솔루션피아

생업이 바빠 풀이가 지지부진한 사이에..

짱세디럭스 — Sun, 3 Aug 2025 17:42:55 +0900

할리데이가 12판이 출간되어 버렸습니다.

좀 애매한데.. 그냥 12판으로 넘어갈까 합니다.

그나마 좀 생업이 괜찮아진 부분이 있어서, 꾸준하고 힘차게 달려보겠습니다.

12-47 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 3 Aug 2025 17:40:25 +0900

$$ \begin{cases} m_A&=46.0\ut{kg}\\ L&=5.00\ut{m}\\ m_B&=250\ut{kg}\\ d&=0.540\ut{m}\\ \end{cases} $$ $$ \begin{cases} x_1&=d\\ x_2&=L-d\\ x_A&=L\\ x_B&={L\over2}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_1+N_2-m_Ag-m_Bg\\ 0&=x_1N_1+x_2N_2-x_Am_Ag-x_Bm_Bg\\ \end{cases} $$ $$\ab{a,b}$$ $$ \begin{cases} N_1&=\cfrac{ m_A (x_A-x_2)+m_B (x_B-x_2)}{x_1-x_2}g\\ N_2&=\cfrac{ m_A (x_1-x_A)+m_B (x_1-x_B)}{x_1-x_2}g \end{cases} $$ $$ \begin{cases} N_1&=\cfrac{ 2 d (m_A+m_B)-L m_B}{4 d-2 L}g\\ N_2&=\cfrac{ 2 d (m_A+m_B)-L (2 m_A+m_B)}{4 d-2 L}g \end{cases} $$ $$ \begin{cases} \vec N_1&=\frac{11629}{98}g\j\\ \vec N_2&=\frac{17379}{98} g\j \end{cases} $$ $$ \begin{cases} \vec N_1&\approx 1.163689110714286\j\times10^3\ut{N}\\ \vec N_2&\approx 1.739079289285714\j\times10^3\ut{N} \end{cases} $$ $$ \begin{cases} \vec N_1&\approx 1.16\j\times10^3\ut{N}\\ \vec N_2&\approx 1.74\j\times10^3\ut{N} \end{cases} $$ $$ \begin{cases} \vec N_1&\approx 1.16\j\ut{kN}\\ \vec N_2&\approx 1.74\j\ut{kN} \end{cases} $$

12-46 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Thu, 1 May 2025 15:22:31 +0900

$$ \begin{cases} m&=95\ut{kg}\\ L&=15\ut{m}\\ 2R&=9.6\ut{mm}\\ \Delta L &= 1.3\ut{cm}\\ \end{cases} $$ $${F\over A}=E{\Delta L\over L},$$ $$ \begin{aligned} E&={F\over A}\cdot{L\over\Delta L}\\ &={mgL\over \pi R^2\Delta L}\\ &=\frac{185546875000 g}{39 \pi }\\ &\approx 1.4851141642012506\times10^{10}\ut{N/m^2}\\ &\approx 1.5\times10^{10}\ut{N/m^2}\\ &\approx 15\ut{GN/m^2}\\ \end{aligned} $$

12-45 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Thu, 1 May 2025 15:04:34 +0900

$$ \begin{cases} m&=300\ut{kg}\\ A&=2.0\times10^{-6}\ut{m^2}\\ L_1=L_3&=2.0000\ut{m}\\ L_2&=L_1+d\\ d&=6.00\ut{mm}\\ g&=9.80665\ut{m/s^2}\\ E_{\text{steel}}&=200\times10^9\ut{N/m^2} \end{cases} $$ $${F\over A}=E{\Delta L\over L},$$ $$ \begin{cases} F_1&=AE\cfrac{\Delta L_1}{L_1}\\ F_2&=AE\cfrac{\Delta L_2}{L_2}\\ \end{cases} $$ $$ \begin{cases} F_1&=AE\cfrac{\Delta L_1}{L_1}\\ F_2&=AE\cfrac{\Delta L_1-d}{L_1+d}\\ \end{cases} $$ $$ \begin{cases} F_1&=2 \Delta L_1\times10^5\\ F_2&=\cfrac{4 (500 \Delta L_1-3)}{1003}\times10^5\\ \end{cases} \taag1$$ $$ \begin{aligned} \Sigma F_{y}&=0\\ &=2F_1+F_2-mg\\ &=2\br{AE\cfrac{\Delta L_1}{L_1}}+\br{AE\cfrac{\Delta L_1-d}{L_1+d}}-mg\\ \end{aligned} $$ $$ \begin{aligned} \Delta L_1&=\frac{L_1 }{A E }\cdot\frac{A d E+ mg (d+L_1)}{ 2 d+3 L_1}\\ &=\frac{276721399}{40080000}\times10^{-3}\ut{m}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} F_1&=AE\cfrac{\Delta L_1-d}{L_1+d}\\ &=\frac{276721399}{200400}\ut{N}\\ &\approx 1380.8453043912175\ut{N}\\ &\approx 1.4\times10^3\ut{N}\\ &\approx 1.4\ut{kN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} F_2&=AE\cfrac{\Delta L_1}{L_1}\\ &=\frac{180665}{1002}\ut{N}\\ &\approx 180.30439121756487\ut{N}\\ &\approx 1.8\times10^2\ut{N}\\ \end{aligned} $$

12-44 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Thu, 1 May 2025 13:24:10 +0900

$$ \begin{cases} L&=2.50\ut{m}\\ mg&=500\ut{N}\\ d&=1.50\ut{m}\\ \end{cases} $$ $$ x=\sqrt{L^2-d^2},$$ $$\ab{a}$$ $$ \begin{aligned} \Sigma \tau&=0\\ &=-\frac{x}{2}mg+LP\\ \end{aligned} $$ $$ \begin{aligned} P&=\frac{mgx}{2L}\\ &=\frac{mg}{2L}\sqrt{L^2-d^2}\\ &=200\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$\put x=\sqrt{L^2-d^2}$$ $$\sin\theta=\frac{d}{L},$$ $$ \begin{aligned} \vec P &=-P\sin\theta\i+P\cos\theta\j\\ &=-\frac{dP}{L}\i+\frac{xP}{L}j\taag1\\ \end{aligned} $$ $$ \begin{aligned} \Sigma \vec F&=0\\ &=m\vec g+\vec P+\vec N\\ \end{aligned} $$ $$ \begin{aligned} N&=\abs{\vec N}\\ &=\abs{-m\vec g-\vec P}\\ &=\abs{-m\br{-g\j}-\br{-\frac{dP}{L}\i+\frac{xP}{L}j}}\\ &=\abs{\frac{dP}{L}\i+\br{mg-\frac{xP}{L}}\j}\\ &=\sqrt{\br{\frac{dP}{L}}^2+\br{mg-\frac{xP}{L}}^2}\\ &=100\sqrt{13}\ut{N}\\ &\approx 360.5551275463989\ut{N}\\ &\approx 361\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} f&=\mu N\\ N_x&=\mu N_y\\ \end{aligned} $$ $$ \begin{aligned} \mu&= \frac{N_x}{N_y}\\ &= \frac{\frac{dP}{L}}{mg-\frac{xP}{L}}\\ &=\frac{d P}{L mg-xP}\\ &=\frac{6}{17}\\ &\approx 0.35294117647058826\\ &\approx 0.353\\ \end{aligned} $$

