有物体落入水中，打捞人员到水下，用一根绳子系住它，试图将其打捞出水．若物体的密度为3× 103 kg/m3，体积为10 dm3，g取10 N/kg.求：
(1)要将物体打捞出水面，所用的绳子至少应能承受多大的拉力？
(2)在从水中向上提升物体的过程中，绳子承受的最小拉力是多大？
解：(1)物体的承力
G =
gV=3×
kg/
×10N／kg×10×
=300 N
当物体离开水面时，物体受到的拉力等于物体受到的重力，所以F=G=300N
(2)物体在水中受到的浮力
F浮=
gV= 1.0×
kg／
×10 N/kg×10×
= 100 N
从水中向上提升物体时，绳子承受的最小拉力为
=G-
= 300 N- 100 N=200 N
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