Wouldn\'t be so lucky! Try calculating the normal force when the level of mercur

Question

Wouldn't be so lucky! Try calculating the normal force when the level of mercury is 4.00 cm. What would happen to the aluminum cube if more mercury were poured into the tank? The cube would have no buoyant force. The cube would float partly submerged in the mercury. The cube would float with its bottom surface at the level of mercury. The cube would float by just touching the surface of the mercury. The cube would float just above the mercury. The cube would remain on the bottom. Use the worked example above to help you solve this problem. A 1.13 x 10^3 kg cube of aluminum is placed in a tank. Water is then added to the tank until half the cube is immersed. What is the normal force on the cube? (See Figure (a).) Mercury is now slowly poured into the tank until the normal force on the cube goes to zero. (See Figure (b).) How deep is the layer of Mercury? A cube of aluminum 1.30 m on a side is immersed one-half in water and one-half in glycerin. What is the normal force on t

Explanation / Answer

here,

density of mercury , pm = 5430 kg/m^3

density of alumunium , pa = 2650 kg/m^3

mass , m = 1.13 * 10^3 kg

the volume of cube , V = m/pa = 0.426 m^3

side , a = 0.752 m

(a)

the normal force on the cube , N = m * g - ( V/2) * pw * g

N = 1130 * 9.8 - 0.426 * 1000 * 9.8

N = 6899.2 N

(b)

V' is the volume immeresed in mercury

let the height of mercury be h

N' = N - V' * pm * g

0 = 6899.2 - V' * 5430 * 9.8

V' = 0.1297 m^3

let the height of mercury be h

h * a^2 = V'

h * 0.752^2 = 0.1297

h = 0.23 m

the height immeresed in mercury is 0.23 m

-----------------------

side of cube , a = 1.3 m

volume , V = 2.197 m^3

mass ,m = V * pa

m = 5822.05 kg

p = 1.26 * 10^3 kg/m^3

the normal force , F = m * g - pw * ( V/2) * g - p * ( V/2 ) * g

F = 5822.05 *9.8 - 1000 * 1.0985 *9.8 - 1260 * 1.0985 * 9.8

F = 32726.512 N

F = 3.27 * 10^4 N

the normal force is 3.27 * 10^4 N

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