12-43 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 30 Apr 2025 19:34:45 +0900

$$ \begin{cases} T&=535\ut{N}\\ \end{cases} $$ $$ \put\begin{cases} T&=\overline{AB}\\ F_v&=\overline{AD},\overline{BC}\\ F_d&=\overline{AC},\overline{BD}\\ F_h&=\overline{CD}\\ \end{cases} $$ $$ \put \begin{cases} \text{Tensile Force} : +\\ \text{Compressive Force} :- \end{cases} $$ $$ \begin{cases} \Sigma \vec F_B&=0\\ \Sigma \vec F_C&=0\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{Bx}&=0\\ \Sigma F_{By}&=0\\ \Sigma F_{Cx}&=0\\ \Sigma F_{Cy}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=T+F_{dx}\\ 0&=F_v+F_{dy}\\ 0&=F_{dx}+F_h\\ 0&=-F_v-F_{dy}\\ \end{cases} $$ $$ \begin{cases} 0&=T+{F_d\over\sqrt2}\\ 0&=F_v+{F_d\over\sqrt2}\\ 0&={F_d\over\sqrt2}+F_h\\ 0&=-F_v-{F_d\over\sqrt2}\\ \end{cases} $$ $$ \begin{cases} F_v&=T\\ F_d&=-\sqrt2 T\\ F_h&=T\\ \end{cases} $$ $$\ab{a}$$ $$ \text{Tensile Force}: F_v, F_h$$ $$\overline{AD},\overline{BC}, \overline{CD} $$ $$\ab{b}$$ $$\text{Tensile Force}: F_v, F_h$$\\$$ $$ \begin{aligned} F_v&=F_h\\ &=T\\ &=535\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$\text{Compressive Force} :-$$ $$ \begin{aligned} F_d&=-\sqrt2 T\\ &=535\sqrt2\ut{N}\\ &\approx 756.6042558696059\ut{N}\\ &\approx 757\ut{N}\\ \end{aligned} $$

12-42 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 30 Apr 2025 15:05:05 +0900

$$ \begin{cases} L&=6.2\ut{m}\\ mg&=400\ut{N}\\ \mu&=0.39\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \put\begin{cases} F=\text{Force Wall}\\ N=\text{Force Floor}\\ h=\text{Wall Distance}\\ x=\text{Floor Distance}\\ \end{cases} $$ $$ h=\sqrt{L^2-x^2} $$ $$ \begin{cases} 0&=F-\mu N\\ 0&=N-mg\\ 0&=-\frac{x}{2}mg+xN-h\mu N\\ \end{cases} $$ $$ \begin{aligned} x&=\frac{2 \mu L}{\sqrt{4 \mu ^2+1}}\\ &=\frac{1209}{5 \sqrt{4021}}\ut{m}\\ &\approx 3.813197151902038\ut{m}\\ &\approx 3.8\ut{m}\\ \end{aligned} $$

12-41 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 30 Apr 2025 14:26:49 +0900

$$ \begin{cases} L_A&=2.40\ut{m}\\ m_A&=54.0\ut{kg}\\ m_B&=68.0\ut{kg}\\ d&=1.60\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{cases} \Sigma \vec F_A&=0\\ \Sigma \vec F_B&=0\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{Ax}&=0\\ \Sigma F_{Bx}&=0\\ \Sigma F_{Ay}&=0\\ \Sigma F_{By}&=0\\ \Sigma \tau_A&=0\\ \Sigma \tau_B&=0\\ \end{cases} $$ $$\put \vec F=\text{Force }A \lrarr B$$ $$ \begin{cases} 0&=N_{Ax}+F_x\\ 0&=N_{Bx}-F_x\\ 0&=N_{Ay}-m_Ag+F_y\\ 0&=N_{By}-m_Bg-F_y\\ 0&=-{L_A\over2}m_Ag+L_AF_y\\ 0&=-{L_A\over2}m_Bg+dF_x \end{cases} $$ $$ \begin{cases} N_{Ax}&=-\cfrac{L_A}{2d}m_Bg\\ N_{Ay}&=\cfrac{1}{2}m_Ag\\ N_{Bx}&=\cfrac{L_A}{2d}m_Bg\\ N_{By}&=\cfrac{m_A+2m_B}{2}g\\ F_x&=\cfrac{L_A}{2d}m_Bg\\ F_y&=\cfrac{1}{2}m_Ag\\ \end{cases} $$ $$ \begin{cases} \vec N_A&=-\cfrac{L_A}{2d}m_Bg\i+\cfrac{1}{2}m_Ag\j\\ \vec N_B&=\cfrac{L_A}{2d}m_Bg\i+\cfrac{m_A+2m_B}{2}g\j\\ \vec F&=\cfrac{L_A}{2d}m_Bg\i+\cfrac{1}{2}m_A\j \\ \end{cases} $$ $$ \begin{cases} \vec N_A&=-51g\i+27g\j\\ \vec N_B&=51g\i+95g\j\\ \vec F&=51g\i +27g\j \\ \end{cases} $$ $$ \begin{cases} \vec N_A&=-500.13915\i+264.77955\j\ut{N}\\ \vec N_B&=500.13915\i+931.63175\j\ut{N}\\ \vec F&=500.13915\i +264.77955\j\ut{N}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec N_A&=-500.13915\i+264.77955\j\ut{N}\\ &\approx \br{-5.00\i+2.65\j}\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec F&=500.13915\i +264.77955\j\ut{N}\\ &\approx \br{5.00\i +2.65\j}\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \vec N_B&=500.13915\i+931.63175\j\ut{N}\\ &\approx \br{5.00\i+9.32\j}\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} -\vec F&=-500.13915\i -264.77955\j\ut{N}\\ &\approx \br{-5.00\i -2.65\j}\times10^2\ut{N}\\ \end{aligned} $$

12-40 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 15 Dec 2024 16:44:22 +0900

$$ \begin{cases} L&=12.0\ut{m}\\ \theta&=50.0\degree\\ T&=400\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=-T+N_x\\ 0&=-mg+N_y\\ 0&=TL\cos\theta-mg{L\over2}\sin\theta\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} m\vec g &={2T\over\tan\theta}\j\\ &=800\tan40\degree\j\ut{N}\\ &\approx 671.279704941824\j\ut{N}\\ &\approx 671\j\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} N_x&=T\\ N_y&=mg\\ \end{cases} $$ $$ \begin{aligned} \vec N&=T\i+mg\j\\ &=400\i+800\tan40\degree\j\ut{N}\\ &\approx 400\i+671\degree\j\ut{N}\\ \end{aligned} $$

12-39 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 15 Dec 2024 16:18:28 +0900

$$ \begin{cases} 2x&=3.44\ut{m}\\ y&=25.0\ut{cm}\\ mg&=3160\ut{N} \end{cases} $$ $$ \begin{aligned} \tan\theta&={y\over x},\\ \end{aligned} $$ $$\Sigma F_y=0,$$ $$ \begin{aligned} 0&=2T_y-mg\\ T_y&={mg\over2} \end{aligned} $$ $$ \begin{aligned} T&={T_y\over \sin\theta}\\ &=m g\cdot\frac{ x }{2 y}\sqrt{\frac{y^2}{x^2}+1}\\ &=\frac{316 }{5}\sqrt{30209}\ut{N}\\ &\approx 1.098462544468404\times 10^4\ut{N}\\ &\approx 1.10\times 10^4\ut{N}\\ &\approx 11.0\ut{kN}\\ \end{aligned} $$

12-38 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 15 Dec 2024 15:58:22 +0900

$$ \begin{cases} H&=59.1\ut{m}\\ 2R&=7.44\ut{m}\\ \Delta x_{\text{top}0} &= 4.01\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$\Delta x_\com = R,$$ $$ \begin{aligned} \Delta x_{\text{top}}&=2\Delta x_\com\\ &=2R\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=2R-\Delta x_{\text{top}0}\\ &=3.43\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta&=\sin^{-1}{2R\over H}\\ &=\sin^{-1}{124\over 985}\\ &\approx 0.12622322910580994\ut{rad}\\ &\approx 0.126\ut{rad}\\ \end{aligned} $$

12-37 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 15 Dec 2024 15:28:25 +0900

$$ T=900\cdot {y\over L} \taag 1$$ $$ \abs{F_h}=300-300\cdot{y\over L} \taag 2$$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=F_a-T\cos\theta+F_h\\ 0&=N-T\sin\theta-mg\\ 0&=LT\cos\theta-yF_a\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \theta&=2\tan^{-1}{1\over\sqrt2}\\ &\approx 1.2309594173407745\ut{rad}\\ &\approx 1.2\ut{rad}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} F_a&=300\ut{N} \end{aligned} $$

