Problem 76 - University Physics with Modern Physics

Question

CP SHM of a Floating Object. An object with height \(h,\) mass \(M,\) and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\) (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7\()\) .

Step-by-Step Solution

Answer

(a) \(x_1 = \frac{M}{\rho A}\). (b) \(\Delta x = \frac{F}{\rho g A}\). (c) \(T = 2\pi \sqrt{\frac{M}{\rho g A}}\).

Step 1: Equilibrium Position Without External Force

At equilibrium, the buoyant force must equal the weight of the object. The buoyant force is given by the weight of the liquid displaced: \( F_b = \rho g V_{\text{displaced}} \), where \( V_{\text{displaced}} = x_1 A \) and \( x_1 \) is the submerged height of the object. Therefore, the equation at equilibrium is:\[\rho g x_1 A = Mg\]Solving for \( x_1 \), we find:\[\x_1 = \frac{M}{\rho A}\]This is the vertical distance from the surface to the bottom of the object at equilibrium without the force.

Step 2: Equilibrium Position With External Force

When an additional downward force, \( F \), is applied, the total force balance becomes the sum of the object's weight and the external force:\[\rho g x_2 A = Mg + F\]where \( x_2 \) is the new submerged height. Solving for \( x_2 \), we get:\[x_2 = \frac{M + \frac{F}{g}}{\rho A}\]The additional submerged depth due to the force \( F \) is:\[\Delta x = x_2 - x_1 = \frac{F}{\rho g A}\]Thus, the bottom of the object is \( \Delta x \) farther below the surface.

Step 3: Calculate the Period of SHM

The object will execute simple harmonic motion (SHM) when the force is removed, oscillating about the equilibrium position without external force. For SHM, the restoring force can be described by Hooke's Law: \( F_{\text{restoring}} = -kx \), where \( k \) is the effective spring constant. Since \( F = \rho g A x \), we have:\[k = \rho g A\]The period of a mass-spring system is given by:\[T = 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M}{\rho g A}}\]This is the period of oscillation.

Key Concepts

Buoyant Force

Understanding buoyant force is essential when discussing floating objects. When an object floats in a fluid, it experiences a force that opposes its weight. This force is called the buoyant force. It is what allows objects to float and maintain stability in fluids. The buoyant force is defined by Archimedes' principle, which states that the force is equal to the weight of the fluid displaced by the object. Here is how the concept works:

• The buoyant force can be calculated using the formula: \( F_b = \rho g V_{\text{displaced}} \), where \( \rho \) is the fluid's density, \( g \) is the acceleration due to gravity, and \( V_{\text{displaced}} \) is the volume of the fluid displaced by the submerged part of the object.
• This force acts upwards, counteracting the object's weight, helping it float.

In equilibrium, when no external forces act, the buoyant force exactly balances the weight of the object. It's crucial for allowing objects to reach a stable floating position where their weight and the buoyant force are perfectly balanced. This equilibrium lets us easily calculate how much of the object remains submerged.

Equilibrium Position

The equilibrium position is where the object's weight is balanced by the buoyant force, resulting in no net force and no acceleration. At this point, the object floats steadily without moving up or down. This position is determined differently depending on whether external forces are applied. Let's explore two scenarios:

• Without external force: The buoyant force equals the object's weight. Mathematically, it is represented as \( \rho g x_1 A = Mg \), leading to an equilibrium depth \( x_1 = \frac{M}{\rho A} \).
• With external force: An additional force shifts the equilibrium position, requiring a new balance of forces: \( \rho g x_2 A = Mg + F \). Here, \( x_2 = \frac{M + \frac{F}{g}}{\rho A} \) shows the change in submerged height due to the force.

In essence, equilibrium positions provide a way to identify how forces and displacements are balanced in a floating object scenario. They are crucial when determining how far an object is submerged under various conditions.

Oscillation Period

The concept of the oscillation period becomes important when discussing simple harmonic motion (SHM) in floating objects. When an external force is removed, the object will oscillate around its equilibrium position. This oscillation has a periodic nature, described by its oscillation period, the time taken for one complete cycle of motion. Here’s what you need to know:

• After removing an added force, the object experiences a restoring force directed toward the equilibrium position. This force is similar to Hooke's Law, defined by \( F_{\text{restoring}} = -kx \), where \( k = \rho g A \) is the spring constant in this context.
• The oscillation period \( T \) can be found using the formula \( T = 2\pi \sqrt{\frac{M}{k}} \), resulting in \( T = 2\pi \sqrt{\frac{M}{\rho g A}} \).

The oscillation period offers key insights into how quickly an object will oscillate in a fluid once displaced and then allowed to return freely to its equilibrium position. Understanding this helps us predict and analyze the dynamic behavior of floating objects under various conditions.