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In fluid mechanics, the concept of buoyant force plays a pivotal role, especially when dealing with objects submerged in a fluid. The buoyant force is the upward force exerted by any fluid upon a body placed in it. It is a direct result of the pressure differences exerted by the fluid at different depths.

The fundamental principle governing buoyant force is Archimedes' Principle. This principle states that the buoyant force experienced by a submerged object is equal to the weight of the fluid that the object displaces. In mathematical terms, if an object displaces a fluid volume, the buoyant force \( F_b \) is calculated as:
- \( F_b = V \times \text{density of fluid} \)
- Here, \( V \) represents the volume of fluid displaced.

For our specific case of the submerged tree, since it is underwater, the displaced volume of water is equivalent to the volume of the tree itself. Given the density of water as 62.4 lb/ft³, the buoyant force can be robustly computed, illustrating the significant force fighting against the gravity. Understanding this force is crucial, especially for scenarios where objects like logs could behave unpredictably when cut loose underwater.

Specific gravity is a dimensionless quantity. It compares the density of a substance with a reference, usually water for liquids and solids. In simple terms, it tells us how dense a substance is in relation to water.

For example, the specific gravity \( SG \) of wood given as 0.6 implies that wood is 60% as dense as the water it is submerged in. It's determined by the ratio:
- \( SG = \frac{\text{density of substance}}{\text{density of water}} \)

Knowing the specific gravity allows us to calculate the density of the wood needed to determine its weight while submerged. For water with a known density of 62.4 lb/ft³, the density of wood becomes 0.6 times this value. This calculation leads us to the understanding that wood will float unless weighed down because its density is less than that of water.

Specific gravity plays a crucial role in buoyancy and is widely used in calculations involving ships, submarines, or any objects intended to float or be neutrally buoyant.

A truncated cone is simply a cone with its tip cut off parallel to its base. Calculating the volume of such a shape is crucial, especially in applications like calculating force and weight for submerged solids.

For a truncated cone with a base radius \( R \), top radius \( r \), and height \( h \), the formula for volume \( V \) is:
- \( V = \frac{1}{3} \pi h (R^2 + Rr + r^2) \)

This formula accounts for the base area, the sloping sides, and the top area of the truncated cone. Using this, we calculate the volume using the given dimensions: base diameter of 8 ft (hence \( R = 4 \) ft), top diameter of 2 ft (\( r = 1 \) ft), and a height of 100 ft.

Understanding the shape as a truncated cone is critical because it provides a realistic model for objects like tree trunks, enabling accurate estimations of displacement, buoyancy, and forces involved when submerged. With this volume, we can straightforwardly determine both the weight and the balancing forces needed against submersion.