12-36 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 15 Dec 2024 14:44:43 +0900

$$ \begin{cases} \theta&=26\degree\\ L&=43\ut{m}\\ T&=2.5\ut{m}\\ W&=12\ut{m}\\ \rho&=3.2\ut{g/cm^3}=3.2\times10^{3}\ut{kg/m^2}\\ \mu_s&=0.39\\ \end{cases} $$ $$ \begin{aligned} m&=\rho V\\ &=\rho LTW\\ &=4.128\times10^6\ut{kg} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \Ans&=mg\sin\theta\\ &=4.128g\sin26\degree \times10^6\ut{N}\\ &\approx 17.74607553468879\times10^6\ut{N}\\ &\approx 18\times10^6\ut{N}\\ &\approx 18\ut{MN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans&=\mu_smg\cos\theta\\ &=1.609920g\cos26\degree\times10^6\ut{N}\\ &\approx 14.190090268274229 \times10^6\ut{N}\\ &\approx 14 \times10^6\ut{N}\\ &\approx 14\ut{MN}\\ \end{aligned} $$ $$\ab{c}$$ $$p_{\max}=3.6\times 10^8 \ut{N/m^2},$$ $$A=6.4\ut{cm^2},$$ $${F\over A} \lt p_{\max},$$ $$ \begin{aligned} F_{\max} &= p_{\max}\cdot A \end{aligned} $$ $$ \begin{aligned} {-mg\sin\theta+\mu_smg\cos\theta}+N\cdot F_{\max} &\gt 0\\ \end{aligned} $$ $$ \begin{aligned} N&\gt{mg\sin\theta-\mu_smg\cos\theta\over {F_{\max}}}\\ &~~~~~=\frac{43}{240} g (100 \sin 26 \degree-39 \cos 26 \degree)\\ &~~~~~\approx 15.433963829924307\\ \end{aligned} $$ $$ \begin{aligned} N_{\min}&=16 \end{aligned} $$

12-35 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Thu, 17 Oct 2024 21:03:36 +0900

$$ \begin{cases} m_A&=430\ut{kg}\\ m_B&=45.0\ut{kg}\\ \phi&=30.0\degree\\ \theta&=45.0\degree\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T&=m_Ag\\ &=430g\\ &=4.2168595\times10^3\ut{N}\\ &\approx 4.22\times10^3\ut{N}\\ &\approx 4.22\ut{kN}\\ \end{aligned} $$ $$\ab{b}$$ $$\Sigma F_{x}=0,$$ $$ \begin{aligned} N_x-T_x&=0 \end{aligned} $$ $$ \begin{aligned} N_x&=T\cos\phi\\ &=430g\cos\phi\\ &=215 \sqrt{3} g\\ &\approx 3.651907451189746\times10^3\ut{N}\\ &\approx 3.65\times10^3\ut{N}\\ &\approx 3.65\ut{kN}\\ \end{aligned} $$ $$\ab{c}$$ $$\Sigma F_{y}=0,$$ $$ \begin{aligned} 0&=N_y-T_{Ly}-T_{Ry}-m_Bg\\ \end{aligned} $$ $$ \begin{aligned} N_y &=T_{Ly}+T_{Ry}+m_Bg\\ &=T\sin\phi+T+m_Bg\\ &=430g\sin\phi+430g+45g\\ &=690g\\ &=6.7665885\times10^3\ut{N}\\ &\approx 6.77\times10^3\ut{N}\\ &\approx 6.77\ut{kN}\\ \end{aligned} $$

12-34 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 16 Oct 2024 17:31:22 +0900

$$ \begin{cases} m_A&=10\ut{kg}\\ m_B&=5.0\ut{kg}\\ \theta&=30\degree \end{cases} $$ $$0=\vec F_A+\vec F_B+\vec T,$$ $$ \begin{aligned} \tan\theta&={F_A\over F_B}\\ &={\mu m_Ag\over m_Bg}\\ \end{aligned} $$ $$ \begin{aligned} \mu&={m_B\over m_A}\tan\theta\\ &={1\over2\sqrt3}\\ &\approx 0.2886751345948129\\ &\approx 0.29\\ \end{aligned} $$

12-33 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 16 Oct 2024 16:54:37 +0900

$$ \begin{cases} M&=m_b+m_p=89.00\ut{kg}\\ T_a&=500\ut{N}\\ T_b&=700\ut{N}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\put {x\over L}=k, $$ $$\Sigma \tau=0,$$ $$ \begin{aligned} 0&=-{L\over2}m_bg-\br{x\over L}Lm_pg+LT_y \\ &={m_bg}+2km_pg-2T\sin\theta \\ \end{aligned} $$ $$ \begin{cases} 0&={m_bg}+2\br{0}m_pg-2\br{500}\sin\theta \\ 0&={m_bg}+2\br{1}m_pg-2\br{700}\sin\theta \\ 89&=m_b+m_p \end{cases} $$ $$\ab{a,b,c}$$ $$ \begin{cases} \theta&=\sin^{-1}\br{89g\over1200}\\ m_b&={445\over6}\ut{kg}\\ m_p&={89\over6}\ut{kg}\\ \end{cases} $$ $$ \begin{cases} \theta&\approx 0.8144183521806214\ut{rad}\\ m_b&\approx 74.16666666666667\ut{kg}\\ m_p&\approx 14.833333333333334\ut{kg}\\ \end{cases} $$ $$ \begin{cases} \theta&\approx 0.814\ut{rad}\\ m_b&\approx 74.2\ut{kg}\\ m_p&\approx 14.8\ut{kg}\\ \end{cases} $$

12-32 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 9 Oct 2024 18:03:45 +0900

$$ \begin{cases} \vec F_1&=8.40\i-5.70\j\ut{N}\\ \vec F_2&=16.0\i+4.10\j\ut{N}\\ \end{cases} $$ $$\ab{a,b}$$ $$ \begin{aligned} \vec F_3&=-\br{\vec F_1+\vec F_2}\\ &=-24.4\i+1.6\j\ut{N}\\ \end{aligned} $$ $$ \begin{cases} F_{3x}&=-24.4\ut{N}\\ F_{3y}&=1.6\ut{N}\\ \end{cases} $$ $$\ab{c}$$ $$ \begin{aligned} \theta&=\tan^{-1}{1.6\over-24.4}\\ &\approx 3.0761126286663285\ut{rad}\\ &\approx 3.08\ut{rad}\\ \end{aligned} $$

12-31 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 9 Oct 2024 17:48:15 +0900

$$ \begin{cases} m_1&=85.0\ut{kg}\\ m_2&=30.0\ut{kg}\\ L_2&=2.00\ut{m}\\ m_3&=20.0\ut{kg}\\ d&=0.500\ut{m}\\ L_1&=L_2+2d\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{1y}&=0\\ \Sigma \tau_1&=0\\ \Sigma F_{2y}&=0\\ \Sigma \tau_2&=0\\ \end{cases} $$ $$ \begin{cases} 0&=T_{1L}+T_{1R}-T_{2L}-T_{2R}-m_1g\\ 0&=-dT_{2L}-{L_1\over2}m_1g-\br{L_1-d}T_{2R}+L_1T_{1R} \\ 0&=T_{2L}+T_{2R}-m_2g-m_3g\\ 0&=-dm_3g-{L_2\over2}m_2g+L_2T_{2R} \\ \end{cases} $$ $$ \begin{aligned} T_{1R}&=\frac{1}{2} \br{m_1 g+m_2g+\frac{4 d }{L_1}m_3 g}\\ &={385\over6}g\\ &={15102241\over24000}\ut{N}\\ &\approx 629.2600416666667\ut{N}\\ &\approx 629\ut{N}\\ \end{aligned} $$

