = 0 = N B − N L − N R − m g = 0 = − 2 l ⋅ m g + b 1 ⋅ N L − l ⋅ N T L = 0 = 2 l ⋅ m g − b 1 ⋅ N R + l ⋅ N TR { N T L = 1 2 m g N T R = 1 2 m g N L = 3 2 m g N R = 3 2 m g N B = 4 m g b 1 = 2 3 l \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} ⎩ ⎨ ⎧ N T L N TR N L N R N B b 1 = 2 1 m g = 2 1 m g = 2 3 m g = 2 3 m g = 4 m g = 3 2 l [About All Block] \title{About All Block} [About All Block] c o m ( All , x ) = l 2 ≤ l − b 2 b 2 ≤ l 2 \begin{aligned} \com(\text{All},x)={l\over2}&\le l-b_2\\ b_2&\le{l\over2} \end{aligned} com ( All , x ) = 2 l b 2 ≤ l − b 2 ≤ 2 l h b = b 1 + b 2 = 2 3 l + l 2 = 7 6 l \begin{aligned} h_b&=b_1+b_2\\ &={2\over3}l+{l\over2}\\ &={7\over6}l\\ \end{aligned} h b = b 1 + b 2 = 3 2 l + 2 l = 6 7 l