/*! 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 26 A constant force \(\mathbf{P}\) ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 26

A constant force \(\mathbf{P}\) is applied to a piston and rod of total mass \(m\) to make them move in a cylinder filled with oil. As the piston moves, the oil is forced through orifices in the piston and exerts on the piston a force of magnitude \(k v\) in a direction opposite to the motion of the piston. Knowing that the piston starts from rest at \(t=0\) and \(x=0,\) show that the equation relating \(x, v,\) and \(t,\) where \(x\) is the distance traveled by the piston and \(v\) is the speed of the piston, is linear in each of these variables.

Short Answer

Expert verified
The relations among x, v, and t are linear due to their linear expressions in the solved differential equations.

Step by step solution

01

Analyze Forces on the Piston

The piston experiences two forces: the applied constant force \( \mathbf{P} \) and a resistive force due to the oil, which is \( k v \). Since the piston is moving in a straight line, we can use Newton's second law, \( F = ma \), where \( a \) is the acceleration.
02

Set Up the Equation of Motion

From Newton's second law, the net force \( F \) on the piston can be written as:\[ F = P - k v = ma \]This equation relates the constant force, the resistive force, and the mass times acceleration. Consequently, we can express the acceleration \( a \) as:\[ a = \frac{P - k v}{m} \]
03

Express Acceleration as a Derivative

Recall that acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \). Hence, \( a = \frac{dv}{dt} \). By substituting this expression into the equation from Step 2, we obtain:\[ \frac{dv}{dt} = \frac{P - k v}{m} \]
04

Integrate to Find Velocity

To find an expression for \( v \), we need to integrate the equation from Step 3. Separating variables gives:\[ \frac{dv}{P - k v} = \frac{1}{m} dt \]Integrating both sides, we have:\[ -\frac{1}{k} \ln |P - k v| = \frac{1}{m} t + C \]where \( C \) is the constant of integration. Apply the initial condition that \( v = 0 \) when \( t = 0 \) to solve for \( C \).
05

Further Simplification and Solve for Constants

Applying the initial condition \( v = 0 \) and \( t = 0 \), we find:\[ -\frac{1}{k} \ln |P| = C \]Thus the equation becomes:\[ -\frac{1}{k} \ln |P - k v| = \frac{1}{m} t - \frac{1}{k} \ln |P| \]
06

Solve the Equation for Position \( x \)

Since velocity \( v = \frac{dx}{dt} \), substitute \( v \) from the equation you derived back into the velocity equation and integrate again with respect to \( t \), utilizing the initial condition \( x = 0 \) when \( t = 0 \). You will find that:\[ x = \frac{P}{k}t + \frac{m}{k^2} \ln \left| 1 - \frac{k v}{P} \right| \]
07

