A weightless box with a 0.4m by 0.4m square base is floating horizontally on the

Question

A weightless box with a 0.4m by 0.4m square base is floating horizontally on the
water surface. It contains a 0.2m diameter steel sphere located at its center. a) How deep will the box sink due to the weight of the steel ball with a density of 8000kg/m3? b) Next, the ball is thrown into the water, but it is held by a weightless wire at a small distance below the center of the floating box. How deep will the box sink in this case?

Here is the link with the photo I couldn't get it to upload. Thanks!

http://s1077.photobucket.com/albums/w462/portland36/?action=view&current=EM3.jpg

Explanation / Answer

Weight of the ball = 4/3 * pi*r^3 * density = 0.03351 m^3 * 8000 = 268.08 kg weight of ball = weight of water displaced = 268.08 kg Volume of water to be displaced = 268.08/1000 kg/m^3 = 0.26808 m^3 (a)depth sunk = 0.26808/(0.4*0.4) = 1.68 m (b) Here, sphere already displaces water of its own volume = 0.003351 m^3 So, volume of water displaced by box = 0.26808 - 0.03351 = 0.23457 kg depth sunk = 0.23457/(0.4*0.4) = 1.47 m

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