A rectangular crate of weight W is floating in water with 2/3 of its volume above water. If you now want to push down on it to hold it lower in the water, so that only 1/3 of its volume is above water, the downward force needed to do this is:

Question

A rectangular crate of weight W is floating in water with 2/3 of its volume above water. If you now want to push down on it to hold it lower in the water, so that only 1/3 of its volume is above water, the downward force needed to do this is:

A. W
B. 2W
C. W/2
D. W/3

I have no idea what equation to use to help solve this problem. Could someone point me in the right direction..
Thanks in advance.

Answer 1

The volume submerged call it x. Now the weight of the water displaced is equal to the weight of the crate...W. So if you increase the weight by W , for a total of 2W, the new volume displaced is 2x

So F must be W.

Answer 2

To solve this problem, we can start by understanding the concept of buoyancy and the forces acting on the crate.

According to Archimedes' principle, an object submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. This buoyant force opposes the weight of the object. When an object is floating, the weight of the object is balanced by the buoyant force.

In this case, the crate is floating with 2/3 of its volume above water, which means that the buoyant force is equal to 2/3 of the weight of the crate. Mathematically, we can represent this as:

Buoyant force = (2/3) * W

Now, we want to push down on the crate so that only 1/3 of its volume is above water. This means that the buoyant force needs to be adjusted accordingly. Since the weight of the crate remains the same, we need to apply a downward force to balance the decreased buoyant force.

Let's denote the downward force needed as F_down. The buoyant force when only 1/3 of the volume is above water would be:

Buoyant force = (1/3) * W

To hold the crate lower in the water, the downward force needed should be equal to the difference between the initial buoyant force and the buoyant force when only 1/3 of the volume is above water. Therefore:

F_down = [(1/3) * W] - [(2/3) * W]

Simplifying the equation:

F_down = (1/3 - 2/3) * W

F_down = (-1/3) * W

So, the downward force needed to hold the crate lower in the water is (-1/3) times the weight of the crate, or equivalently, -W/3.

Therefore, the correct answer is option D: W/3.