/*! 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 67 A box sits on a horizontal woo... [FREE SOLUTION] | 91影视

91影视

Log out

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掳.

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 = 碌_s * N where 碌_s is the coefficient of static friction (0.30), and N is the normal force, which is equal to mg * cos(胃).
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(胃)
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: 碌_s * N = mg * sin(胃)
05

Simplify the force balance equation with N

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

Calculate the angle 胃 between the ramp and the horizontal direction

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

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 crow perches on a clothesline midway between two poles. Each end of the rope makes an angle of \(\theta\) below the horizontal where it connects to the pole. If the weight of the crow is \(W,\) what is the tension in the rope? Ignore the weight of the rope.
A 6.0 -kg block, starting from rest, slides down a frictionless incline of length \(2.0 \mathrm{m} .\) When it arrives at the bottom of the incline, its speed is \(v_{\mathrm{f}} .\) At what distance from the top of the incline is the speed of the block \(0.50 v_{\mathrm{f}} ?\)
Two of Robin Hood's men are pulling a sledge loaded with some gold along a path that runs due north to their hideout. One man pulls his rope with a force of \(62 \mathrm{N}\) at an angle of \(12^{\circ}\) east of north and the other pulls with the same force at an angle of \(12^{\circ}\) west of north. Assume the ropes are parallel to the ground. What is the sum of these two forces on the sledge?
In Fig. 4.15 an astronaut is playing shuffleboard on Earth. The puck has a mass of \(2.0 \mathrm{kg} .\) Between the board and puck the coefficient of static friction is 0.35 and of kinetic friction is \(0.25 .\) (a) If she pushes the puck with a force of \(5.0 \mathrm{N}\) in the forward direction, does the puck move? (b) As she is pushing, she trips and the force in the forward direction suddenly becomes \(7.5 \mathrm{N} .\) Does the puck move? (c) If so, what is the acceleration of the puck along the board if she maintains contact between puck and stick as she regains her footing while pushing steadily with a force of \(6.0 \mathrm{N}\) on the puck? (d) She carries her game to the Moon and again pushes a moving puck with a force of \(6.0 \mathrm{N}\) forward. Is the acceleration of the puck during contact more, the same, or less than on Earth? Explain. (tutorial: rough table)
An \(85-\mathrm{kg}\) skier is sliding down a ski slope at a constant velocity. The slope makes an angle of \(11^{\circ}\) above the horizontal direction. (a) Ignoring any air resistance, what is the force of kinetic friction acting on the skier? (b) What is the coefficient of kinetic friction between the skis and the snow?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electromagnetism

Read Explanation

Astrophysics

Read Explanation

Fluids

Read Explanation

Torque and Rotational Motion

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.