/*! 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 99 Block \(A,\) with weight \(3 w,\... [FREE SOLUTION] | 91Ó°ÊÓ

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Block \(A,\) with weight \(3 w,\) slides down an inclined plane \(S\) of slope angle \(36.9^{\circ}\) at a constant speed while plank \(B\) with weight \(w,\) rests on top of A. The plank is attached by a cord to the wall (Fig. P5.99). (a) Draw a diagram of all the forces acting on block A. (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A,\) determine its value.

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.33.

Step by step solution

01

Understand the Problem Setup

Block A slides down an inclined plane with a constant speed, which means the net force acting along the plane is zero. Block B rests on top of Block A and is held in place by a cord attached to a wall. This tells us that static and kinetic friction play essential roles in maintaining the system's dynamics.
02

Identify the Forces on Block A

For Block A, identify the forces: weight of Block A (\[3w\], acting vertically downwards), normal force (\[N_A\], perpendicular to the incline), gravitational component parallel to the incline (\[3w \sin(36.9^\circ)\]), kinetic friction opposing motion (\[f_k\]), and tension from Block B inhibited by the cord.
03

Calculate Forces Perpendicular to Plane

Calculate the normal force \(N_A\) acting on Block A using the resolved component of gravity perpendicular to the incline.\[\begin{align*}N_A &= 3w \cos(36.9^\circ)\end{align*}\]
04

Calculate Forces Parallel to Plane

The force due to gravity parallel to the incline is balanced by the frictional force.\[\begin{align*}3w \sin(36.9^\circ) &= f_k + w\end{align*}\]Where kinetic friction \(f_k\) is \(\mu_k N_A\).
05

Express Friction in Terms of Kinetic Coefficient

Substitute \(f_k = \mu_k N_A\) into the previous equation:\[\begin{align*}3w \sin(36.9^\circ) &= \mu_k \cdot 3w \cos(36.9^\circ) + w\end{align*}\]This relationship allows us to solve for \(\mu_k\).
06

Solve for the Coefficient of Kinetic Friction

Rearrange the equation from Step 5 to solve for \(\mu_k\):\[\begin{align*}3 \sin(36.9^\circ) - 1 &= \mu_k \cdot 3 \cos(36.9^\circ) \\mu_k &= \frac{3 \sin(36.9^\circ) - 1}{3 \cos(36.9^\circ)}\end{align*}\]
07

Calculate and Finalize the Coefficient Value

Insert the trigonometric values \(\sin(36.9^\circ) = 0.6\) and \(\cos(36.9^\circ) = 0.8\) into the equation:\[\begin{align*}\mu_k &= \frac{3 \, \times \, 0.6 - 1}{3 \, \times \, 0.8} \\mu_k &= \frac{1.8 - 1}{2.4} \\mu_k &= \frac{0.8}{2.4} \\mu_k &= \frac{1}{3}\end{align*}\]Thus, the coefficient of kinetic friction is approximately 0.33.

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Key Concepts

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

Inclined Plane Physics
Inclined plane physics is a fascinating area of study that explores how objects behave when placed on slopes. An inclined plane, like the one mentioned in the problem, is simply a flat surface tilted at an angle compared to the horizontal ground. This setup allows us to see forces at work, particularly gravity and friction. When a block is placed on such an incline, gravity helps to pull it down the slope.

The slope's angle is crucial because it affects how the forces are split into parallel and perpendicular components relative to the plane. For example, in our exercise, the angle of the incline is 36.9°. This angle impacts how much of the gravitational force attempts to slide the block down, and how much presses against the plane. The gravitational component pulling the block downward is calculated using \(3w \sin(36.9^\circ)\), and the component pushing the block into the incline is \(3w \cos(36.9^\circ)\).

This scenario is a great opportunity to observe Newton's laws of motion at work, as the forces must balance for the block to slide at a constant speed.
Kinetic Friction Coefficient
Friction is a force that resists the motion of objects sliding against each other. The kinetic friction coefficient, denoted by \(\mu_k\), quantifies this resistance when the surfaces are in motion. In the context of our problem, friction is essential as it keeps block A moving at a constant speed down the incline. Frictional force works against the gravitational pull, ensuring there's no acceleration.

The kinetic friction force \(f_k\) is calculated using the formula: \(f_k = \mu_k N_A\), where \(N_A\) is the normal force pressing the object into the plane. This frictional force modifies the net movement of block A, so it doesn't accelerate. By balancing the forces parallel to the plane, we can solve for the kinetic friction coefficient. Using the given values and calculations, \(\mu_k\) is determined to be approximately 0.33.
  • The normal force varies based on the block’s weight and the incline angle.
  • The friction coefficient depends on the materials' surfaces in contact.
Understanding \(\mu_k\) is crucial for engineers and physicists when predicting how objects move across different surfaces.
Newton's Laws of Motion
Newton's laws of motion are foundational principles that describe the behavior of objects in motion. Let's unpack how they apply to our inclined plane scenario.

  • First Law (Law of Inertia): An object in motion stays in motion with the same speed and in the same direction unless acted upon by a net external force. In this problem, block A slides down at a constant speed, indicating balanced forces.
  • Second Law (Law of Acceleration): The acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass. Here, since Block A moves at a constant speed, the net force parallel to the incline is zero \((F = ma = 0)\). This confirms the balance between the gravitational component down the slope and the kinetic frictional force.
  • Third Law (Action and Reaction): For every action, there is an equal and opposite reaction. The normal force exerted by the incline on the block is equal and opposite to the gravitational force component pressing against the hill.
This problem helps illustrate Newton's laws by showing how friction and gravity can work together to create stable motion on an incline. Understanding these laws allows us to predict and analyze motion in various physical scenarios.

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Most popular questions from this chapter

A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass \(m\) is suspended from the other end. When the blocks are released, the tension in the rope is 10.0 \(\mathrm{N}\) (a) Draw two free-body diagrams, one for the \(4.00-\) kg block and one for the block with mass \(m\) . (b) What is the acceleration of either block? (c) Find the mass \(m\) of the hanging block. (d) How does the tension compare to the weight of the hanging block?

You observe a 1350 -kg sports car rolling along flat pavement in a straight line. The only horizontal forces acting on it are a constant rolling friction and air resistance (proportional to the square of its speed). You take the following data during a time interval of \(25 \mathrm{s} :\) When its speed is \(32 \mathrm{m} / \mathrm{s},\) the car slows down at a rate of \(-0.42 \mathrm{m} / \mathrm{s}^{2},\) and when its speed is decreased to \(24 \mathrm{m} / \mathrm{s},\) it slows down at \(-0.30 \mathrm{m} / \mathrm{s}^{2} .\) (a) Find the coefficient of rolling friction and the air drag constant \(D\) . (b) At what constant speed will this car move down an incline that makes a \(2.2^{\circ}\) angle with the horizontal? (c) How is the constant speed for an incline of angle \(\beta\) related to the terminal speed of this sports car if the car drops off a high cliff? Assume that in both cases the air resistance force is proportional to the square of the speed, and the air drag constant is the same.

\(\mathrm{A} 75.0\) -kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.0 \(\mathrm{kg}\) . (a) Find the maximum and minimum tension in the chain. (b) What is the tension at a point three-fourths of the way up from the bottom of the chain?

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.)

A 750.0 -kg boulder is raised from a quarry 125 \(\mathrm{m}\) deep by a long uniform chain having a mass of 575 kg. This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking. (a)What is the maximum acceleration the boulder can have and still get out of the quarry, and (b) how long does it take to be lifted out at maximum acceleration if it started from rest?

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