A 3000 pound cylindrical pontoon having a radius of 6 feet floats in a body of fluid. A driver exerts a harmonic force of magnitude \(500 \mathrm{lb}\) at a rate of \(200 \mathrm{cy}-\) cles per minute at the center of the up- per surface of the float as indicated. \((a)\) Determine the density of the fluid if the pontoon is observed to bob with an amplitude of 1 foot. (b) What is the magnitude of the bobbing motion of the pontoon when the excitation fre- quency is reduced to \(5 \mathrm{rad} / \mathrm{sec}\) ?

Short Answer

Expert verified
Answer: To find the density of the fluid, use the following formula: $$ ρ_\text{fluid} = \frac{3000~\text{lb} /(32.2~\text{ft} / \text{s}^2)}{\pi(6~\text{ft})^2(2~\text{ft})(32.2~\text{ft}/\text{s}²)} $$ Plugging in the given values and calculating the fluid density will give you the answer. Question: Can we find the magnitude of the bobbing motion of the pontoon when the excitation frequency is reduced to 5 rad/sec? Answer: No, we cannot find the new amplitude at the reduced excitation frequency, as we do not have enough information about the mass-spring system parameters, such as the spring constant (k) and damping coefficient (b).

Step by step solution

01

Calculate the Volume of the Pontoon

First, we need to find the volume of the cylindrical pontoon. The formula for the volume of a cylinder is: $$ V_\text{cylinder} = \pi r^2 h $$ Since we do not have height (h), we have to assume it as a variable ‘h’ for now.
02

Calculate the Buoyant Force

The buoyant force on the pontoon is equal to the weight of the fluid displaced by the pontoon. The formula for buoyant force is: $$ F_\text{buoyant} = V_\text{displaced} \times \rho_\text{fluid} \times g $$ Where \(V_\text{displaced}\) is the volume of fluid displaced, \(\rho_\text{fluid}\) is the density of the fluid, and \(g\) is the acceleration due to gravity. Since the submerged volume of the pontoon is the same as the volume of the fluid displaced, we can write: $$ F_\text{buoyant} = V_\text{cylinder} \times \rho_\text{fluid} \times g = \pi r^2 h \rho_\text{fluid} g $$
03

Establish Equilibrium of Forces

At equilibrium, the weight of the pontoon will be balanced by the buoyant force: $$ W_\text{pontoon} = F_\text{buoyant} $$ $$ 3000~\text{lb} = \pi r^2 h \rho_\text{fluid} g $$ We can use this equation to find the density of the fluid (\(\rho_\text{fluid}\)).
04

Solve for Fluid Density

Rearrange the equation to solve for the fluid density: $$ \rho_\text{fluid} = \frac{3000~\text{lb}}{\pi (6~\text{ft})^2 h g} $$ Since we are given that the pontoon bobs with an amplitude of 1 foot, it means that \(\frac{1}{2} h = 1~\text{ft}\). Therefore, \(h=2~\text{ft}\). Now, we can plug the value of \(h\) and convert lb to slug (~32.2 ft/s²) to get the fluid density: $$ \rho_\text{fluid} = \frac{3000~\text{lb} /(32.2~\text{ft} / \text{s}^2)}{\pi(6~\text{ft})^2(2~\text{ft})(32.2~\text{ft}/\text{s}²)} $$ Calculate the value of the fluid density.
05

Calculate the Amplitude at Reduced Excitation Frequency

We are given that the excitation frequency is reduced to 5 rad/sec. We know that amplitude and excitation frequency are related through the equation: $$ A = \frac{F_0}{\sqrt{(k - m\omega^2)^2 + (b\omega)^2}} $$ Where \(A\) is the amplitude, \(F_0\) is the force, \(k\) is the stiffness, \(m\) is the mass, \(\omega\) is the frequency, and \(b\) is the damping coefficient. In this case, we have to assume that the whole mass of the pontoon is acting as a mass-spring system with a spring constant \(k\) and damping coefficient \(b\). Since we do not know these parameters and only have the fluid density, we cannot find the amplitude at the new excitation frequency.

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