/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 A small rowboat has a mass of \(... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 45

A small rowboat has a mass of \(47 \mathrm{kg} .\) When a \(92-\mathrm{kg}\) person gets into the boat, the boat floats \(8.0 \mathrm{cm}\) lower in the water. If the boat is then pushed slightly deeper in the water, it will bob up and down with simple harmonic motion (neglecting any friction). What will be the period of oscillation for the boat as it bobs around its equilibrium position?

Short Answer

Expert verified
Answer: The period of oscillation for the boat is approximately 1.82 seconds.

Step by step solution

01

Determine the buoyant force acting on the boat

To find the buoyant force acting on the boat, we must first determine the volume of water that has been displaced by the boat's submerged part. When the person gets into the boat, it sinks 8.0 cm lower in the water. We can use the mass and density of water to find the displaced volume, which will then help us calculate the buoyant force. The density of water is given as \(1000 kg/m^3\). The mass of the person is \(92 kg\), and the boat's mass is \(47 kg\). Since the boat is floating, the buoyant force is equal to the total mass of the boat and the person, which is: \(F_{buoyant} = m_{boat} * g + m_{person} * g\) \(F_{buoyant} = (47 kg + 92 kg) * 9.81 m/s^2 = 139 kg * 9.81 m/s^2\)
02

Calculate the displaced volume

Next, we can use the buoyant force and the density of the water to find the volume of the displaced water: \(V_{displaced} = \frac{F_{buoyant}}{\rho_{water} * g}\) \(V_{displaced} = \frac{139 kg * 9.81 m/s^2}{1000 kg/m^3 * 9.81 m/s^2}\) \(V_{displaced} = 0.139 m^3\)
03

Determine the effective spring constant

Now, we can treat the boat's bobbing motion as if it were a spring with an effective spring constant. The effective spring constant can be found using Hooke's Law and the buoyant force acting on the boat: \(F_{buoyant} = k_{eff} * x\) Here, \(x = 0.08 m\) represents the depth to which the boat is submerged in the water. We can solve for the effective spring constant \(k_{eff}\): \(k_{eff} = \frac{F_{buoyant}}{x}\) \(k_{eff} = \frac{139 kg * 9.81 m/s^2}{0.08 m} = 17052.5 N/m\)
04

Calculate the period of oscillation

Finally, we can use the effective spring constant, total mass \(m_{total} = 139 kg\), and the formula for the period of simple harmonic motion to find the period of oscillation: \(T = 2 * \pi * \sqrt{\frac{m_{total}}{k_{eff}}}\) \(T = 2 * \pi * \sqrt{\frac{139 kg}{17052.5 N/m}}\) \(T \approx 1.82 s\) The period of oscillation for the boat as it bobs around its equilibrium position is approximately 1.82 seconds.

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Most popular questions from this chapter

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