2024/10/16 2

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} 0=FA+FB+T,0=\vec F_A+\vec F_B+\vec T,tanθ=FAFB=μmAgmBg \begin{aligned}\tan\theta&={F_A\over F_B}\\&={\mu m_Ag\over m_Bg}\\\end{aligned} μ=mBmAtanθ=1230.28867513459481290.29 \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

{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} put xL=k,\put {x\over L}=k, Στ=0,\Sigma \tau=0,0=L2mbg(xL)Lmpg+LTy=mbg+2kmpg2Tsinθ \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=mbg+2(0)mpg2(500)sinθ0=mbg+2(1)mpg2(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} $$\ab{a,b,..