11판/5. 힘과 운동 I

5-36 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 19. 16:11
{m=120[kg]θ=35.0° \begin{cases} m&=120\ut{kg}\\ \theta&=35.0\degree\\ \end{cases} ΣF=0Δv=0,\Sigma \vec F = 0 \Harr \Delta \vec v=0, {ΣFx=0ΣFy=0 \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} {Fcosθmgsinθ=0FNFsinθmgcosθ=0 \begin{cases} F\cos\theta-mg\sin\theta &= 0\\ F_N-F\sin\theta -mg\cos\theta&= 0\\ \end{cases} (a)\ab{a} F=mgsinθcosθ=mgtanθ=120gtan35°824.0028305501099[N]824[N] \begin{aligned} F&=\frac{mg\sin\theta}{\cos\theta}\\ &=mg\tan\theta\\ &=120g\tan35\degree\\ &\approx 824.0028305501099\ut{N}\\ &\approx 824\ut{N}\\ \end{aligned} (b)\ab{b} FN=Fsinθ+mgcosθ=mgtanθsinθ+mgcosθ=mgcosθ=120gcos35°1436.605094505304[N]1.44×103[N]1.44[kN] \begin{aligned} F_N &= F\sin\theta+mg\cos\theta\\ &= mg\tan\theta\sin\theta+mg\cos\theta\\ &= \frac{mg}{\cos\theta}\\ &= \frac{120g}{\cos35\degree}\\ &\approx 1436.605094505304\ut{N}\\ &\approx 1.44\times 10^3\ut{N}\\ &\approx 1.44\ut{kN}\\ \end{aligned}