滑轮组机械效率有什么规律?

就是钩码数越多,机械效率越大之类的
2025-04-08 10:40:14
推荐回答(2个)
回答1:

1.当滑轮组沿水平方向拉物体时,忽略对动滑轮的重力做功:
利用滑轮组沿水平方向拉物体时,总功为W总=F*S ,有用功为W有=F拉*L ,若物体沿水平方向做匀速直线运动,则物体受平衡力作用,应有:F拉=f,而s=nL, 这时滑轮组的机械效率为 ¶=W有/ W总*100%= f/(N*f)*100%.
2.当使用滑轮组沿竖直方向拉物体时:
(1)考虑动滑轮自重,摩擦阻力及绳重等所有额外功的情形 W总=F*S. 有用功为 W有=F拉*L,因为物体沿竖直方向被匀速提起,则物体受平衡力作用,则有:F拉=G物,s=nh.所以有 ¶=W有/ W总*100%=G物/(N*F)*100%
(2)考虑动滑轮自重,但不计轮轴间摩擦阻力及绳重等额外功的情形。此时由于物体沿竖直方向被匀速提起,物体受平衡力作用,即:因此滑轮组做的有用功为W有= G物*h ,总功为 W总= W有+W额,所以有 ¶=W有/ W总*100%=W有/(W有+W额)*100%=G物/(G物+G动)*100%.
(3)不计动滑轮自重,轮轴间摩擦阻力及绳重所做的额外功的情形: 这时滑轮组做的有用功 W有= G物*h,W额=0,W总=F*s=n*F*h,故有W总=W有,N*F=G物,F=G物/h,
滑轮组的机械效率为100%

注意:(1) 当不同滑轮组提升相同的物重时,由于绳子股数越多时,G动 越大,将使滑轮组的机械效率降低
(2) 用同一滑轮组提升不同的物重时,由于额外功一定,当所提升的物重越大时,滑轮组所做的有用功越多,机械效率越高

(3)绳子的自由端和动滑轮移动的距离之间有一定的几何关系。一般情况下为:S=nL 或s=nh(其中:s是指绳的自由端移动的距离,L是指动滑轮或物体在水平方向移动的距离,n是动力拉动滑轮绳子的股数,h是物体上升的高度)。

回答2:

滑轮组的机械效率

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