Since the object is floating, I know to usegV=g(V-Vu), and from there I think wh
ID: 1727946 • Letter: S
Question
Since the object is floating, I know to usegV=g(V-Vu), and from there I think what Ineed to do is solve for Vu because that will be the volume ofthe block that is submersed in the water. However,I am not sure, and I would really like to understand thisproblem. So if you could illustrate every step I wouldgreatly appreciate it. Thank You. A block of wood has a density of 850kg/m^3, and water hensityis 1000 kg/^3. a. What portion of the volume of a wood cube .15 meter on eachside will be above the water line as it floats? b. what weight will the block support such that it just sinkslevel with the surface of the water? Since the object is floating, I know to usegV=g(V-Vu), and from there I think what Ineed to do is solve for Vu because that will be the volume ofthe block that is submersed in the water. However,I am not sure, and I would really like to understand thisproblem. So if you could illustrate every step I wouldgreatly appreciate it. Thank You. A block of wood has a density of 850kg/m^3, and water hensityis 1000 kg/^3. a. What portion of the volume of a wood cube .15 meter on eachside will be above the water line as it floats? b. what weight will the block support such that it just sinkslevel with the surface of the water?Explanation / Answer
Density of wood, = 850 kg/m^3 Density of water, ' = 1000 kg/m^3 Side of the wooden block, s = 0.15 m (a) Weight of the wooden block = weight of theliquid displaced by immersed portion V g = ' V' g V = ' V' Volume the wooden block under water, V' = ( / ' ) V = ( 850 / 1000 ) * V = 0.85 * s3 = 0.85 * 0.15 3 = 0.00287 m^3 Volume of the wooden block above water = V -V' = V - 0.85 V = 0.15 V = 0.15 * 0.15 3 = 0.00051 m^3 (b) Weight of the wooden block + weight of thesupport = Weight of the liquid displaced by whole woodenblock V g + W = ' V g Weight to be supported, W = ( ' - ) V g = ( 1000 - 850 ) * 0.153 * 9.8 = 4.96 N