12-30 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 9 Oct 2024 16:57:12 +0900

$$ \begin{cases} m_1&=40.0\ut{kg}=4m\\ m_2&=10.0\ut{kg}=m\\ L&=0.800\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_A+N_B-m_1g-m_2g\\ 0&=-r_1m_1g-r_2m_2g+r_BN_B\\ \end{cases} $$ $$ \begin{cases} N_A&=\cfrac{r_B-r_1}{r_B}m_1g+\cfrac{r_B-r_2}{r_B}m_2g\\ N_B&=\cfrac{r_1}{r_B}m_1g+\cfrac{r_2}{r_B}m_2g\\ \end{cases} $$ $$ \begin{cases} r_B&={L\over2}\\ r_1&=r_2-{L\over4} \end{cases} $$ $$ \begin{cases} N_A&=\br{7-10\cdot\cfrac{r_2}{L}}mg\\ N_B&=\br{-2+10\cdot\cfrac{r_2}{L}}mg\\ \end{cases} $$ $$\ab{a,b}$$ $$ r_2={L\over2}, $$ $$ \begin{cases} N_A&=2mg=196.133\ut{N}\approx 196\ut{N}\\ N_B&=3mg=294.1995\ut{N}\approx 294\ut{N}\\ \end{cases} $$ $$\ab{c,d}$$ $$ r_2={L\over4}, $$ $$ \begin{cases} N_A&={9\over2}mg=441.29925\ut{N}\approx 441\ut{N}\\ N_B&={1\over2}mg=49.03325\ut{N}\approx 49.0\ut{N}\\ \end{cases} $$ $$\ab{e}$$ $$ \begin{aligned} N_A&=\br{7-10\cdot\cfrac{r_2}{L}}mg=0,\\ \end{aligned} $$ $$ \begin{aligned} r_2&={7\over10}L \end{aligned} $$ $$ \begin{aligned} \therefore \overline{ m_2 N_B}&=r_2-r_B\\ &={7\over10}L-{1\over2}L\\ &={L\over5}\\ &={4\over25}\ut{m}\\ &=0.160\ut{m}\\ &=16.0\ut{cm}\\ \end{aligned} $$

12-29 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 9 Oct 2024 15:52:12 +0900

$$ \begin{cases} L&=6.10\ut{m}\\ mg&=510\ut{N}\\ h&=4.00\ut{m}\\ \theta_0&=70\degree\\ \end{cases} $$ $$ \begin{cases} r_f&=h\\ r_{\text{floor}}\tan\theta&=h\\ \br{r_{mg}+\frac{L}{2}\cos\theta}\tan\theta&=h \end{cases} $$ $$ \begin{cases} N_{\text{edge }x}&=N_{\text{edge}}\sin\theta\\ N_{\text{edge }y}&=N_{\text{edge}}\cos\theta\\ \end{cases} $$ $$ \begin{aligned} f&=\mu N_{\text{floor}}, \end{aligned} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=f-N_{\text{edge }x}\\ 0&=N_{\text{edge }y}+N_{\text{floor}}-mg\\ 0&=r_ff-r_{\text{floor}}N_{\text{floor}}+r_{mg}mg\\ \end{cases} $$ $$ \begin{cases} 0&=\mu N_{\text{floor}}-N_{\text{edge}}\sin\theta\\ 0&=N_{\text{edge}}\cos\theta+N_{\text{floor}}-mg\\ 0&=h\mu N_{\text{floor}}-{h\over\tan\theta}N_{\text{floor}}+\br{{h\over \tan\theta}-{L\over2}\cos\theta}mg\\ \end{cases} $$ $$ \begin{aligned} \therefore \mu&=\frac{2 L \sin\theta \sin2 \theta}{8 h-L \sin\theta -L \sin3 \theta}\\ &=\frac{244 \sin40 \degree \cos 20 \degree}{701-122 \cos 20 \degree}\\ &\approx 0.2513510216566625\\ &\approx 0.251\\ \end{aligned} $$

12-28 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Tue, 8 Oct 2024 21:40:26 +0900

$$ \begin{cases} 2R&=2.4\ut{cm}\\ m&=670\ut{kg}\\ E_{\text{steel}}&=200\times10^9\ut{N/m^2}\\ g&= 9.80665\ut{m/s^2} \end{cases} $$ $$ {F\over A}= E{\Delta L\over L}, $$ $$ \begin{aligned} \Delta L &={ FL \over A E} \\ &={ mgL \over \pi R^2 E} \\ &={ 67g \over 2880000\pi } L\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \Delta L &={ 67g \over 2880000\pi } L\\ &=\frac{13140911}{4800000000 \pi }\ut{m}\\ &\approx 8.714337259919424\times10^{-4}\ut{m}\\ &\approx 8.7\times10^{-4}\ut{m}\\ &\approx 0.87\ut{mm}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Delta L &={ 67g \over 2880000\pi } L\\ &=\frac{2378504891}{28800000000 \pi }\ut{m}\\ &\approx 0.02628825073409026\ut{m}\\ &\approx 0.026\ut{m}\\ &\approx 2.6\ut{cm}\\ \end{aligned} $$

12-27 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Mon, 7 Oct 2024 18:22:45 +0900

$$ \begin{cases} \theta&=36.9\degree\\ \phi&=53.1\degree\\ L&=4.35\ut{m}\\ \end{cases} $$ $$L=x+y,$$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=-T_1\sin\theta+T_2\sin\phi\\ 0&=T_1\cos\theta+T_2\cos\phi-mg\\ 0&=-xT_1\cos\theta+yT_2\cos\phi\\ \end{cases} $$ $$ \begin{aligned} x&={\cos\phi\sin\theta\over\sin\br{\theta+\phi}}L\\ &=L \sin ^236.9\degree\\ &\approx 1.568194344364676\ut{m}\\ &\approx 1.57\ut{m}\\ \end{aligned} $$

12-26 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Mon, 7 Oct 2024 17:51:36 +0900

$$ \begin{cases} mg&=60\ut{N}\\ L&=3.2\ut{m}\\ F&=50\ut{N}\\ \theta&=25\degree\\ h&=2.0\ut{m}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=F+N_x-T\cos\theta\\ 0&=N_y-mg-T\sin\theta\\ 0&=hT\cos\theta-LF\\ \end{cases} $$ $$ \begin{cases} T&={F L\over h\cos\theta}\\ N_x&={F(L-h)\over h}\\ N_y&={FL\over h}\tan\theta+mg\\ \end{cases} $$ $$\ab{a}$$ $$T={F L\over h\cos\theta},$$ $$ \begin{aligned} \vec T &=-{F L\over h\cos\theta}\cos\theta\i-{F L\over h\cos\theta}\sin\theta\j\\ &=-{F L\over h}\i-{F L\over h}\tan\theta\j\\ &=-80\i-80\tan25\degree\j\ut{N}\\ &\approx -80\i-37.30461265239989\j\ut{N}\\ &\approx -80\i-37\j\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} N_x&={F(L-h)\over h}\\ N_y&={FL\over h}\tan\theta+mg\\ \end{cases} $$ $$ \begin{cases} N_x&=30\ut{N}\\ N_y&=\br{80\tan25\degree+60}\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \vec N&={F(L-h)\over h}\i+\br{{FL\over h}\tan\theta+mg}\j\\ &=30\i+\br{80\tan25\degree+60}\j\ut{N}\\ &\approx 30\i+97.3046126523999\j\ut{N}\\ &\approx 30\i+97\j\ut{N}\\ \end{aligned} $$

12-25 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Mon, 7 Oct 2024 17:18:34 +0900

$$ \begin{cases} 2R&=4.8\ut{cm}\\ L&=5.3\ut{cm}\\ m&=1500\ut{kg}\\ G&=3.0\times10^{10}\ut{N/m^2}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} {F\over A}&={mg\over \pi R^2}\\ &=\frac{7812500 g}{3 \pi }\\ &\approx 8.129045951417372\times10^6\ut{N/m^2}\\ &\approx 8.1\ut{MN/m^2}\\ \end{aligned} $$ $$\ab{b}$$ $${F\over A}=G{\Delta x\over L},$$ $$ \begin{aligned} \Delta x&={FL\over AG}\\ &={Lmg\over \pi R^2 G}\\ &=\frac{53 g}{11520000 \pi }\ut{m}\\ &\approx 1.436131451417069\times10^{-5}\ut{m}\\ &\approx 1.4\times10^{-5}\ut{m}\\ &\approx 14\ut{\mu m}\\ \end{aligned} $$

