11판/12. 평형과 탄성

12-9 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 6. 28. 21:22
put {A:RodB:Box \put \begin{cases} A : \text{Rod}\\ B : \text{Box}\\ \end{cases} {L=1.82[m]WA=200[N]WB=300[N]θ=30.0°Tmax=412[N] \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} {ΣFx=0ΣFy=0Στ=0 \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} {0=NxTcosθ0=NyWAWB+Tsinθ0=xWBL2WA+TLsinθ \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} (a)\ab{a} T=12sinθ(2xLWB+WA) T=\frac{1}{2\sin\theta}\left(\frac{2x }{L}W_B +W_A\right)\\ TTmax T\le T_{\max}\\ x2TmaxsinθWA2WBL x\le\frac{2 T_{\max} \sin \theta - W_A}{2 W_B}L\\ x48237500[m] \begin{aligned} x&\le{4823\over7500}\ut{m}\\ \end{aligned} xmax=48237500[m]0.6430666666666667[m]0.643[m]64.3[cm] \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} (b)\ab{b} Nx=TcosθN_x=T\cos\theta max(Nx)=Tmaxcosθ=2063[N]356.8024663591887[N]357[N] \begin{aligned} \max(N_x)&=T_{\max}\cos\theta\\ &=206\sqrt3\ut{N}\\ &\approx 356.8024663591887\ut{N}\\ &\approx 357\ut{N}\\ \end{aligned} (c)\ab{c} Ny=WA+WBTsinθN_y=W_A+W_B-T\sin\theta max(Ny)=WA+WBTsinθ=294[N] \begin{aligned} \max(N_y)&=W_A+W_B-T\sin\theta\\ &=294\ut{N}\\ \end{aligned}