谁能帮忙解答一下这道物理题，要详细的过程，谢谢了！ 请帮忙解答一下这道物理题！ 要有过程！
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\u518d\u6839\u636eG=mg\u6709\uff1amA\uff1amB=GA\uff1aGB=FA\uff1aFB=2\uff1a3\uff1b
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\u7b80\u5355
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分析：1.光滑指不计摩擦力。
2.设小物块对斜面的正压力为N，小物重力为G，N/G=COSθ，
N=GCOSθ=mgCOSθ
对M：Nsina=Ma a为光滑斜面体加速度
对m：以M为参考系:N+masina=mgcosa ma为惯性力
消去a，得N=(自己算吧)
(你错了，光滑斜面体不是固定的，有加速度，不是惯性参考系，不能这么算)
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绛旓細(2)鈭慚A=0 -2r.cos胃.2m.g+2r.sin胃.NB=0 (3) (A涓哄崐鐞冨渾蹇冿級鏈澶ч潤鎽╂摝鍏崇郴寮忥細 F=0.5NA (4)浠ヤ笂鍥涗釜鏂圭▼鑱旂珛姹傚緱锛 NB=F=1.5m.g 銆丯A=3m.g tan胃=4/3 L=2r.cos胃+r=2r.(3/5)+r=11r/5 *鍘熼鏃犳閫夐」锛
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