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物理
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身高相同的甲、乙俩人,用长1.6m的扁担共同抬一桶重为1800N的货物,甲在前、乙在后,货物放在扁担的中间.走了一会儿,乙为了照顾甲,将货物向自己这一边移动了0.3m,问现在甲、乙肩上对扁担所施加的力各多大.
人气:366 ℃ 时间:2020-03-23 21:58:06
解答
设甲受力为F1,乙受力为F2.甲力臂长为L1,乙力臂长为L2.
则 F1+F2=1800 (N)
L1=1.6/2+0.3=1.1(m)
L2=1.6/2-0.3=0.5(m)
F1L1=F2L2
F2=F1L1/L2=F1×1.1/0.5=2.2F1
F1+2.2F1=1800
F1=1800÷(1+2.2)=562.5(N)
F2=1800-F1=1800-562.5=1237.5(N)
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