Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 76

A \(2.00-\mathrm{kg}\) block is on a plane inclined at \(20.0^{\circ}\) with respect to the horizontal. The coefficient of static friction between the block and the plane is \(0.60 .\) a) How many forces are acting on the block? b) What is the normal force? c) Is this block moving? Explain.

Short Answer

Expert verified
There are 3 forces acting on the block: gravitational force, normal force, and frictional force. b) What is the normal force? The normal force is 18.51 N. c) Is this block moving? Explain. The block is not moving because the maximum static friction force (11.11 N) is greater than the parallel force component of gravity acting on the block (6.68 N).

Step by step solution

01

Identify forces acting on the block

To determine the number of forces acting on the block, we must first recognize the types of forces involved. There are three forces acting on the block: gravitational force (weight), the normal force (perpendicular to the inclined plane), and frictional force (opposing any motion). a) So, there are 3 forces acting on the block.
02

Resolve gravitational force into components

To analyze the normal force and frictional force, we need to resolve the gravitational force of the block into components parallel and perpendicular to the inclined plane: 1. Parallel component: \(F_{g \parallel} = mg \sin{(\theta)}\) 2. Perpendicular component: \(F_{g \perp} = mg \cos{(\theta)}\) Here, \(m = 2.00\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), and \(\theta = 20.0^{\circ}\). First, let's find \(F_{g\parallel}\) and \(F_{g\perp}\): 1. \(F_{g \parallel} = (2.00\,\text{kg})(9.81\,\text{m/s}^2)\sin{(20.0^{\circ})} = 6.68\,\text{N}\) 2. \(F_{g \perp} = (2.00\,\text{kg})(9.81\,\text{m/s}^2)\cos{(20.0^{\circ})} = 18.51\,\text{N}\)
03

Calculate the normal force

The normal force (\(F_N\)) acts perpendicular to the inclined plane and has the same magnitude as \(F_{g\perp}\), but in the opposite direction. Thus, b) The normal force is \(F_N = 18.51\,\text{N}\).
04

Determine the maximum static friction force

The maximum static friction force (\(F_{s\,max}\)) can be calculated using: \(F_{s\,max} = \mu_{s}F_N\) where, \(\mu_{s}=0.60\) (coefficient of static friction) and \(F_N = 18.51\,\text{N}\) (normal force). Now, let's calculate \(F_{s\,max}\): \(F_{s\,max} = (0.60)(18.51\,\text{N}) = 11.11\,\text{N}\)
05

Determine if the block is moving

To determine if the block will move, compare \(F_{s\,max}\) and \(F_{g\parallel}\). If the maximum static friction force (\(F_{s\,max}\)) is greater than or equal to the parallel gravitational force component (\(F_{g\parallel}\)), the block remains stationary. If \(F_{s\,max} < F_{g\parallel}\), the block will slide down the incline. We have calculated \(F_{g\parallel} = 6.68\,\text{N}\) and \(F_{s\,max} = 11.11\,\text{N}\). Here, \(F_{s\,max} > F_{g\parallel}\). c) So, the block is not moving, as the maximum static friction force is greater than the parallel force component of gravity acting on the block.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A box of books is initially at rest a distance \(D=0.540 \mathrm{~m}\) from the end of a wooden board. The coefficient of static friction between the box and the board is \(\mu_{s}=0.320\), and the coefficient of kinetic friction is \(\mu_{k}=0.250 .\) The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board. -4.55 A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

A bosun's chair is a device used by a boatswain to lift himself to the top of the mainsail of a ship. A simplified device consists of a chair, a rope of negligible mass, and a frictionless pulley attached to the top of the mainsail. The rope goes over the pulley, with one end attached to the chair, and the boatswain pulls on the other end, lifting himself upward. The chair and boatswain have a total mass \(M=90.0 \mathrm{~kg}\). a) If the boatswain is pulling himself up at a constant speed, with what magnitude of force must he pull on the rope? b) If, instead, the boatswain moves in a jerky fashion, accelerating upward with a maximum acceleration of magnitude \(a=2.0 \mathrm{~m} / \mathrm{s}^{2},\) with what maximum magnitude of force must he null on the rone?

Leonardo da Vinci discovered that the magnitude of the friction force is usuzlly simply proportional to the magnitude of the normal force; that is, the friction force does not depend on the width or length of the contact area. Thus, the main reason to use wide tires on a race car is that they a) Iook cool. b) have more apparent contact area. c) cost more. d) can be made of softer materials.

Four weights, of masses \(m_{1}=6.50 \mathrm{~kg}_{3}\) \(m_{2}=3.80 \mathrm{~kg}, m_{3}=10.70 \mathrm{~kg},\) and \(m_{4}=\)

A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Scientific Method Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Linear Momentum

Read Explanation

Physics of Motion

Read Explanation

Kinematics Physics

Read Explanation
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.

Sign-up for free