11판/5. 힘과 운동 I

5-35 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 19. 15:46
{m=4.00[kg]F1=18.0i^[N]a=14.0[m/s2],θa=20.0° \begin{cases} m&=4.00\ut{kg}\\ \vec F_1 &= 18.0\i\ut{N}\\ a&=14.0\ut{m/s^2},\theta_a=20.0\degree\\ \end{cases} (a)\ab{a} ΣF=ma,\Sigma \vec F = m\vec a, {F1+F2=maF2=maF1=4(14cos20°i^+14sin20°)18i^=(56cos20°18)i^+56sin20°j^[N]34.6i^+19.2j^[N] \begin{cases} \vec F_1 +\vec F_2&= m \vec a\\ \vec F_2&= m \vec a-\vec F_1\\ &=4(14\cos20\degree\i+14\sin20\degree)-18\i\\ &=(56\cos20\degree-18)\i+56\sin20\degree\j\ut{N}\\ &\approx 34.6\i+19.2\j\ut{N} \end{cases} (b)\ab{b} F2=((56cos20°18)2+(56sin20°)2=2865504cos20°[N]39.56740674463779[N]39.6[N] \begin{aligned} F_2 &=\sqrt{\((56\cos20\degree-18\)^2+\(56\sin20\degree\)^2}\\ &=2 \sqrt{865-504\cos20\degree}\ut{N}\\ &\approx 39.56740674463779\ut{N}\\ &\approx 39.6\ut{N}\\ \end{aligned} (c)\ab{c} θ=tan156sin20°56cos20°180.5052923343710929[rad]0.505[rad] \begin{aligned} \theta&=\tan^{-1}\frac{56\sin20\degree}{56\cos20\degree-18}\\ &\approx 0.5052923343710929\ut{rad}\\ &\approx 0.505\ut{rad}\\ \end{aligned}