物体在同一平面内受F1、F2、F3三个共点力的作用处于平衡状态,当把

物体在同一平面内受F1F2F3三个共点力的作用处于平衡状态,当把F3的方向在同平面内旋转60°时(F3大小不变,F1F2大小方向都不变),则三力的合力大小为(  )

A0       BF3     C F       D2F3

答案

解:根据共点力平衡的特点可知,

F1F2、的合力与F3大小相等,方向相反,

当把F3的方向在同平面内旋转60°时,

就相当于计算两个大小相等的力,在夹角为120°时的合力的大小,

根据平行四边形定则可知,此时合力的大小为F3

故选:B

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