/*! 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 5 The coefficient of static fricti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 5

The coefficient of static friction between a box and a ramp is \(0.5 .\) The ramp's incline angle is \(30^{\circ} .\) If the box is placed at rest on the ramp, the box will do which of the following? (A) Accelerate down the ramp (B) Accelerate briefly down the ramp but then slow down and stop (C) Move with constant velocity down the ramp (D) Not move

Short Answer

Expert verified
The box will not move (D).

Step by step solution

01

Identify the forces acting on the box

There are two main forces acting on the box: the force of gravity acting vertically downwards and the normal force acting perpendicular to the inclined surface of the ramp. The force of gravity can be further divided into two components - a force acting perpendicular to the incline (\(mg\cos\theta\)) and a force acting parallel to the incline (\(mg\sin\theta\)), where \(m\) is the mass of the box, \(g\) is the acceleration due to gravity, and \(\theta\) is the incline angle of the ramp.
02

Calculate the maximum static friction force

The static friction force (\(f_s\)) that can act on the box is a product of the coefficient of static friction (\(\mu_s\)) and the normal force. Here, \(\mu_s = 0.5\) and the normal force is equal to the perpendicular component of weight, there by \(f_s = \mu_s*mg\cos\theta\).
03

Compare the static friction force and gravitational force component parallel to the incline

If the static frictional force is larger or equal to the parallel component of gravity (\(mg\sin\theta\)), the box will not move, because the frictional force is sufficient to prevent sliding. Otherwise, there will be an net acceleration down the ramp.
04

Compute

Without knowing the exact weight of the box, we can compare the forces qualitatively. We can see that because \(\mu_s \cos\theta\) (\(0.5 * \cos30^{\circ}\)) is larger than \(\sin\theta\) (\(\sin 30^{\circ}\)), theoretically the frictional force will be larger than the force trying to move the box down the incline. It means the box will not move.

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 91影视!

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 crate of mass \(100 \mathrm{kg}\) is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is \(0.4,\) and the coefficient of kinetic friction is \(0.3 .\) A force \(\mathbf{F}\) of magnitude \(344 \mathrm{N}\) is then applied to the crate, parallel to the floor. Which of the following is true? (A) The crate will accelerate across the floor at \(0.5 \mathrm{m} / \mathrm{s}^{2}\) (B) The static friction force, which is the reaction force to \(\mathbf{F}\) as guaranteed by Newton's Third Law, will also have a magnitude of \(344 \mathrm{N}\) (C) The crate will slide across the floor at a constant speed of 0.5 \(\mathrm{m} / \mathrm{s}\) (D) The crate will not move.

1\( is on the bottom, and Crate \)\\# 2\( is on the top. Both crates have the same mass. Compare鈥� # Two crates are stacked on top of each other on a horizontal floor; Crate \)\\# 1\( is on the bottom, and Crate \)\\# 2\( is on the top. Both crates have the same mass. Compared to the strength of the force \)\mathbf{F}_{1}\( necessary to push only Crate \)\\# 1\( at a constant speed across the floor, the strength of the force \)\mathbf{F}_{2}\( necessary to push the stack at the same constant speed across the floor is greater than \)F_{1}\( because (A) the normal force on Crate \)\\# 1\( is greater (B) the coefficient of kinetic friction between Crate \)\\# 1\( and the floor is greater (C) the coefficient of static friction between Crate \)\\# 1\( and the floor is greater (D) the weight of Crate \)\\# 1$ is greater

A person who weighs \(800 \mathrm{N}\) steps onto a scale that is on the floor of an elevator car. If the elevator accelerates upward at a rate of \(5 \mathrm{m} / \mathrm{s}^{2}\), what will the scale read? (A) \(400 \mathrm{N}\) (B) \(800 \mathrm{N}\) (C) \(1,000 \mathrm{N}\) (D) \(1,200 \mathrm{N}\)

A block of mass \(m\) is at rest on a frictionless, horizontal table placed in a laboratory on the surface of the Earth. An identical block is at rest on a frictionless, horizontal table placed on the surface of the Moon. Let \(\mathbf{F}\) be the net force necessary to give the Earth-bound block an acceleration of a across the table. Given that \(g_{\text {Moon } ~}\) is one- sixth of \(g_{\text {Earth }},\) the force necessary to give the Moon-bound block the same acceleration a across the table is (A) \(\mathbf{F} / 6\) (B) \(\mathbf{F} / 3\) (C) \(\mathbf{F}\) (D) \(6 \mathbf{F}\)

A 20 \(\mathrm{N}\) block is being pushed across a horizontal table by an \(18 \mathrm{N}\) force. If the coefficient of kinetic friction between the block and the table is \(0.4,\) find the acceleration of the block. (A) \(0.5 \mathrm{m} / \mathrm{s}^{2}\) (B) \(1 \mathrm{m} / \mathrm{s}^{2}\) (C) \(5 \mathrm{m} / \mathrm{s}^{2}\) (D) \(7.5 \mathrm{m} / \mathrm{s}^{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Modern Physics

Read Explanation

Engineering Physics

Read Explanation

Fluids

Read Explanation

Further Mechanics and Thermal Physics

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.