12-24 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Mon, 7 Oct 2024 16:55:01 +0900

$$ \begin{cases} \theta_1&=51.0\degree\\ \theta_2&=66.0\degree\\ m&=704\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T_A&=mg\\ &=6.9038816\times10^3\ut{N}\\ &\approx 6.90\times10^3\ut{N}\\ &\approx 6.90\ut{kN}\\ \end{aligned} $$ $$\ab{b,c}$$ $$ \begin{cases} \theta&=\theta_1+\theta_2-90\degree&=27.0\degree\\ \phi&=90\degree-\theta_2&=24.0\degree\\ \end{cases} $$ $$ \begin{aligned} 0&=\vec T_A+\vec T_B+\vec T_C\\ &=\br{-T_A\cos\phi\i-T_A\sin\phi\j}+\br{T_B\sin\theta\i+T_B\cos\theta\j}+T_C\i\\ \end{aligned} $$ $$ \begin{cases} 0&=-T_A\cos\phi+T_B\sin\theta+T_C\\ 0&=-T_A\sin\phi+T_B\cos\theta\\ \end{cases} $$ $$ \begin{cases} T_B&=\frac{\sin\phi}{\cos\theta}mg\\ T_C&=\frac{\cos\br{\theta+\phi}}{\cos\theta}mg\\ \end{cases} $$ $$ \begin{cases} T_B&\approx 3.1515612399524152\times10^3\ut{N}\\ T_C&\approx 4.876230813496548\times10^3\ut{N}\\ \end{cases} $$ $$ \begin{cases} T_B&\approx 3.15\times10^3\ut{N}\\ T_C&\approx 4.88\times10^3\ut{N}\\ \end{cases} $$ $$ \begin{cases} T_B&\approx 3.15\ut{kN}\\ T_C&\approx 4.88\ut{kN}\\ \end{cases} $$

12-23 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Fri, 4 Oct 2024 17:28:08 +0900

(풀이자 주:축에서 가해지는 힘이 없다고 가정했습니다. 그러지 않을 경우 각각의 힘을 특정할 수 없습니다.)

$$ \begin{cases} r_A&=7.0\ut{cm}\\ r_B&=4.0\ut{cm}\\ R&=4.0\ut{cm}\\ F&=220\ut{N}\\ \end{cases} $$ $$ \begin{cases} 0&=F-N_A+N_B\\ 0&=RF-r_AN_A-r_BN_B \end{cases} $$ $$ \begin{cases} N_A&=\cfrac{R+r_B}{2r_B}F\\ N_B&=\cfrac{R-r_B}{2r_B}F\\ \end{cases} $$ $$\ab{a,b}$$ $$ \begin{cases} N_A&=220\ut{N}\\ N_B&=0\\ \end{cases} $$

12-22 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 30 Jun 2024 20:00:45 +0900

$$ \put \begin{cases} A : \text{Top Ball}\\ B : \text{Bottom Ball}\\ L : \text{Left Wall}\\ R : \text{Right Wall}\\ G : \text{Ground}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{Ax}&=0\\ \Sigma F_{Ay}&=0\\ \Sigma F_{Bx}&=0\\ \Sigma F_{By}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_{ABx}-N_R\\ 0&=N_{ABy}-mg\\ 0&=N_L-N_{ABx}\\ 0&=N_G-N_{ABy}-mg\\ \end{cases} $$ $$\theta=45\degree\Rarr N_{ABx}=N_{ABy}$$ $$\ab{a,b,c,d}$$ $$ \begin{cases} N_G&=2mg\\ N_L&=mg\\ N_R&=mg\\ N_{AB}&=\sqrt2 mg \end{cases} $$

12-21 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 30 Jun 2024 19:44:20 +0900

$$ \begin{cases} L&=2.00\ut{m}\\ m&=78.0\ut{kg}\\ d_h&=3.00\ut{m}\\ d_v&=4.00\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$x_1=d_h-{L\over2},$$ $$\theta=\tan^{-1}{d_h\over d_v}$$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_x-T\sin\theta\\ 0&=N_y+T\cos\theta-mg\\ 0&=-x_1 mg+Td_h\cos\theta\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T &= {x_1 mg\over d_h\cos\theta}\\ &={5\over6 }mg\\ &=65g\\ &=637.43225\ut{N}\\ &\approx 637\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} N_x &=T\sin\theta\\ &= {x_1 mg\over d_h}\tan\theta\\ &={1\over2}mg\\ &=39g\\ &=382.45935\ut{N}\\ &\approx 382\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$N_x\gt0 \Rarr\text{Right}$$ $$\ab{d}$$ $$ \begin{aligned} N_y &=mg-T\cos\theta\\ &=mg\br{1-{x_1\over d_h}}\\ &={1\over3}mg\\ &=26g\\ &=254.9729\ut{N}\\ &\approx 255\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$N_y\gt0 \Rarr\text{Up}$$

12-20 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 30 Jun 2024 18:42:03 +0900

$$ \begin{cases} \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=3F-mg\\ 0&=x\cdot 2F+L F-{L\over2}mg\\ \end{cases} $$ $$ \begin{aligned} x&={L\over4} \end{aligned} $$

12-19 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 30 Jun 2024 18:14:32 +0900

$$ \begin{cases} m&=1.05\ut{kg}\\ r&=4.2\ut{cm}\\ L&=6.0\ut{cm}\\g&=9.80665\ut{m/s^2} \end{cases} $$ $$\theta=\tan^{-1}{r\over L}$$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N-T\sin\theta\\ 0&=T\cos\theta-mg\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T &={mg\over \cos\theta}\\ &={mg\over {L\over \sqrt{L^2+r^2}}}\\ &=\frac{21 \sqrt{149} g}{200}\\ &\approx 12.569068956048666\ut{N}\\ &\approx 13\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} N &=mg\tan\theta\\ &=mg{r\over L}\\ &=\frac{147 g}{200}\\ &=7.20788775\ut{N}\\ &\approx 7.2\ut{N}\\ \end{aligned} $$

12-18 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 30 Jun 2024 13:08:22 +0900

$$ \begin{cases} m&=13\ut{kg}\\ \theta&=55\degree\\ \phi&=25\degree\\g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=T\cos\phi-mg\sin\theta\\ 0&=-T\sin\phi-mg\cos\theta+N\\ \end{cases} $$ $$ \begin{aligned} T&={\sin\theta\over\cos\phi}mg \\ &\approx 1.1522662348306987\times10^2\ut{N}\\ &\approx 1.2\times10^2\ut{N}\\ \end{aligned} $$

12-17 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 30 Jun 2024 12:33:15 +0900

$$ \begin{cases} d&=60\ut{m}\\ L&=150\ut{m}\\ H&=7.2\ut{m}\\ z&=5.8\ut{m}\\ A&=885\ut{cm^2}\\ \rho&=2.8\ut{g/cm^3}\\ S_{\text{steel}}&=400\times10^6\ut{N/m^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Ans&=mg\\ &=\rho V g\\ &=\rho L d z g\\ &=1.4616\times10^8\ut{N}\\ &\approx 1.5\times10^8\ut{N}\\ &\approx 150\ut{MN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} {F \over NA}&={1\over2}S_{\text{steel}}\\ \end{aligned} $$ $$ \begin{aligned} N &={2F \over AS_{\text{steel}}}\\ &={2436g \over 295}\\ &\approx 80.97965898305085\\ &\approx 81\\ \end{aligned} $$

