Homework 3 Assignment Print View

1. Award: 33.33 out of 33.33 points It is estimated that 92 percent of an iceberg’s volume is below the surface, while only 8 percent is visible above the surface. For seawater with a density of 1025 kg/m3, estimate the density of the iceberg. The density of the iceberg is 943 kg/m3. References Worksheet Difficulty: Medium It is estimated that 92 percent of an iceberg’s volume is below the surface, while only 8 percent is visible above the surface. For seawater with a density of 1025 kg/m3, estimate the density of the iceberg. The density of the iceberg is 943 ± 10% kg/m3. Explanation: The following assumptions have been made here: 1. The buoyancy force in air is negligible. 2. Density changes in the water are negligible with depth. The weight of a body floating in a fluid is equal to the buoyant force acting on it (a consequence of vertical force balance from static equilibrium). Therefore, in this case, the average density of the body must be equal to the density of the seawater since W = FB Vsubmerged Vtotal = → ρiceberg gVtotal = ρseawater gVsubmerged ρicberg ρseawater In the present problem, the submerged volume is 92% of the total volume. Thus, ρicberg = ρseawater Vsubmerged Vtotal 92 = (1025 m3 ) ( 100 ) = 943 kg/m 3 kg 2. Award: 33.33 out of 33.33 points A water tank is being towed on an uphill road that makes 14° with the horizontal with a constant acceleration of 3 m/s2 in the direction of motion. Determine the angle that the free surface of water makes with the horizontal. What would your answer be if the direction of motion were downward on the same road with the same acceleration? The angle that the free surface makes with the horizontal when the direction of motion is upward is 16.2 °. when the direction of motion is downward is -17.8 °. References Worksheet Difficulty: Medium A water tank is being towed on an uphill road that makes 14° with the horizontal with a constant acceleration of 3 m/s2 in the direction of motion. Determine the angle that the free surface of water makes with the horizontal. What would your answer be if the direction of motion were downward on the same road with the same acceleration? The angle that the free surface makes with the horizontal when the direction of motion is upward is 15.4 ± 10% °. when the direction of motion is downward is -17.8 ± 10% °. Explanation: The following assumptions have been made here: 1. Effects of splashing, braking, driving over bumps, and climbing hills are assumed to be secondary and are not considered. 2. The acceleration remains constant. We take the x- and z-axes as shown in the figure. From geometrical considerations, the horizontal and vertical components of acceleration are ax = acosα az = asinα The tangent of the angle the free surface makes with the horizontal is (3 m/s )cos14° 2 tanθ = ax g+az = acosα g+asinα = 9.81 m/s +(3 m/s )sin14° 2 2 = 0.2762 Therefore, θ = 15.4° When the direction of motion is reversed, both ax and az are in the negative x- and z-direction, respectively, and thus become negative quantities, ax = −acosα az = −asinα Then, the tangent of the angle the free surface makes with the horizontal becomes −(3 m/s )cos14° 2 tanθ = ax g+az Therefore, θ = -17.8° = acosα g+asinα = 9.81 m/s −(3 m/s )sin14° 2 2 = -0.3203 3. Award: 33.34 out of 33.34 points A 10-cm-radius, 90-cm-high vertical cylindrical container is partially filled with 60-cm-high water. Now, the cylinder is rotated at a constant angular speed of 180 rpm. Determine how much the liquid level at the center of the cylinder will drop as a result of this rotational motion. The drop in the liquid level due to the rotational motion is 0.09 m. References Worksheet Difficulty: Medium A 10-cm-radius, 90-cm-high vertical cylindrical container is partially filled with 60-cm-high water. Now, the cylinder is rotated at a constant angular speed of 180 rpm. Determine how much the liquid level at the center of the cylinder will drop as a result of this rotational motion. The drop in the liquid level due to the rotational motion is 0.091 ± 10% m. Explanation: The following assumptions have been made here: 1. The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2. The bottom surface of the container remains covered with liquid during rotation (no dry spots). Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0 and z = 0), the equation for the free surface of the liquid is given as zs (r) = h0 − ω2 (R 2 4g − 2r 2 ) where h0 = 0.6 m is the original height of the liquid before rotation, and ω = 2πṅ = 2π(180 rev/min) ( 160min ) = 18.85 rad/s s Then, the vertical height of the liquid at the center of the container where r = 0 becomes zs (0) = h0 - ω2 R2 4g (18.85 rad/s) (0.1 m) 2 = (0.60 m) - 4(9.81 m/s 2 ) 2 = 0.509 m Therefore, the drop in the liquid level at the center of the cylinder is Δhdrop,center = h0 - zs (0) = 0.60 m - 0.509 m = 0.091 m