/*! 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 23 A uniform board is leaning again... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A uniform board is leaning against a smooth vertical wall. The board is at an angle \(\theta\) 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
\( \theta \approx 52.1^\circ \) is the smallest angle.

Step by step solution

01

Understand the Forces

We have a board leaning against a wall, making an angle \( \theta \) with the horizontal ground. The forces acting on the board include gravitational force (weight), normal force from the ground, frictional force from the ground, and the normal force from the wall.
02

Determine the Forces

Let \( W \) be the weight of the board, \( N_1 \) the normal force from the ground, and \( N_2 \) the normal force from the wall. The static frictional force \( f_s \) is given by \( f_s = \mu_s N_1 \), where \( \mu_s = 0.650 \).
03

Set Up Equilibrium Equations

For the board not to slide, solve the equilibrium equations. Balance vertical forces: \( N_1 = W \cos(\theta) \). Balance horizontal forces: \( N_2 = W \sin(\theta) \). Balance torque (calculated about the base): \( f_s L = N_2 H/2 \).
04

Substitute and Simplify

Known \( f_s = \mu_s N_1 \). Substitute into the torque equation: \( \mu_s W \cos(\theta) L = (W \sin(\theta) H)/2 \). Cancel common terms and express \( \mu_s \cos(\theta) = (\sin(\theta))/2 \).
05

Solve for Angle \( \theta \)

Rearrange the equation: \( 2\mu_s \cos(\theta) = \sin(\theta) \). Prepare to solve for \( \theta \) using trigonometric identities: \( 2\mu_s = \tan(\theta) \).
06

Calculate Smallest Angle \( \theta \)

Substitute \( \mu_s = 0.650 \), yields \( \tan(\theta) = 1.3 \). Use \( \theta = \arctan(1.3) \), calculating \( \theta \approx 52.1^\circ \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Equilibrium in Physics
Equilibrium is a concept in physics where an object is in a state of balance. This means that all the forces acting on the object are perfectly balanced, and the object does not accelerate.

In this exercise, for the board to be in equilibrium, the forces must satisfy certain conditions:
  • The sum of all vertical forces must be zero.
  • The sum of all horizontal forces must also be zero.
  • The sum of the torques about any axis must be zero.
For the board leaning against the wall, these conditions ensure it does not slide or topple over. The vertical forces are the weight of the board and the normal force from the ground, while the horizontal forces include the friction force between the ground and the board, and the normal force from the wall.

To achieve equilibrium, particularly for avoiding sliding, the frictional force at the base must be sufficient to resist the weight component trying to slide the board downwards. This is crucial when considering static friction's role.
Torque and Rotational Motion
Torque is a measure of the rotational force on an object. It causes objects to rotate around a pivot point or axis.

In this context, the board resting on the wall is subject to torques that determine its rotation or lack thereof. The point where the board touches the ground is a natural choice for calculating torques because this is the pivot point around which the board may start to rotate if it does not remain in equilibrium.
  • The torque due to the board's weight attempts to rotate it over its base.
  • Frictional forces and the normal force from the wall provide counteracting torques.
The torque balance equation in this problem accounts for these opposing torques and ensures that the clockwise and counterclockwise torques are equal, keeping the board from rotating.

When the torques are balanced, the board remains stable, and by solving the torque equation, we help find the critical angle where this balance is precisely maintained.
Trigonometric Functions
Trigonometric functions help relate the angles and sides of triangles, crucial in calculating the forces and torques in this problem.

When dealing with forces acting at angles, such as the ones here, trigonometric functions like sine, cosine, and tangent become indispensable. Here's how they come into play:
  • The cosine function relates to the component of the weight acting perpendicular to the board's length, which affects the normal force from the ground.
  • The sine function deals with the component of weight parallel to the board's length, influencing the horizontal force.
  • The tangent function directly relates the vertical and horizontal components of the forces, helping to solve for the angle. In this problem, it manifests as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} \).
By rearranging and solving these trigonometric relationships, we can find the minimum angle \( \theta \), ensuring the lower end of the board does not slide. Understanding these functions thus allows us to relate the geometry of the board's position to the physical forces at play.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of \(6.6 \mathrm{m} / \mathrm{s}\). Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{c m}+M h^{2},\) where \(I_{c m}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

A platform is rotating at an angular speed of 2.2 rad/s. A block is resting on this platform at a distance of \(0.30 \mathrm{m}\) from the axis. The coefficient of static friction between the block and the platform is \(0.75 .\) Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates.

A rotating door is made from four rectangular sections, as indicated in the drawing. The mass of each section is \(85 \mathrm{kg} .\) A person pushes on the outer edge of one section with a force of \(F=68 \mathrm{N}\) that is directed perpendicular to the section. Determine the magnitude of the door's angular acceleration.

The drawing shows an outstretched arm \((0.61 \mathrm{m}\) in length) that is parallel to the floor. The arm is pulling downward against the ring attached to the pulley system, in order to hold the 98 -N weight stationary. To pull the arm downward, the latissimus dorsi muscle applies the force \(\overrightarrow{\mathbf{M}}\) in the drawing, at a point that is \(0.069 \mathrm{m}\) from the shoulder joint and oriented at an angle of \(29^{\circ} .\) The arm has a weight of \(47 \mathrm{N}\) and a center of gravity (cg) that is located \(0.28 \mathrm{m}\) from the shoulder joint. Find the magnitude of \(\overrightarrow{\mathbf{M}}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.