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Chapter 6: Problem 4

A slide-loving pig slides down a certain \(35^{\circ}\) slide in twice the time it would take to slide down a frictionless \(35^{\circ}\) slide. What is the coefficient of kinetic friction between the pig and the slide?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.525.

Step by step solution

01

Identify the Key Variables

Let's identify the variables: the slide angle \( \theta = 35^{\circ} \), time on the frictionless slide \( t_0 \), and the time on the actual slide with friction \( t = 2t_0 \). We need to find the coefficient of kinetic friction \( \mu_k \).
02

Understand the Forces Involved

When the pig moves down the slide, two main forces act parallel to the slide's surface: the gravitational component \( mg \sin\theta \) and the frictional force \( \mu_k mg \cos\theta \).
03

Equations for Frictionless Slide

On a frictionless slide, the only force is the gravitational component. Using the equation of motion: \[ s = \frac{1}{2} a_0 t_0^2 \] where \( a_0 = g \sin\theta \).
04

Equations for Real Slide with Friction

On the real slide, the net acceleration \( a \) is given by:\[ a = g \sin\theta - \mu_k g \cos\theta \] Thus, the equation becomes:\[ s = \frac{1}{2} a (2t_0)^2 \]
05

Setting the Equations Equal

Since the slides cover the same distance \( s \), equate the two expressions:\[ \frac{1}{2} g \sin\theta \cdot t_0^2 = \frac{1}{2} (g \sin\theta - \mu_k g \cos\theta) \cdot (2t_0)^2 \]
06

Simplify and Solve for \( \mu_k \)

Simplifying:\[ g \sin\theta = 4(g \sin\theta - \mu_k g \cos\theta) \]This results in:\[ g \sin\theta = 4g \sin\theta - 4\mu_k g \cos\theta \]Rearrange to find \( \mu_k \):\[ 3g \sin\theta = 4\mu_k g \cos\theta \]\[ \mu_k = \frac{3 \tan\theta}{4} \]Substituting \( \theta = 35^\circ \), we get:\[ \mu_k = \frac{3 \tan(35^\circ)}{4} \]
07

Compute the Coefficient of Friction

Calculate \( \tan(35^\circ) \approx 0.7002 \). Then:\[ \mu_k \approx \frac{3 \times 0.7002}{4} \approx 0.525 \]

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Key Concepts

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

forces on an incline
When dealing with forces on an inclined plane, it's important to understand how gravity is at play. Imagine a pig sliding down a slope at an angle of 35 degrees. Gravity pulls the pig directly downward, but the slide is at an angle. This means we need to split the gravitational force into two components:

  • Parallel Component: This is the force pulling the pig down the slide. It is given by the formula \(mg \sin \theta\), where \(m\) is the mass of the pig, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle of the slide.
  • Perpendicular Component: This is the force pushing the pig into the surface of the slide. It is given by \(mg \cos \theta\).
These forces are crucial for understanding why a pig can slide down in the first place! When friction comes into play, these components help us calculate the frictional force acting against the slide motion.
coefficient of friction
The coefficient of kinetic friction, denoted as \(\mu_k\), is a key factor in determining how easily one object slides over another. It is a dimensionless value that represents the ratio between the force of friction and the normal (perpendicular) force. In our pig's journey down the slide, the friction force is what slows the pig down compared to a frictionless slide.

To calculate \(\mu_k\), you need to know:
  • The frictional force: Which is \(\mu_k \cdot mg \cos \theta\), based on the normal force.
  • The net forces acting: The difference between the gravitational component \(mg \sin \theta\) pulling the pig down the slope and the frictional force holding it back.
In our problem, we found that the pig slides down the slide twice as slowly due to friction, which allows us to derive that \(\mu_k\) is approximately 0.525 using the relationship described in the step-by-step solution.
equations of motion
The equations of motion are used to describe how an object moves. They take into account the initial velocity, acceleration, and distance travelled over time. In the context of sliding down a slide, the main equation to consider is \[ s = \frac{1}{2} at^2 \]Where:
  • \(s\) is the distance travelled down the slide.
  • \(a\) is the acceleration, which for our pig is affected by both gravity and friction.
  • \(t\) is the time taken.
For a frictionless slide, acceleration \(a = g \sin \theta\). However, with friction, this changes to \(a = g \sin \theta - \mu_k g \cos \theta\). By comparing these equations, we can find the relationship between the time it takes to slide down with and without friction, ultimately helping to define the coefficient of friction.
inclined plane dynamics
Inclined plane dynamics delve into how objects behave when placed on slopes rather than flat surfaces. This involves analyzing all forces acting on an object to understand its motion. In our example, the pig's traversal of an inclined plane with friction requires:
  • Recognizing the incline effects: Gravity pulls differently on a slope than on a level surface.
  • Understanding net force: Both gravitational force and friction influence the pig's slide performance.
  • Utilizing mathematical models: Use the equations described to solve for unknowns, like the coefficient of friction.
The key dynamics here are how gravity and friction interact on an inclined plane, impacting the acceleration and motion, and how these relationships can be modeled mathematically to predict or understand real-world behavior like that of our slide-loving pig.

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Most popular questions from this chapter

Calculate the magnitude of the drag force on a missile \(53 \mathrm{~cm}\) in diameter cruising at \(250 \mathrm{~m} / \mathrm{s}\) at low altitude, where the density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^{3} .\) Assume \(C=0.75\)

A cat dozes on a stationary merry-go-round in an amusement park, at a radius of \(5.4 \mathrm{~m}\) from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every \(6.0 \mathrm{~s}\). What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding (or the cat clinging with its claws)?

The terminal speed of a sky diver is \(160 \mathrm{~km} / \mathrm{h}\) in the spreadeagle position and \(310 \mathrm{~km} / \mathrm{h}\) in the nosedive position. Assuming that the diver's drag coefficient \(C\) does not change from one position to the other, find the ratio of the effective cross-sectional area \(A\) in the slower position to that in the faster position.

A sling-thrower puts a stone \((0.250 \mathrm{~kg})\) in the sling's pouch \((0.010 \mathrm{~kg})\) and then begins to make the stone and pouch move in a vertical circle of radius \(0.650 \mathrm{~m}\). The cord between the pouch and the person's hand has negligible mass and will break when the tension in the cord is \(33.0 \mathrm{~N}\) or more. Suppose the slingthrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur?

You must push a crate across a floor to a docking bay. The crate weighs \(165 \mathrm{~N}\). The coefficient of static friction between crate and floor is 0.510 , and the coefficient of kinetic friction is 0.32 . Your force on the crate is directed horizontally. (a) What magnitude of your push puts the crate on the verge of sliding? (b) With what magnitude must you then push to keep the crate moving at a constant velocity? (c) If, instead, you then push with the same magnitude as the answer to (a), what is the magnitude of the crate's acceleration?
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