11판/4. 2차원 운동과 3차원 운동

4-36 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 7. 14:23
{θ1=30.0°θ2=50.0°θ3=80.0°d1=5.00[m]d2=9.20[m]d3=13.5[m] \begin{cases} \theta_1=30.0\degree\\ \theta_2=50.0\degree\\ \theta_3=80.0\degree\\ d_1=5.00\ut{m}\\ d_2=9.20\ut{m}\\ d_3=13.5\ut{m}\\ \end{cases} ϕ2=180°(θ2θ1)=180°(50°30°)=160° \begin{aligned} \phi_2 &= 180\degree-(\theta_2-\theta_1)\\ &=180\degree-(50\degree-30\degree)\\ &=160\degree\\ \end{aligned} ϕ3=ϕ2θ3+180°=160°80°+180°=260° \begin{aligned} \phi_3 &= \phi_2-\theta_3+180\degree\\ &=160\degree-80\degree+180\degree\\ &=260\degree \end{aligned} {d1=5cos30°i^+5sin30°j^d2=9.2cos160°i^+9.2sin160°j^d3=13.5cos260°i^+13.5sin260°j^ \begin{cases} \vec d_1 &= 5\cos30\degree\i+5\sin30\degree\j\\ \vec d_2 &= 9.2\cos160\degree\i+9.2\sin160\degree\j\\ \vec d_3 &= 13.5\cos260\degree\i+13.5\sin260\degree\j\\ \end{cases} {d1=532i^+52j^d2=465cos20°i^+465sin20°j^d3=272sin10°i^+272cos10°j^ \begin{cases} \vec d_1 &= \cfrac{5\sqrt3}{2}\i+\cfrac{5}{2}\j\\ \vec d_2 &= -\cfrac{46}{5}\cos20\degree\i+\cfrac{46}{5}\sin20\degree\j\\ \vec d_3 &= -\cfrac{27}{2}\sin10\degree\i+\cfrac{27}{2}\cos10\degree\j\\ \end{cases} Σd6.659295490811724i^     +18.94148998426096j^6.66i^+18.9j^ \begin{aligned} \Sigma \vec d&\approx -6.659295490811724\i\\ &~~~~~+18.94148998426096\j\\ &\approx -6.66\i+18.9\j\\ \end{aligned}