12-16 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 30 Jun 2024 10:29:01 +0900

(풀이자 주:문제의 맥락이 경첩과 잠금쇠를 잇는 1차원 막대문제로 풀라는 의도로 보입니다. 즉, 정사각형이라는 조건은 무시합니다. 만일 아니라면 답은 달라집니다.) $$ \put \begin{cases} A : \text{Hinge}\\ B : \text{Lock}\\ \end{cases} $$ $$ \begin{cases} m&=15\ut{kg}\\ L&=0.91\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$x_\com={L\over2}-0.1=0.355\ut{m}$$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_A+N_B-mg\\ 0&=-x_\com mg+LN_B\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} N_B&={x_\com \over L}mg\\ &={1065\over182}g\\ &\approx 57.38506730769231\ut{N}\\ &\approx 57\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} N_A&={L-x_\com \over L}mg\\ &={1665\over182}g\\ &\approx 89.71468269230769\ut{N}\\ &\approx 90\ut{N}\\ \end{aligned} $$

12-15 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sat, 29 Jun 2024 16:18:12 +0900

$$ \begin{cases} a&=2.5\ut{m}\\ b&=3.0\ut{m}\\ c&=1.0\ut{m}\\ F_1&=20\ut{N}\\ F_2&=10\ut{N}\\ F_3&=8.0\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=F_h-F_3\\ 0&=F_v-F_1-F_2\\ 0&=dF_v-bF_2-aF_3\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} F_h&=F_3=8.0\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} F_v&=F_1+F_2\\ &=30\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} d&={bF_2+aF_3\over F_v}\\ &={5\over3}\ut{m}\\ &\approx 1.6666666666666667\ut{m}\\ &\approx 1.7\ut{m}\\ \end{aligned} $$

12-14 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sat, 29 Jun 2024 16:10:07 +0900

$$ \begin{cases} F_h&=13.4\ut{N}\\ F_v&=162.4\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \Sigma \tau&=-{d\over3}F_t\sin45\degree+d F_v \sin10\degree+dF_h\sin80\degree\\ &=0 \end{aligned} $$ $$ \begin{aligned} F_t &=3\sqrt2\br{F_h\cos10\degree+F_v\sin10\degree}\\ &={3\sqrt2\over5}\br{67\cos10\degree+812\sin10\degree}\\ &\approx 175.63212110909666\ut{N}\\ &\approx 176\ut{N}\\ \end{aligned} $$

12-13 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sat, 29 Jun 2024 15:29:36 +0900

$$ \begin{cases} h&=3.00\ut{cm}\\ r&=6.00\ut{cm}\\ m&=0.415\ut{kg}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{cases} x&=\sqrt{r^2-y^2}\\ y&=r-h\\ \end{cases} $$ $$ \begin{aligned} \Sigma \tau&=-yF+xmg=0\\ F&={x\over y} mg\\ &=\frac{ \sqrt{h (2r-h)}}{r-h}mg\\ &={83\sqrt3\over200}g\\ &=\frac{16279039 \sqrt{3}}{4000000}\ut{N}\\ &\approx 7.049030661598812\ut{N}\\ &\approx 7.05\ut{N}\\ \end{aligned} $$

12-12 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sat, 29 Jun 2024 13:50:57 +0900

$$ \begin{cases} \theta_1&=40\degree\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=-T_1\cos\theta_1+T_2\cos\theta_2\\ 0&=T_1\sin\theta_1+T_2\sin\theta_2-mg\\ \end{cases} $$ $$ T_2={mg\cos\theta_1\over\sin\br{\theta_1+\theta_2}}$$ $$\ab{a}$$ $$ \begin{aligned} \min (T_2),\\ {\sin\br{\theta_1+\theta_2}}&=1\\ \theta_1+\theta_2&=90\degree\\ \theta_2&=50\degree\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} T_2 &={\cos\theta_1\over\sin\br{\theta_1+\theta_2}}mg\\ &=\cos40\degree mg\\ &\approx 0.766044443118978 mg \\ &\approx 0.77 mg \\ \end{aligned} $$

12-11 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sat, 29 Jun 2024 12:16:36 +0900

$$ \begin{cases} y&=2.1\ut{m}\\ x&=0.91\ut{m}\\ h&=0.25\ut{m}\\ m&=27\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} p&=y-h\\ q&=h\\ \end{cases} $$ $$ \begin{cases} A&=(0,0)\\ B&=(x,0)\\ C&=(x,y)\\ D&=(0,y)\\ P&=(0,p)\\ Q&=(0,q)\\ \end{cases} $$ $$ \begin{cases} N_{Py}&={mg\over2}\\ N_{Qy}&={mg\over2}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_{Px}+N_{Qx}\\ 0&=-pN_{Px}-qN_{Qx}-{x\over2}mg\\ \end{cases} $$ $$ \begin{cases} N_{Px}&=-{x \over 2(p-q)}mg\\ N_{Qx}&={x \over 2(p-q)}mg\\ \end{cases} $$ $$ \begin{cases} N_{Px}&=-{x \over 2(y-2h)}mg\\ N_{Qx}&={x \over 2(y-2h)}mg\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec N_P &=-{x \over 2(y-2h)}mg\i+{1\over2}mg\j\\ &=-{2457 \over 320}g\i+{27\over2}g\j\\ &=-\br{7.529668453125\i+13.2389775\j}\times10\ut{N}\\ &\approx \br{7.5\i+13\j}\times10\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec N_Q &={x \over 2(y-2h)}\i+{1\over2}mg\j\\ &={2457 \over 320}g\i+{27\over2}g\j\\ &=\br{7.529668453125\i+13.2389775\j}\times10\ut{N}\\ &\approx \br{7.5\i+13\j}\times10\ut{N}\\ \end{aligned} $$

12-10 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sat, 29 Jun 2024 09:58:41 +0900

$$ \begin{cases} S_{\text{wood}}&=50\times10^6\ut{N/m^2}\\ \rho_{\text{wood}}&=525\ut{kg/m^3}\\ A_{\text{wood}}&=d^2\\ M&=430\ut{kg}\\ a&=1.9\ut{m}\\ b&=2.5\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_x-T_x\\ 0&=N_y-Mg-mg\\ 0&=aT_x-bMg-{b\over2}mg\\ \end{cases} $$ $$ \begin{cases} N_x&={b\over2a}\br{m+2M}g\\ N_y&=\br{m+M}g\\ \end{cases} $$ $$m=\rho AL=\rho A(a^2+b^2),$$ $$ \begin{cases} N_x&=\frac{25}{76} \left(10353 A+1720\right) g\\ N_y&=\left(\frac{10353 }{2}A+430\right) g\\ \end{cases} $$ $$ \begin{aligned} N&\lt{1\over6}N_{\max}\\ \sqrt{{N_x}^2+{N_y}^2}&\lt{1\over6}AS_{\max}\\ \end{aligned} $$ $$A \gtrsim 0.0008420800400491018\ut{m^2}$$ $$ \begin{aligned} d &\gtrsim 0.029018615405444516\ut{m}\\ &\gtrsim 0.029\ut{m}\\ &\gtrsim 2.9\ut{cm}\\ \end{aligned} $$

