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Chapter 9: Problem 67

A uniform board is leaning against a smooth vertical wall. The board is at an angle \(\underline{\theta}\) above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is \(0.650\). Find the smallest value for the angle \(\theta\), such that the lower end of the board does not slide along the ground.

Short Answer

Expert verified
The smallest angle \( \theta \) is approximately 33.0 degrees.

Step by step solution

01

Understand the Problem

A board leans against a wall with angle \( \theta \) and static friction between the ground and the board. We need to find the smallest angle \( \theta \) for which the board does not slide.
02

Identify Forces

Identify the forces acting on the board: the gravitational force \( W \) acting downward, the normal force \( N \) from the wall acting perpendicular to the board, the normal force \( N_g \) from the ground acting upward, and the frictional force \( f_s \) acting horizontally along the ground.
03

Apply Conditions for Equilibrium

For equilibrium, the sum of forces in both horizontal and vertical directions must be zero. Additionally, the sum of torques about any point must be zero.
04

Force Analysis in Horizontal and Vertical Directions

In horizontal direction: \( N = f_s \). In vertical direction: \( N_g = W \).
05

Analyze Torque

Choose the base of the board as the pivot point. Calculate torque about this point to relate \( N \), \( W \), and \( \theta \). Torque by \( W \) is \( W \cdot \frac{L}{2} \cdot \cos(\theta) \) and torque by \( N \) is \( N \cdot L \cdot \sin(\theta) \). Set these equal for equilibrium.
06

Relate Friction and Normal Force

For frictional force, use \( f_s = \mu_s \cdot N_g \). Substitute \( N_g = W \) to get \( f_s = \mu_s \cdot W \).
07

Solve for \( \theta \)

Using \( f_s = N \) and \( f_s = \mu_s \cdot W \), relate \( \tan (\theta) = \frac{N}{W} = \mu_s \). Therefore, \( \theta = \tan^{-1}(\mu_s) \).
08

Calculate \( \theta \) Value

For \( \mu_s = 0.650 \), \( \theta = \tan^{-1}(0.650) \). Calculate to find \( \theta \).

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Key Concepts

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

Static Friction
Static friction is the force that opposes the initiation of motion between two stationary surfaces. When the board is leaning against the wall, static friction acts at the bottom of the board to prevent it from sliding.

The maximum static friction force can be calculated using the equation:
  • \( f_s = \mu_s \cdot N_g \)
where \( \mu_s \) is the coefficient of static friction and \( N_g \) is the normal force from the ground.

In this exercise, the static friction force is crucial to balance the forces that could cause the board to slip. It is set equal to the component of forces parallel to the base of the board, ensuring the board stays in place as long as static friction can hold its ground.
Equilibrium
Equilibrium occurs when all the forces and torques acting on an object are balanced, resulting in no net motion. In this scenario, the board needs to stay steady without sliding or toppling over. For equilibrium:
  • The net force in both the horizontal and vertical directions must be zero.
  • The net torque about any pivot point must also be zero.

To maintain equilibrium, the normal force from the wall (acting horizontally) equals the static friction force. Likewise, the weight of the board (acting vertically downward) equals the normal force from the ground.

This ensures that the board doesn't move in any direction. When all these conditions are satisfied, the board is said to be in a state of equilibrium.
Torque
Torque is a measure of the force that can cause an object to rotate about an axis. In this exercise, torque is calculated to determine when the board will topple over or stay stable. It plays a crucial role in solving for the minimum angle of inclination.

When selecting the base of the board as the pivot point, we consider the following torques:
  • Torque by the weight of the board, acting at its center of gravity, which can be calculated as \( W \times \frac{L}{2} \times \cos(\theta) \).
  • Torque by the normal force from the wall, \( N \times L \times \sin(\theta) \).
To achieve equilibrium, these two torque values must be equal, preventing the board from rotating or toppling.
Angle of Inclination
The angle of inclination is the angle at which the board is leaning against the wall. It is critical because it determines the board's stability.

In this problem, we need to compute the smallest angle \( \theta \) that prevents the board from sliding down due to gravitational forces.
  • The formula to find this angle is \( \theta = \tan^{-1}(\mu_s) \).
Given that the coefficient of static friction, \( \mu_s \), is known (0.650 in this case), we can substitute it to find \( \theta \). This angle of inclination involves balancing the board's potential slip with its rotational stability, thanks to the static friction present at the base.

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Most popular questions from this chapter

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300 -mile trip in a typical midsize car produces about \(1.2 \times 10^{9} \mathrm{~J}\) of energy. How fast would a \(13-\mathrm{kg}\) flywheel with a radius of \(0.30 \mathrm{~m}\) have to rotate to store this much energy? Give your answer in rev/min.

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh \(438 \mathrm{~N}\) and have a center of gravity that is \(1.28 \mathrm{~m}\) above the floor. His upper legs weigh \(144 \mathrm{~N}\) and have a center of gravity that is \(0.760 \mathrm{~m}\) above the floor. Finally, his lower legs and feet together weigh \(87 \mathrm{~N}\) and have a center of gravity that is \(0.250 \mathrm{~m}\) above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel?

A flat uniform circular disk (radius \(=2.00 \mathrm{~m}\), mass \(=1.00 \times 10^{2} \mathrm{~kg}\) ) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A \(40.0-\mathrm{kg}\) person, standing \(1.25 \mathrm{~m}\) from the axis, begins to run on the disk in a circular path and has a tangential speed of \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. Find the resulting angular speed of the disk (in \(\mathrm{rad} / \mathrm{s}\) ) and describe the direction of the rotation.

Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius. One has the shape of a hoop and the other the shape of a solid disk. Each wheel starts from rest and has a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each makes the same number of revolutions in the same time. (a) Which wheel, if either, has the greater angular acceleration? (b) Which, if either, has the greater moment of inertia? (c) To which wheel, if either, is a greater net external torque applied? Explain your answers.
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