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Chapter 3: Problem 11

An object weighing \(16 \mathrm{lb}\) is dropped from rest on the surface of a calm lake and thereafter starts to sink. While its weight tends to force it downward, the buoyancy of the object tends to force it back upward. If this buoyancy force is one of \(6 \mathrm{lb}\) and the resistance of the water (in pounds) is numerically equal to twice the square of the velocity (in feet per second), find the formula for the velocity of the sinking object as a function of the time.

Short Answer

Expert verified
The formula for the velocity of the sinking object as a function of time is \(v(t) = 10v^2 + v^4\).

Step by step solution

01

Determine the forces acting on the object

There are three forces acting on the object: 1. The weight force (downward) = 16 lb 2. The buoyancy force (upward) = 6 lb 3. The water resistance (upward) = 2v^2, where v is the velocity of the object in feet per second.
02

Calculate the net force acting on the object

To find the net force, we will subtract the upward forces (buoyancy force and water resistance) from the downward force (weight force). Net force = Weight force - (Buoyancy force + Water resistance) = 16 - (6 + 2v^2)
03

Apply Newton's second law of motion

According to Newton's second law, the net force on the object is equal to its mass times its acceleration. The mass of the object can be found by dividing the weight by the acceleration due to gravity (32 ft/s^2). Mass = Weight / Acceleration due to gravity = 16 lb / 32 ft/s^2 = 0.5 slugs Now, we can apply Newton's second law: Net force = Mass 脳 Acceleration 16 - (6 + 2v^2) = 0.5 脳 a
04

Solve for the acceleration a as a function of velocity v

We need to solve the equation from Step 3 for the acceleration a in terms of velocity v: 16 - (6 + 2v^2) = 0.5 脳 a a = 2 * (16 - 6 - 2v^2) a = 20 - 4v^2
05

Integrate the acceleration to find the velocity function

To find the velocity function v(t), we need to integrate the acceleration function a(t): v(t) = 鈭�(20 - 4v^2) dt To do this, we perform a substitution: let u = -4v^2, so du = -8v dv. Then we rewrite the integral in terms of u and du: v(t) = -(1/8)鈭�(20 - u) du = -(1/8)(20u - (1/2)u^2) + C鈧�, where C鈧� is the constant of integration Now we substitute v back into the equation: v(t) = -(1/8)(20(-4v^2) - (1/2)(-4v^2)^2) + C鈧� v(t) = -(1/8)(-80v^2 - 8v^4) + C鈧� v(t) = 10v^2 + v^4 + C鈧� Since the object is dropped from rest initially at t=0, its initial velocity is 0. Thus, we can find the value of C鈧�. 0 = 10(0) + (0)^4 + C鈧� C鈧� = 0
06

Express the velocity function v(t)

Now we have found the velocity function v(t) as a function of time: v(t) = 10v^2 + v^4 This is the formula for the velocity of the sinking object as a function of time.

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