A cube of aluminum 0.80 m on a side is immersed one-half in water and one-half i

Question

A cube of aluminum 0.80 m on a side is immersed one-half in water and one-half in glycerin. What is the normal force on the cube? (The density of glycerin is 1.26 103 kg/m3.) I keep getting this wrong A cube of aluminum 0.80 m on a side is immersed one-half in water and one-half in glycerin. What is the normal force on the cube? (The density of glycerin is 1.26 103 kg/m3.) I keep getting this wrong A cube of aluminum 0.80 m on a side is immersed one-half in water and one-half in glycerin. What is the normal force on the cube? (The density of glycerin is 1.26 103 kg/m3.) I keep getting this wrong

Explanation / Answer

Assuming that by 'normal force' you mean the net downward force, that will be the weight of the aluminum minus the buoyant forces.

weight of aluminium = 0.748^3 * density of aluminum * g = 0.748^3 x 2700 g/m^3 x 9.8 m/s^2 = 1107374 N
Buoyant force from water will be = 1/3 * 0.748^3 * 1000 g.m^3 * g = 1367N
Buoyant force of glycerin will be = 2/3 *0.748^3 * 1260 g/m^3 * g = 3444.8N

Total Buoyant force = 1367 + 3444.8 = 4811.8 N

Normal force = 1107374 - 4811.8 = 1102562.2 N

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