審題分別對(duì)重物和懸掛點(diǎn)A做受力分析如圖乙、丙所示,其主要方程是牛頓第二定律.切入點(diǎn)應(yīng)先求出懸掛重物繩子的拉力T,令FAB=Fm=1000N,由正弦定理,得∴T=(sin75°)/(sin45°)·FAB=(sin(45°+35°))/(sin45°)×1000N=1365N拐彎點(diǎn)根據(jù)牛頓第二定律 (共 157 字) [閱讀本文] >>
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 審題分別對(duì)重物和懸掛點(diǎn)A做受力分析如圖乙、丙所示,其主要方程是牛頓第二定律.切入點(diǎn)應(yīng)先求出懸掛重物繩子的拉力T,令FAB=Fm=1000N,由正弦定理,得∴T=(sin75°)/(sin45°)·FAB=(sin(45°+35°))/(sin45°)×1000N=1365N拐彎點(diǎn)根據(jù)牛頓第二定律 (共 157 字) [閱讀本文] >>