12-9 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Fri, 28 Jun 2024 21:22:20 +0900

$$ \put \begin{cases} A : \text{Rod}\\ B : \text{Box}\\ \end{cases} $$ $$ \begin{cases} L&=1.82\ut{m}\\ W_{A}&=200\ut{N}\\ W_{B}&=300\ut{N}\\ \theta&=30.0\degree\\ T_{\max}&=412\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_x-T\cos\theta\\ 0&=N_y-W_A-W_B+T\sin\theta\\ 0&=-xW_B-{L\over2}W_A+TL\sin\theta\\ \end{cases} $$ $$\ab{a}$$ $$ T=\frac{1}{2\sin\theta}\left(\frac{2x }{L}W_B +W_A\right)\\ $$ $$ T\le T_{\max}\\ $$ $$ x\le\frac{2 T_{\max} \sin \theta - W_A}{2 W_B}L\\ $$ $$ \begin{aligned} x&\le{4823\over7500}\ut{m}\\ \end{aligned} $$ $$ \begin{aligned} x_{\max}&={4823\over7500}\ut{m}\\ &\approx 0.6430666666666667\ut{m}\\ &\approx 0.643\ut{m}\\ &\approx 64.3\ut{cm}\\ \end{aligned} $$ $$\ab{b}$$ $$N_x=T\cos\theta$$ $$ \begin{aligned} \max(N_x)&=T_{\max}\cos\theta\\ &=206\sqrt3\ut{N}\\ &\approx 356.8024663591887\ut{N}\\ &\approx 357\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$N_y=W_A+W_B-T\sin\theta$$ $$ \begin{aligned} \max(N_y)&=W_A+W_B-T\sin\theta\\ &=294\ut{N}\\ \end{aligned} $$

12-8 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Wed, 26 Jun 2024 16:28:28 +0900

$$\ab{a}$$ $$\com(1)={L\over2}\le L-a_1,$$ $$ \begin{aligned} a_1\le{L\over2}\taag1 \end{aligned} $$ $$\therefore \max(a_1)={L\over2}$$ $$\ab{b}$$ $$\com(12)={L+a_1\over2}\le L-a_2,$$ $$ \begin{aligned} a_2\le{L-a_1\over2}\taag2 \end{aligned} $$ $$\therefore \max(a_2)={L\over2}\\(a_1=0)$$ $$\ab{c}$$ $$\com(123)={L+a_1+a_2\over2}\le L-a_3,$$ $$ \begin{aligned} a_3\le{L-a_1-a_2\over2}\taag3 \end{aligned} $$ $$\therefore \max(a_3)={L\over2}\\(a_1=0, a_2=0)$$ $$\ab{d}$$ $$\com(1234)={L+a_1+a_2+a_3\over2}\le L-a_4,$$ $$ \begin{aligned} a_4\le{L-a_1-a_2-a_3\over2}\taag4 \end{aligned} $$ $$\therefore \max(a_4)={L\over2}\\(a_1=0, a_2=0,a_3=0)$$ $$\ab{e}$$ $$h=a_1+a_2+a_3+a_4,$$ $$h\le a_1+{L-a_1\over2}+{L-a_1-a_2\over2}+{L-a_1-a_2-a_3\over2}\\ $$ $$h\le{a_1+7L\over8}$$ $$ h\le{15\over16}L\\ $$ $$\therefore \max(h)={15\over16}L\\\br{a_1={L\over2}, a_2={L\over4},a_3={L\over8},a_4={L\over16}}$$

12-7 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Tue, 25 Jun 2024 21:29:00 +0900

$$ \begin{cases} E_\text{Iron}&=200\times10^9\ut{N/m^2}\\ m&=210\ut{kg}\\ R&=1.20\ut{mm}\\ p&=2.00\ut{mm}\\ L_{A}&=2.50\ut{m}\\ L_{B}&=L_{A}+p\\ \Delta L_B&=\Delta L_A-p\\ \end{cases} $$ $${F\over A}=E\cdot{\Delta L\over L},$$ $$ \begin{cases} 0&=\Sigma F=T_A+T_B-mg\\ T_A&=E (\pi R^2)\cdot{\Delta L_A\over L_A}\\ T_B&=E (\pi R^2)\cdot{\Delta L_B\over L_B}\\ \end{cases} $$ $$ \begin{aligned} \Delta L_A&=\frac{{p E\pi R^2+mg(L_A+p)}}{ E \pi R^2(2 L_A+p)}L_A,\\ &=\br{\frac{2919 g}{3201280 \pi }+\frac{5}{5002}}\ut{m}\\ &\approx 3.845903854320749\times10^{-3}\ut{m}\\ &\approx 3.85\times10^{-3}\ut{m}\\ &\approx 3.85\ut{mm}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} T_A&=E (\pi R^2)\cdot{\Delta L_A\over L_A}\\ &=\frac{90 (2919 g+3200 \pi )}{2501}\ut{N}\\ &\approx 1.391876731600904\times10^3\ut{N}\\ &\approx 1.39\times10^3\ut{N}\\ &\approx 1.39\ut{kN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} T_B &=E (\pi R^2)\cdot{\Delta L_B\over L_B}\\ &=E (\pi R^2)\cdot{\Delta L_A-p\over L_{A}+p}\\ &=\frac{1500 (175 g-192 \pi )}{2501}\ut{N}\\ &\approx 667.5197683990962\ut{N}\\ &\approx 668\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} 0&=\Sigma \tau\\ &=d_A T_A-d_BT_B, \end{aligned} $$ $$ \begin{aligned} {d_A\over d_B} &={T_B\over T_A}\\ &=\frac{\frac{1500 (175 g-192 \pi )}{2501}}{\frac{90 (2919 g+3200 \pi )}{2501}}\\ &=\frac{8750 g-9600 \pi }{8757 g+9600 \pi }\\ &\approx 0.4795825328808613\\ &\approx 0.480\\ \end{aligned} $$

12-6 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Tue, 25 Jun 2024 19:33:24 +0900

$$ \begin{cases} m_A&=11.0\ut{kg}\\ \theta&=30.0\degree\\ m_B&=7.00\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$T_A=m_Ag\sin\theta,$$ $$ \begin{aligned} \vec T_A &=-m_Ag\sin\theta\cos\theta\i-m_Bg\sin^2\theta\j\\ &=-{11\sqrt3g\over4}\i-{11g\over4}\j\taag1\\ \end{aligned} $$ $$T_B=m_Bg,$$ $$ \begin{aligned} \vec T_B &=-m_Bg\j\\ &=-7g\j\taag2\\ \end{aligned} $$ $$ \begin{aligned} \vec T_S &=-\br{\vec T_A+\vec T_B}\\ &={{11\sqrt3g\over4}\i+{39g\over4}\j}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} T_S &=\abs{{11\sqrt3g\over4}\i+{39g\over4}\j}\\ &={g\over2}\sqrt{471}\\ &\approx 106.4145795565597\ut{N}\\ &\approx 106\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \phi&=\tan^{-1}{{39g\over4}\over{11\sqrt3g\over4}}\\ &=\tan^{-1}{{13\sqrt3}\over11}\\ &\approx 1.1163690504087669\ut{rad}\\ &\approx 1.12\ut{rad}\\ \end{aligned} $$

12-5 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Tue, 25 Jun 2024 18:25:09 +0900

$$ \begin{cases} mg&=413\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_x&=0\\ \Sigma F_y&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_x-T\sin\theta\\ 0&=N_y+T\cos\theta-mg\\ 0&=mg\br{L\over2}\sin\br{2\theta}-TL\sin\theta\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T &=mg \cos\theta\\ &={413\sqrt2\over2}\ut{N}\\ &\approx 357.6684917629731\ut{N}\\ &\approx 358\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} N_x &= {1\over2} mg \sin(2\theta)\\ &={413\over4}\sqrt3\ut{N}\\ &\approx 178.8342458814866\ut{N}\\ &\approx 179\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} N_y &=mg \sin^2\theta\\ &=-{413\over4}\ut{N}\\ &= 103.25\ut{N}\\ &\approx 103\ut{N}\\ \end{aligned} $$

12-4 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Tue, 25 Jun 2024 17:45:51 +0900

