题干

2010年广东亚运会,我国运动员陈一冰勇夺吊环冠军,为中国体育军团勇夺第一金,其中有一个高难度的动作就是先双手撑住吊环(设开始时两绳与肩同宽),然后身体下移,双臂缓慢张开到如图所示位置,则在两手之间的距离增大过程中,吊环的两根绳的拉力FT(两个拉力大小相等)及它们的合力F的大小变化情况为(  )

A:FT增大,F不变

B:FT增大,F增大

C:FT增大,F减小

D:FT减小,F不变

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答案(点此获取答案解析)

A

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