12-35 할리데이 11판 솔루션 일반물리학 {mA=430[kg]mB=45.0[kg]ϕ=30.0°θ=45.0° \begin{cases} m_A&=430\ut{kg}\\m_B&=45.0\ut{kg}\\\phi&=30.0\degree\\\theta&=45.0\degree\\\end{cases} ⎩⎨⎧mAmBϕθ=430[kg]=45.0[kg]=30.0°=45.0°{Tx=TcosϕTy=Tsinϕ \begin{cases} T_x&=T\cos\phi\\T_y&=T\sin\phi\\\end{cases} {TxTy=Tcosϕ=Tsinϕ{ΣFx=0ΣFy=0Στ=0 \begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases} ⎩⎨⎧ΣFxΣFyΣτ=0=0=0{0=Nx−Tx0=Ny−Ty−mAg0=TxLsinθ−(Ty+mAg)Lcosθ−mBg(L2)cosθ \begin{cases} 0&=N_x-T_x\\0&=N_y-T_y-m_Ag\\0&=T_xL\sin\theta-\br{T_y+m_Ag}L\cos\theta-m_Bg\br{L\over2}\cos\theta\\\end{cases} ⎩⎨⎧000=Nx−Tx=Ny−Ty−mAg=TxLsinθ−(Ty+mAg)Lcosθ−mBg(2L)cosθ(a)\ab{a}(a)$$ \beg.. 11판/12. 평형과 탄성 2024.10.17
12-34 할리데이 11판 솔루션 일반물리학 {mA=10[kg]mB=5.0[kg]θ=30° \begin{cases} m_A&=10\ut{kg}\\m_B&=5.0\ut{kg}\\\theta&=30\degree\end{cases} ⎩⎨⎧mAmBθ=10[kg]=5.0[kg]=30°0=F⃗A+F⃗B+T⃗,0=\vec F_A+\vec F_B+\vec T,0=FA+FB+T,tanθ=FAFB=μmAgmBg \begin{aligned}\tan\theta&={F_A\over F_B}\\&={\mu m_Ag\over m_Bg}\\\end{aligned} tanθ=FBFA=mBgμmAgμ=mBmAtanθ=123≈0.2886751345948129≈0.29 \begin{aligned}\mu&={m_B\over m_A}\tan\theta\\&={1\over2\sqrt3}\\&\approx 0.2886751345948129\\&\approx 0.29\\\end{aligned} μ=mAmBtanθ=231≈0.2886751345948129≈0.29 11판/12. 평형과 탄성 2024.10.16
12-33 할리데이 11판 솔루션 일반물리학 {M=mb+mp=89.00[kg]Ta=500[N]Tb=700[N]g=9.80665[m/s2] \begin{cases} M&=m_b+m_p=89.00\ut{kg}\\T_a&=500\ut{N}\\T_b&=700\ut{N}\\g&=9.80665\ut{m/s^2}\end{cases} ⎩⎨⎧MTaTbg=mb+mp=89.00[kg]=500[N]=700[N]=9.80665[m/s2]put xL=k,\put {x\over L}=k, put Lx=k,Στ=0,\Sigma \tau=0,Στ=0,0=−L2mbg−(xL)Lmpg+LTy=mbg+2kmpg−2Tsinθ \begin{aligned}0&=-{L\over2}m_bg-\br{x\over L}Lm_pg+LT_y \\&={m_bg}+2km_pg-2T\sin\theta \\\end{aligned} 0=−2Lmbg−(Lx)Lmpg+LTy=mbg+2kmpg−2Tsinθ{0=mbg+2(0)mpg−2(500)sinθ0=mbg+2(1)mpg−2(700)sinθ89=mb+mp \begin{cases} 0&={m_bg}+2\br{0}m_pg-2\br{500}\sin\theta \\0&={m_bg}+2\br{1}m_pg-2\br{700}\sin\theta \\89&=m_b+m_p\end{cases} ⎩⎨⎧0089=mbg+2(0)mpg−2(500)sinθ=mbg+2(1)mpg−2(700)sinθ=mb+mp$$\ab{a,b,.. 11판/12. 평형과 탄성 2024.10.16
