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2010年广州亚运会上,“吊环王”陈一冰成功捍卫荣誉,以16.075分摘得金牌成功卫冕,其中有一个高难度的动作就是先双手撑住吊环,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的过程中,吊环的两根绳的拉力F T(两个拉力大小相等)及它们的合力F的大情况为( )
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
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAACAAAAAgCAYAAABzenr0AAAA5ElEQVRYR2NkGGDAOMD2M+B0wKpVq9huv/jbxPD/fxwDA4MkhQ59zsDIuEhVgrkuLCzsF7JZOB3QOml5B8P//+UUWoyqnZGxszovsoI4B0xc9gyfzxmZmb2rcsK3IRvWMWm599///7fgcfTz6vwoKWId8B+f75kZGX0q8iK3kugAhur8KJRQxx0FE5fhdQC5UTPqgNEQGA2B0RAYDYHREBgNgaETAuTW96TqG7yt4tbRFhGRIQBqHTP9+8dIoDUMTxqk5AK8zXKYiaDWMYhNpANIaJYPdMdkwLtmpOZnctUPeDkAAP6tmiH1ggTLAAAAAElFTkSuQmCC)
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