如图所示,在动力小车上固定一直角硬杆ABC,分别系在水平直杆AB两端的轻弹簧和细线将小球P悬吊起来,轻弹簧的劲度系数为k
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设细线的拉力为T,则有

Tsinθ=ma,Tcosθ+F=mg,F=kx

联立解得:x=m(g-acotθ)/k

讨论:

(1)若a<g/cotθ,则弹簧伸长x=m(g-acotθ)/k

(2)若a=g/cotθ,则弹簧伸长x=0

(3)若a>g/cotθ,则弹簧压缩x=m(acotθ-g)/k