Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 61

A wedge of mass \(m=36.1 \mathrm{~kg}\) is located on a plane that is inclined by an angle \(\theta=21.3^{\circ}\) with respect to the horizontal. A force \(F=302.3 \mathrm{~N}\) in the horizontal direction pushes on the wedge, as shown in the figure. The coefficient of kinetic friction between the wedge and the plane is 0.159 What is the acceleration of the wedge along the plane?

Short Answer

Expert verified
Answer: The acceleration of the wedge along the inclined plane is approximately 2.95 m/s².

Step by step solution

01

Diagram and Forces

Draw a diagram of the wedge on the inclined plane, with the horizontal force F acting to the right. Label the angle of inclination (θ), and indicate the weight of the wedge (mg) acting vertically downward.
02

Find the weight components

The weight of the wedge can be broken down into components parallel and perpendicular to the inclined plane: - Parallel component: \(mg \sin \theta\) - Perpendicular component: \(mg \cos \theta\)
03

Calculate the friction force

The friction force acting on the wedge is given by: - \(F_f = \mu F_n\) Where F_f is the friction force and F_n is the normal force. In this case, the normal force is equal to the perpendicular component of the weight (mg cos(θ)): \(F_f = \mu (mg \cos \theta)\)
04

Resolve the horizontal force

The horizontal force F can also be broken down into components parallel and perpendicular to the inclined plane: - Parallel component: \(F \cos \theta\) - Perpendicular component: \(F \sin \theta\)
05

Apply Newton's Second Law

Now we apply Newton's second law along the parallel direction, taking into account the horizontal force component, weight component, and the friction force: - \(ma_{\parallel} = F_{\parallel} - mg_{\parallel} - F_f\) Substitute the known values and the expressions derived in Steps 2, 3, and 4: \(m a_{\parallel} = (F \cos \theta) - (mg \sin \theta) - (\mu (mg \cos \theta))\)
06

Solve for acceleration

Now, solve for the acceleration along the plane (a_𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙). \(a_{\parallel} = \frac{(F \cos \theta) - (mg \sin \theta) - (\mu (mg \cos \theta))}{m}\) Plug in the given values of m, θ, F, and μ: \(a_{\parallel} = \frac{(302.3 \mathrm{~N} \cos 21.3^{\circ}) - (36.1 \mathrm{~kg} \cdot 9.81 \mathrm{~m/s^2} \sin 21.3^{\circ}) - (0.159 (36.1 \mathrm{~kg} \cdot 9.81 \mathrm{~m/s^2} \cos 21.3^{\circ}))}{36.1 \mathrm{~kg}}\) After calculating the expression, we find: \(a_{\parallel} \approx 2.95 \mathrm{~m/s^2}\) The acceleration of the wedge along the inclined plane is approximately \(2.95 \mathrm{~m/s^2}\).

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

Two blocks \(\left(m_{1}=1.23 \mathrm{~kg}\right.\) and \(m_{2}=2.46 \mathrm{~kg}\) ) are glued together and are moving downward on an inclined plane having an angle of \(40.0^{\circ}\) with respect to the horizontal. Both blocks are lying flat on the surface of the inclined plane. The coefficients of kinetic friction are 0.23 for \(m_{1}\) and 0.35 for \(m_{2}\). What is the acceleration of the blocks?

A person stands on the surface of the Farth. The mass of the person is \(m\) and the mass of the Earth is \(M\). The person jumps upward, reaching a maximum height \(h\) above the Earth. When the person is at this height \(h,\) the magnitude of the force exerted on the Earth by the person is a) \(m \mathrm{~g}\). c) \(M^{2} g / m\) e) zero. b) \(\mathrm{Mg}\) d) \(m^{2} g / M\).

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 \(m_{4}=\)

4.40 A store sign of mass \(4.25 \mathrm{~kg}\) is hung by two wires that each make an angle of \(\theta=42.4^{\circ}\) with the ceiling. What is the tension in each wire?
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Electrostatics

Read Explanation

Atoms and Radioactivity

Read Explanation

Circular Motion and Gravitation

Read Explanation

Wave Optics

Read Explanation

Quantum 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