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24. (10分)跨学科 曹冲称象 (2024·浙江绍兴期末)某动物园根据杠杆原理$G_{1}\cdot L_{1}= G_{2}\cdot L_{2}$上演了一幕现代版“曹冲称象”,具体做法如下：如图所示,在一根已经水平地挂在起重机上的钢梁的左右两边分别挂上一根弹簧秤(质量可以忽略不计)和装有大象的铁笼,其中弹簧秤与钢梁之间的距离为$L_{1}= 6m$,装有大象的铁笼与钢梁之间的距离为$L_{2}= 0.2m$,已知当钢梁又呈水平状态(铁笼已经离地)时,弹簧秤显示的读数为$G_{1}= 1200N$,装有大象的铁笼及其挂钩的总质量为$G_{2}$.
(1)求装有大象的铁笼及其挂钩的总质量$G_{2}$；
(2)若装大象的铁笼固定不动,装有大象的铁笼及其挂钩的总质量不变,用式子表示$G_{1}与L_{1}$的关系；
(3)当$L_{1}= 8m$时,求弹簧秤的显示读数$G_{1}$,当弹簧秤的显示读数$G_{1}= 1800N$时,求$L_{1}$.
(1)求装有大象的铁笼及其挂钩的总质量$G_{2}$；
(2)若装大象的铁笼固定不动,装有大象的铁笼及其挂钩的总质量不变,用式子表示$G_{1}与L_{1}$的关系；
(3)当$L_{1}= 8m$时,求弹簧秤的显示读数$G_{1}$,当弹簧秤的显示读数$G_{1}= 1800N$时,求$L_{1}$.
答案：
(1)把L_{1}=6m,L_{2}=0.2m,G_{1}=1200N代入G_{1}·L_{1}=G_{2}·L_{2},得1200×6=0.2G_{2},解得G_{2}=36000N.故装有大象的铁笼及其挂钩的总质量G_{2}为36000N.
(2)
∵G_{1}·L_{1}=G_{2}·L_{2}=1200×6,
∴$G_{1}=\frac{7200}{L_{1}}(L_{1}＞0).(3)$把L_{1}=8m代入$G_{1}=\frac{7200}{L_{1}},$得$G_{1}=\frac{7200}{8}=900(N);$把G_{1}=1800N代入$G_{1}=\frac{7200}{L_{1}},$得$1800=\frac{7200}{L_{1}},$解得L_{1}=4m.
(1)把L_{1}=6m,L_{2}=0.2m,G_{1}=1200N代入G_{1}·L_{1}=G_{2}·L_{2},得1200×6=0.2G_{2},解得G_{2}=36000N.故装有大象的铁笼及其挂钩的总质量G_{2}为36000N.
(2)
∵G_{1}·L_{1}=G_{2}·L_{2}=1200×6,
∴$G_{1}=\frac{7200}{L_{1}}(L_{1}＞0).(3)$把L_{1}=8m代入$G_{1}=\frac{7200}{L_{1}},$得$G_{1}=\frac{7200}{8}=900(N);$把G_{1}=1800N代入$G_{1}=\frac{7200}{L_{1}},$得$1800=\frac{7200}{L_{1}},$解得L_{1}=4m.
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