12-32 할리데이 11판 솔루션 일반물리학 {F⃗1=8.40i^−5.70j^[N]F⃗2=16.0i^+4.10j^[N] \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} {F1F2=8.40i^−5.70j^[N]=16.0i^+4.10j^[N](a,b)\ab{a,b}(a,b)F⃗3=−(F⃗1+F⃗2)=−24.4i^+1.6j^[N] \begin{aligned}\vec F_3&=-\br{\vec F_1+\vec F_2}\\&=-24.4\i+1.6\j\ut{N}\\\end{aligned} F3=−(F1+F2)=−24.4i^+1.6j^[N]{F3x=−24.4[N]F3y=1.6[N] \begin{cases} F_{3x}&=-24.4\ut{N}\\F_{3y}&=1.6\ut{N}\\\end{cases} {F3xF3y=−24.4[N]=1.6[N](c)\ab{c}(c)$$ \begin{aligned}\theta&=\tan^{-1}{1.6\over-24.4}\\&\approx 3.0761126286663285\ut{rad}\\&\approx 3.08\ut{rad}\\\end{ali.. 11판/12. 평형과 탄성 2024.10.09
12-31 할리데이 11판 솔루션 일반물리학 {m1=85.0[kg]m2=30.0[kg]L2=2.00[m]m3=20.0[kg]d=0.500[m]L1=L2+2d \begin{cases} m_1&=85.0\ut{kg}\\m_2&=30.0\ut{kg}\\L_2&=2.00\ut{m}\\m_3&=20.0\ut{kg}\\d&=0.500\ut{m}\\L_1&=L_2+2d\\\end{cases} ⎩⎨⎧m1m2L2m3dL1=85.0[kg]=30.0[kg]=2.00[m]=20.0[kg]=0.500[m]=L2+2d{ΣF1y=0Στ1=0ΣF2y=0Στ2=0 \begin{cases} \Sigma F_{1y}&=0\\\Sigma \tau_1&=0\\\Sigma F_{2y}&=0\\\Sigma \tau_2&=0\\\end{cases} ⎩⎨⎧ΣF1yΣτ1ΣF2yΣτ2=0=0=0=0$$ \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.. 11판/12. 평형과 탄성 2024.10.09
12-30 할리데이 11판 솔루션 일반물리학 {m1=40.0[kg]=4mm2=10.0[kg]=mL=0.800[m]g=9.80665[m/s2] \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} ⎩⎨⎧m1m2Lg=40.0[kg]=4m=10.0[kg]=m=0.800[m]=9.80665[m/s2]{ΣFy=0Στ=0 \begin{cases} \Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases} {ΣFyΣτ=0=0{0=NA+NB−m1g−m2g0=−r1m1g−r2m2g+rBNB \begin{cases} 0&=N_A+N_B-m_1g-m_2g\\0&=-r_1m_1g-r_2m_2g+r_BN_B\\\end{cases} {00=NA+NB−m1g−m2g=−r1m1g−r2m2g+rBNB{NA=rB−r1rBm1g+rB−r2rBm2gNB=r1rBm1g+r2rBm2g \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} ⎩⎨⎧NANB=rBrB−r1m1g+rBrB−r2m2g=rBr1m1g+rBr2m2g$$ \begin{.. 11판/12. 평형과 탄성 2024.10.09
12-29 할리데이 11판 솔루션 일반물리학 {L=6.10[m]mg=510[N]h=4.00[m]θ0=70° \begin{cases} L&=6.10\ut{m}\\mg&=510\ut{N}\\h&=4.00\ut{m}\\\theta_0&=70\degree\\\end{cases} ⎩⎨⎧Lmghθ0=6.10[m]=510[N]=4.00[m]=70°{rf=hrfloortanθ=h(rmg+L2cosθ)tanθ=h \begin{cases} r_f&=h\\r_{\text{floor}}\tan\theta&=h\\\br{r_{mg}+\frac{L}{2}\cos\theta}\tan\theta&=h\end{cases} ⎩⎨⎧rfrfloortanθ(rmg+2Lcosθ)tanθ=h=h=h{Nedge x=NedgesinθNedge y=Nedgecosθ \begin{cases} N_{\text{edge }x}&=N_{\text{edge}}\sin\theta\\N_{\text{edge }y}&=N_{\text{edge}}\cos\theta\\\end{cases} {Nedge xNedge y=Nedgesinθ=Nedgecosθf=μNfloor, \begin{aligned}f&=\mu N_{\text{floor}},\end{aligned} f=μNfloor,$$ \.. 11판/12. 평형과 탄성 2024.10.09
