Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 67

A box sits on a horizontal wooden ramp. The coefficient of static friction between the box and the ramp is 0.30 You grab one end of the ramp and lift it up, keeping the other end of the ramp on the ground. What is the angle between the ramp and the horizontal direction when the box begins to slide down the ramp? (tutorial: crate on ramp)

Short Answer

Expert verified
Answer: The angle between the ramp and the horizontal direction is approximately 73.3°.

Step by step solution

01

Identify the forces acting on the box

Determine the forces that are acting on the box while it is on the ramp. The main forces are the gravitational force acting downward (mg) and the frictional force (f) acting along the ramp. Note that as the box is on the verge of moving, it is still in equilibrium, meaning that the forces along the ramp and perpendicular to the ramp must balance out.
02

Calculate the maximum static friction

The maximum static friction (F_max) is given by the following formula: F_max = µ_s * N where µ_s is the coefficient of static friction (0.30), and N is the normal force, which is equal to mg * cos(θ).
03

Write down the balance of forces along the ramp

The box is still in equilibrium, so the forces acting along the ramp must be equal: Frictional force = Gravitational force component along the ramp. F_max = mg * sin(θ)
04

Substitute F_max in the force balance equation

Now, we will substitute the formula for F_max that we found in step 2 into the force balance equation: µ_s * N = mg * sin(θ)
05

Simplify the force balance equation with N

As N = mg * cos(θ), we can rewrite the force balance equation: µ_s * (mg * cos(θ)) = mg * sin(θ) Now, we can simplify the equation further by dividing both sides by (mg): µ_s * cos(θ) = sin(θ)
06

Calculate the angle θ between the ramp and the horizontal direction

Now, we can manipulate the equation in step 5 to solve for θ: tan(θ) = sin(θ) / cos(θ) = 1 / µ_s To find θ, we will use the inverse tangent function: θ = atan(1 / µ_s) θ = atan(1 / 0.30) θ ≈ 73.3° The angle between the ramp and the horizontal direction when the box begins to slide down the ramp is approximately 73.3°.

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

An \(80.0-\mathrm{N}\) crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at \(20.0^{\circ}\) to the ground. (a) What is the normal force exerted on the crate by the ramp? (b) The interaction partner of this normal force has what magnitude and direction? It is exerted by what object on what object? Is it a contact or a long-range force? (c) What is the static frictional force exerted on the crate by the ramp? (d) What is the minimum possible value of the coefficient of static friction? (e) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. Find the magnitude and direction of the contact force.
A cord, with a spring balance to measure forces attached midway along, is hanging from a hook attached to the ceiling. A mass of \(10 \mathrm{kg}\) is hanging from the lower end of the cord. The spring balance indicates a reading of \(98 \mathrm{N}\) for the force. Then two people hold the opposite ends of the same cord and pull against each other horizontally until the balance in the middle again reads \(98 \mathrm{N} .\) With what force must each person pull to attain this result?
A truck driving on a level highway is acted on by the following forces: a downward gravitational force of \(52 \mathrm{kN}\) (kilonewtons); an upward contact force due to the road of \(52 \mathrm{kN} ;\) another contact force due to the road of \(7 \mathrm{kN},\) directed east; and a drag force due to air resistance of \(5 \mathrm{kN}\), directed west. What is the net force acting on the truck?
(a) If a spacecraft moves in a straight line between the Earth and the Sun, at what point would the force of gravity on the spacecraft due to the Sun be as large as that due to the Earth? (b) If the spacecraft is close to, but not at, this equilibrium point, does the net force on the spacecraft tend to push it toward or away from the equilibrium point? [Hint: Imagine the spacecraft a small distance \(d\) closer to the Earth and find out which gravitational force is stronger.]
At what altitude above the Earth's surface would your weight be half of what it is at the Earth's surface?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Medical Physics

Read Explanation

Oscillations

Read Explanation

Solid State Physics

Read Explanation

Space Physics

Read Explanation

Wave Optics

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