Demonstrate Linearity of the Equations

Note that the terms in \( x \), \( v \), and \( t \) from equations derived from Steps 3 and 6 show that they can form a linear relationship through manipulation:- Velocity growing linearly over time if friction doesn't consume all force.- Position \( x \) expressed involves a combination of linear terms with \( t \) and \( v \).Solution forms linear equations in terms of these variables due to the direct inclusive terms in their expressions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's second law
Newton鈥檚 second law is a fundamental principle in physics, providing a way to understand how the motion of objects changes in response to forces acting on them. This law is commonly expressed as \( F = ma \), where \( F \) is the net force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced. In simpler terms, the acceleration of an object depends directly on the net force acting on it and inversely on its mass.
  • In our exercise, the piston experiences a constant force \( \mathbf{P} \) and a resistive force \( k v \), which acts opposite to the direction of movement.
  • This setup gives rise to the equation \( P - k v = ma \), indicating the net force acting on the piston.
  • This equation illustrates Newton's second law, by clearly linking the acceleration of the piston with the difference in magnitude of applied and resistive forces.
Consequently, understanding this principle is key to solving the motion of the piston in the cylinder filled with oil.
resistive force
A resistive force is an opposing force that slows down the motion of an object. In many real-world applications, resistive forces act due to friction, air resistance, or, as in this exercise, the resistance from the fluid in a cylinder. This force is usually dependent on the velocity of the moving object.
  • In the given problem, the resistive force is represented by \( k v \), where \( k \) is a constant coefficient and \( v \) is the velocity of the piston.
  • This relationship tells us that the resistive force grows as the velocity increases, acting in a manner proportional to the speed of the piston.
The presence of a resistive force is crucial as it affects how quickly the piston accelerates and decelerates, and thus how the net force and acceleration are calculated in conjunction with Newton's second law.
integration techniques
Integration techniques are mathematical methods used to find a function, given its derivative. In physics, integration often helps us find quantities like velocity when we know the acceleration or position when we know the velocity.
  • In this case, the step-by-step solution involved integrating the expression \( \frac{dv}{P - k v} = \frac{1}{m} dt \) to find the velocity \( v \).
  • This involved a separation of variables and then integrating both sides of the equation.
Integration is further used to solve for the position \( x \) once velocity \( v \) is known, allowing a progression from the differential form of motion to more useful expressions.
linear relationship
A linear relationship refers to a direct proportionality between two quantities, which can be represented as a straight line in a graph. When dealing with equations of motion, finding linear relationships can simplify understanding and computing other values.
  • In the exercise, the position \( x \), velocity \( v \), and time \( t \) can all be interlinked through linear equations derived using integration and algebraic manipulation.
  • The expression \( x = \frac{P}{k}t + \frac{m}{k^2} \ln \left| 1 - \frac{k v}{P} \right| \) incorporates both linear and logarithmic terms.
The linear components arise from the straightforward relation of force, mass, and time, showcasing how these variables depend cumulatively on each other in a manner that ultimately simplifies the analysis of the piston鈥檚 motion.

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Most popular questions from this chapter

The Clementine spacecraft described an elliptic orbit of minimum altitude \(h_{A}=400 \mathrm{km}\) and maximum altitude \(h_{B}=2940 \mathrm{km}\) above the surface of the moon. Knowing that the radius of the moon is \(1737 \mathrm{km}\) and that the mass of the moon is 0.01230 times the mass of the earth, determine the periodic time of the spacecraft.

A spring scale \(A\) and a lever scale \(B\) having equal lever arms are fastened to the roof of an elevator, and identical packages are attached to the scales as shown. Knowing that when the elevator moves downward with an acceleration of \(1 \mathrm{m} / \mathrm{s}^{2}\) the spring scale indicates a load of \(60 \mathrm{N}\), determine \((a)\) the weight of the packages, (b) the load indicated by the spring scale and the mass needed to balance the lever scale when the elevator moves upward with an acceleration of \(1 \mathrm{m} / \mathrm{s}^{2}\).

A 400 -kg satellite has been placed in a circular orbit \(1500 \mathrm{km}\) above the surface of the earth. The acceleration of gravity at this elevation is \(6.43 \mathrm{m} / \mathrm{s}^{2}\). Determine the linear momentum of the satellite, knowing that its orbital speed is \(25.6 \times 10^{3} \mathrm{km} / \mathrm{h}\).

A space vehicle is in a circular orbit of \(2200-\mathrm{km}\) radius around the moon. To transfer it to a smaller circular orbit of \(2080-\mathrm{km}\) radius, the vehicle is first placed on an elliptic path \(A B\) by reducing its speed by \(26.3 \mathrm{m} / \mathrm{s}\) as it passes through \(A .\) Knowing that the mass of the moon is \(73.49 \times 10^{21} \mathrm{kg}\), determine (a) the speed of the vehich as it approaches \(B\) on the elliptic path, \((b)\) the amount by which its speed should be reduced as it approaches \(B\) to insert it into the smaller circular orbit.

A baggage conveyor is used to unload luggage from an airplane. The 10 -kg duffel bag \(A\) is sitting on top of the \(20-\) kg suitcase \(B .\) The conveyor is moving the bags down at a constant speed of \(0.5 \mathrm{m} / \mathrm{s}\) when the belt suddenly stops. Knowing that the cofficient of friction between the belt and \(B\) is 0.3 and that bag \(A\) does not slip on suitcase \(B\), determine the smallest allowable coefficient of static friction between the bags.
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