12-28 할리데이 11판 솔루션 일반물리학 {2R=2.4[cm]m=670[kg]Esteel=200×109[N/m2]g=9.80665[m/s2] \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} ⎩⎨⎧2RmEsteelg=2.4[cm]=670[kg]=200×109[N/m2]=9.80665[m/s2]FA=EΔLL, {F\over A}= E{\Delta L\over L}, AF=ELΔL,ΔL=FLAE=mgLπR2E=67g2880000πL \begin{aligned}\Delta L&={ FL \over A E} \\&={ mgL \over \pi R^2 E} \\&={ 67g \over 2880000\pi } L\\\end{aligned} ΔL=AEFL=πR2EmgL=2880000π67gL(a)\ab{a}(a)$$ \begin{aligned}\Delta L&={ 67g \over 2880000\pi } L\\&=\frac{13140911}{4800000000 \pi }\ut{m}\\&\approx.. 11판/12. 평형과 탄성 2024.10.08
12-27 할리데이 11판 솔루션 일반물리학 {θ=36.9°ϕ=53.1°L=4.35[m] \begin{cases} \theta&=36.9\degree\\\phi&=53.1\degree\\L&=4.35\ut{m}\\\end{cases} ⎩⎨⎧θϕL=36.9°=53.1°=4.35[m]L=x+y,L=x+y,L=x+y,{ΣFx=0ΣFy=0Στ=0 \begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases} ⎩⎨⎧ΣFxΣFyΣτ=0=0=0{0=−T1sinθ+T2sinϕ0=T1cosθ+T2cosϕ−mg0=−xT1cosθ+yT2cosϕ \begin{cases} 0&=-T_1\sin\theta+T_2\sin\phi\\0&=T_1\cos\theta+T_2\cos\phi-mg\\0&=-xT_1\cos\theta+yT_2\cos\phi\\\end{cases} ⎩⎨⎧000=−T1sinθ+T2sinϕ=T1cosθ+T2cosϕ−mg=−xT1cosθ+yT2cosϕ$$ \begin{aligned}x&={\cos\phi\sin\theta\over\sin\br{\theta+\phi}}L\\&=L \sin ^236.9\degree\\.. 11판/12. 평형과 탄성 2024.10.07
12-26 할리데이 11판 솔루션 일반물리학 {mg=60[N]L=3.2[m]F=50[N]θ=25°h=2.0[m] \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} ⎩⎨⎧mgLFθh=60[N]=3.2[m]=50[N]=25°=2.0[m]{ΣFx=0ΣFy=0Στ=0 \begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases} ⎩⎨⎧ΣFxΣFyΣτ=0=0=0{0=F+Nx−Tcosθ0=Ny−mg−Tsinθ0=hTcosθ−LF \begin{cases} 0&=F+N_x-T\cos\theta\\0&=N_y-mg-T\sin\theta\\0&=hT\cos\theta-LF\\\end{cases} ⎩⎨⎧000=F+Nx−Tcosθ=Ny−mg−Tsinθ=hTcosθ−LF{T=FLhcosθNx=F(L−h)hNy=FLhtanθ+mg \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} ⎩⎨⎧TNxNy=hcosθFL=hF(L−h)=hFLtanθ+mg$$\ab.. 11판/12. 평형과 탄성 2024.10.07