A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft

Question

11 days ago

11

A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft

must hold a 730 kg steel cube with edges of length 45.0 cm without sinking. Assume the density of water in the pool is 1000 kg/m3.
The students use a crane to gently place the cube on the raft but accidentally place it off center. The cube remains on the raft for a few moments and then the raft tilts, causing the cube to slide off and sink to the bottom of the pool. The raft remains floating in the pool. In a coherent paragraph-length response, indicate whether the water level in the pool when the cube is on the bottom of the pool is higher than, lower than, or the same as when the cube is on the raft, and explain your reasoning. For both cases, assume that there is no motion of the water.

Physics

1 answer:

Sav [1.1K] · Answer 1 · 2026-02-22T11:11:52+03:00

Answer:

The water level increases more when the cube is above the raft before it sinks.

Explanation:

The principle involved is based on Archimedes' theory, which states that immersing an object in water will raise the initial water level. This means that the height of the water in the container increases. The increase in water volume corresponds to the volume of the submerged object.

We can consider two scenarios: when the steel cube rests on the raft and when it settles at the pool's bottom.

When the cube rests at the pool’s bottom, the volume will indeed rise, and we can ascertain this using the cube's volume.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

When an object floats, it's because the densities of the object and water are in equilibrium.

$Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]$

The formula for density is:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyant force can be calculated with this equation:

$F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]$

Vs > Vc indicates that the combined volume of the raft and the cube exceeds that of the cube alone resting at the bottom of the pool. Hence, when the cube is positioned above the raft, the water level rises more before it becomes submerged.