1.3.20 answer is 3670 dyne Last people did it wrong but helped explained. Please

Question

1.3.20 answer is 3670 dyne
Last people did it wrong but helped explained. Please help me I'm so confused ccelerating Flu (1.3.10) (a) A fish tank filled with water and open on the top is accelerating upward at 3.00 m/s Starting from first principles, derive an equation which gives the pressure in the tank as a function of the depth h below the liquid surface. Assume that the pressure at the liquid surface is 0. (b) A cylinder of uniform cross section is floating in the fish tank, partially submerged as shown in the picture. The cylinder has the same upward acceleration as the water. Starting from a free- body diagram drawn for the cylinder, derive the density of the cylinder 3.00 cm Watcr surface 5.00 cm (1.3.20) In the diagram below, a tank of water open at the top surface is accelerating to the right at 2.50 m/s2, resulting in a sloped upper surface, as we saw in class. At the bottom of the tank is a steel sphere of radius R = 0.80 cm, which is connected to the right side of the tank by a light cord. The contact between the steel sphere and the bottom of the tank is frictionless. (a) What is the net horizontal force exerted on the steel sphere by the pressure of the surrounding water? (Hint: What would be your answer if the steel sphere was replaced by a bag of water with the same radius?) (b) Starting from a free-body diagram analysis, determine the tension in the cord. Only g/cm horizontal forces need to be included in the free-body diagram. The density of steel is 7.85 air water

Explanation / Answer

1.3.10

Pressure at a depth of h,

consider a column of liquid of height h, and are a,

volume of the liquid column = ah

if d is the density of the liquid then mass of the liquid column = ahd

now the force at a depth h due to the weight of this column of liquid = ahdg

pressure is Force per unit area and area of the column we have considered is a

Pressure ata depth h due to the lquid column = hdg,

If the pressure at the top is Po then total pressure at dpeth h = Po + hdg

b) as the water and the cylinder are accelrating at the same rate there will be no excess force on the cylinder from acceleration of the liquid.

It is simple floating principle. the forces on the cylinder are

its own weight due to gravity and

the force of bouyance from the water.

Let V be the volume of the cylinder and d is its density the 5/8 of the volume is immersed in water and 3/8 is floating above the surface

Buoyancy force is equal to the wieght of the volume of liquid (water) displaced

                                                   = 5V/8 * 1g ( density of water = 1gm/cc)

weight of the cylinder = Vdg

now equating the forces Vdg = 5V/8 *g

density of the cylinder d = 5/8 = 0.625 gm/cc

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