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

12-3 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Tue, 18 Jun 2024 21:28:12 +0900

$$ \begin{cases} m_A&=85\ut{kg}\\ m_B&=10\ut{kg}\\ M&=\Sigma m= 95\ut{kg}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau_{D}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_C-f_D\\ 0&=N_D-Mg\\ 0&=\br{x_{AD}}m_Ag+\br{x_{BD}}m_Bg-\br{y_{CD}}N_C\\ \end{cases} $$ $$ \begin{cases} x_{AD}&={L_{AD}\over L_{CD}}x_{OD}\\ x_{BD}&={1\over 2}x_{OD}\\ y_{CD}&=\sqrt{{L_{CD}}^2-{x_{OD}}^2}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} N_C &={x_{OD}\over y_{CD}}\cdot\br{{L_{AD}\over L_{CD}}m_A+{m_B\over 2}}g\\ &={x_{OD}\over \sqrt{{L_{CD}}^2-{x_{OD}}^2}}\cdot\br{{L_{AD}\over L_{CD}}m_A+{m_B\over 2}}g\\ &={1475\over8\sqrt{231}}g\\ &\approx 118.96431792914245\ut{N}\\ &\approx 1.2\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{b,c}$$ $$ \begin{cases} f_D&=N_C\\ N_D&=Mg\\ \end{cases} $$ $$ \begin{aligned} \vec F_D &=f_D\i+N_D\j\\ &=N_C\i+Mg\j\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} F_D &=\sqrt{{N_C}^2+\br{Mg}^2}\\ &=g \sqrt{M^2+\br{x_{OD}\over y_{CD}}^2\br{{L_{AD}\over L_{CD}}m_A+{m_B\over2}}^2}\\ &=\frac{5 }{8}\sqrt{\frac{5424049}{231}}g\\ &\approx 939.1965856775718\ut{N}\\ &\approx 9.4\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \theta &=\tan^{-1}{Mg\over N_C}\\ &=\tan^{-1}{M\over {x_{OD}\over y_{CD}}\cdot\br{{L_{AD}\over L_{CD}}m_A+{m_B\over 2}}}\\ &=\tan^{-1}{152\sqrt{231}\over295}\\ &\approx 1.4437891006369301\ut{rad}\\ &\approx 1.4\ut{rad}\\ \end{aligned} $$

12-2 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Mon, 17 Jun 2024 21:30:44 +0900

$$ \begin{cases} m&=6.40\ut{kg}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{cases} F&=T_1\\ T_2&=2T_1\\ T_3&=2T_2\\ 0&=T_1+T_2+T_3-mg\\ T&=2T_3\\ \end{cases} $$ $$ \begin{aligned} T&={8\over7}mg\\ &=71.72864\ut{N}\\ &\approx 71.7\ut{N}\\ \end{aligned} $$

12-1 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Mon, 17 Jun 2024 21:23:35 +0900

$$ \begin{cases} m_Ag&=32\ut{N}\\ m_Bg&=45\ut{N}\\ \phi&=35\degree\\ \end{cases} $$ $$ \begin{cases} \Sigma \vec F_A&=0\\ \Sigma \vec F_B&=0\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{Ax}&=0\\ \Sigma F_{Bx}&=0\\ \Sigma F_{Ay}&=0\\ \Sigma F_{By}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=T_2-T_1\sin\phi\\ 0&=-T_2+T_3\sin\theta\\ 0&=T_1\cos\phi-m_Ag\\ 0&=T_3\cos\theta-m_Bg\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T_1&= {m_A g \over \cos\phi}\\ &={32 \over \cos35\degree}\\ &\approx 39.064786840366594\ut{N}\\ &\approx 39\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} T_2&= {m_A g \tan\phi}\\ &={32 \tan 35\degree}\\ &\approx 22.40664122271071\ut{N}\\ &\approx 22\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} T_3&= \sqrt{{T_2}^2+\br{m_Bg}^2}\\ &\approx 50.269847531927915\ut{N}\\ &\approx 50\ut{N}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \theta &=\tan ^{-1}\br{{T_2\over m_Bg}}\\ &\approx 0.4619865204671019\ut{rad}\\ &\approx 0.46\ut{rad}\\ \end{aligned} $$

11-60 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Mon, 17 Jun 2024 19:50:27 +0900

(풀이자주:풀이에 도형의 회전관성이 필요합니다. 관련한 내용은 별도의 링크로 분리했습니다. 해당 내용에 관한 이해는 현재과정에서의 필수는 아니니 결론만 보고 건너뛰어도 무방합니다.)

https://solutionpia.tistory.com/794

속이 채워진 얇은 원반의 회전 관성

$$\title{Rotational Inertia of Solid Disk}$$ $$ \sigma=\frac{\dd m}{\dd a}=\frac{M}{A}=\frac{M}{\pi R^2} \taag1$$ $$ \begin{aligned} l&=r\theta,\\ \dd l&=r \dd \theta\\ \end{aligned} $$ $$ \begin{aligned} \dd a&=\dd r \cdot \dd l\\ &=\dd r \cdot (r \dd\the

solutionpia.tistory.com

$$I_{\text{Solid Disk}}=\frac{1}{2}MR^2,$$ $$ \begin{cases} m_B&=m\\ m_D&=4.0m\\ R_{Bi}&=R_D\\ R_{Bf}&={1\over2}R_D\\ \omega_i&=0.345\ut{rad/s} \end{cases} $$ $$\ab{a}$$ $$\Delta \Sigma \vec L=0,$$ $$ \begin{aligned} 0 &=\Delta \Sigma \br{I \omega}\\ &=\Delta \br{ \omega\cdot\Sigma I}\\ &={ \omega_f\Sigma I_f}-{ \omega_i\Sigma I_i}\\ \end{aligned} $$ $$ \begin{aligned} \vec \omega_{f} &={\Sigma I_i\over \Sigma I_f}\vec \omega_{i}\\ &={I_{Bi}+I_{D}\over I_{Bf}+I_{D}}\vec \omega_{i}\\ &={m_B{R_{Bi}}^2+\frac{1}{2}m_D{R_D}^2\over m_B{R_{Bf}}^2+\frac{1}{2}m_D{R_D}^2}\vec \omega_{i}\\ &={2(m){R_D}^2+(4m){R_D}^2\over 2(m)\br{R_D\over2}^2+(4m){R_D}^2}\vec \omega_{i}\\ &={4\over3}\vec\omega_i\\ &=0.46\ut{rad/s} \end{aligned} $$ $$\ab{b}$$ $$ \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\ \end{cases} $$ $$ \begin{aligned} \Ans &={\Sigma E_f\over\Sigma E_i}\\ &={\Sigma \RE_f\over\Sigma \RE_i}\\ &={\Sigma \br{{1\over2}I_f{\omega_f}^2}\over\Sigma \br{{1\over2}I_i{\omega_i}^2}}\\ &={{1\over2}{\omega_f}^2\Sigma I_f\over{1\over2}{\omega_i}^2\Sigma I_i}\\ &=\br{\omega_f\over\omega_i}^2\cdot{\Sigma I_f\over \Sigma I_i}\\ &=\br{\omega_f\over\omega_i}^2\cdot{\omega_i\over\omega_f}\\ &={\omega_f\over\omega_i}\\ &={4\over3}\\ &\approx 1.3333333333333333\\ &\approx 1.3\\ \end{aligned} $$ $$\ab{c}$$ $$\text{Bug's move}$$

11-59 할리데이 11판 솔루션 일반물리학

짱세디럭스 — Sun, 16 Jun 2024 19:47:10 +0900

$$ \begin{cases} {I_f\over I_i}&=85\%\\ \end{cases} $$ $$\Delta \Sigma \vec L=0,$$ $$ \begin{aligned} 1 &={L_f\over L_i}\\ &={I_f\omega_f\over I_i\omega_i}\\ \end{aligned} $$ $${\omega_f\over \omega_i}={I_i\over I_f}\taag1$$ $$ \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\ \end{cases} $$ $$ \begin{aligned} Ans &={\RE_f\over\RE_i}\\ &={{1\over2}I_f{\omega_f}^2\over{1\over2}I_i{\omega_i}^2}\\ &={I_f\over I_i}\cdot\br{\omega_f\over \omega_i}^2\\ &={I_f\over I_i}\cdot\br{I_i\over I_f}^2\\ &={I_i\over I_f}\\ &={100\over85}\\ &={20\over17}\\ &\approx 1.1764705882352942\\ &\approx 1.2\\ \end{aligned} $$