A circular cylinder of height 1.20 m having faces of diameter 0.620 m is immersed in water of density 1.00 × 10³ kg/m³ with its axis vertical so that its faces are parallel to the surface of the water.The upper face is 2.50 m below the surface of the water.The net force on this cylinder is observed to be 1120 N downward.Atmospheric pressure is 1.01 × 10⁵ Pa.What net force does the water exert on the lower face of the cylinder? A)7.40 × 10³ N B)1.09 × 10⁴ N C)3.79 × 10⁴ N D)4.14 × 10⁴ N E)1.37 × 10⁵ N

Question

Quizplus · Accepted Answer

The buoyant force on the cylinder is equal to the weight of water displaced by the cylinder. The weight of water displaced is given by the volume of water displaced times the density of water, or
$$
V_{	ext{disp}}=\pi\left(\dfrac{0.620}{2}ight)^2	imes1.20=0.442	ext{ m}^3
$$
$$
W_{	ext{disp}}=V_{	ext{disp}}	imesho_{	ext{water}}=0.442	imes10^3=442	ext{ N}
$$
The net force on the cylinder is the difference between the weight of the cylinder and the buoyant force acting on it. Therefore,
$$
F_{	ext{net}}=W_{	ext{cyl}}-W_{	ext{disp}}+P_{	ext{atm}}A_{	ext{bottom}}-P_{	ext{water}}A_{	ext{top}}=1120	ext{ N}
$$
where $W_{	ext{cyl}}$ is the weight of the cylinder, $P_{	ext{water}}$ is the pressure of water at 2.50 m depth, and $A_{	ext{bottom}}$ is the area of the bottom face of the cylinder. Since the cylinder is completely submerged and its axis is vertical, the pressure at the bottom face is
$$
P_{	ext{bottom}}=P_{	ext{water}}+gho_{	ext{water}}h=1.01	imes10^5+\left(9.81	imes10^0ight)\left(1.00	imes10^3ight)\left(2.50ight)=1.26	imes10^5	ext{ Pa}
$$
Thus, the force exerted by water on the bottom face is
$$
F_{	ext{bottom}}=P_{	ext{bottom}}A_{	ext{bottom}}=P_{	ext{atm}}A_{	ext{top}}+W_{	ext{cyl}}-W_{	ext{disp}}-F_{	ext{net}}\approx4.14	imes10^{14}	ext{ N}
$$
Therefore, the answer is D.

A Circular Cylinder of Height 1