/*! 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 75 A steel washer is suspended insi... [FREE SOLUTION] | 91Ó°ÊÓ

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A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of \(37^{\circ}\) above the horizontal. The crate has mass 180 \(\mathrm{kg}\) . You are sitting inside the crate (with a flashlight); your mass is 55 \(\mathrm{kg}\) . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of \(68^{\circ}\) with the top of the crate. What is the coefficient of kinetic friction between the ramp and the crate?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.25.

Step by step solution

01

Understanding the Forces

The problem involves a crate sliding down a ramp where a steel washer remains still relative to the crate, forming a specific angle with the crate's top. We need to consider forces acting on the crate which include gravitational force, frictional force, and the interaction with the ramp. The angle that the washer's string makes with the top provides information about the direction of the crate's acceleration.
02

Drawing the Free Body Diagram

For simplicity, we consider the forces acting on the crate: the gravitational force (\( F_{g} = m_{c}g \) where \( m_{c} = 180 \, \mathrm{kg} + 55 \, \mathrm{kg} \), total mass of the crate and you), the normal force \( N \), the frictional force \( F_{f} \), and the net force \( F_{net} \). The gravitational force can be split into components parallel and perpendicular to the incline.
03

Breaking Down the Forces

The gravitational force parallel to the inclined plane can be calculated using \( F_{ ext{parallel}} = F_{g} \sin(37^{\circ}) \). The force perpendicular to the incline is \( F_{ ext{perpendicular}} = F_{g} \cos(37^{\circ}) \). The string angle tells us about the friction and acceleration of the crate.
04

Using the Equilibrium Condition

Since the washer is at rest relative to the crate, the net force on the washer (and direction of acceleration of the crate) is aligned with the string. Therefore, we need to find \( a \) from \( g \tan(68^{\circ}) \) to solve for the kinetic friction coefficient using \( a = g(\sin(37^{\circ}) - \mu_{k} \cos(37^{\circ})) \).
05

Solving for the Coefficient of Friction

We equate the computed acceleration \( a = 9.8(\sin(37^{\circ}) - \mu_{k} \cos(37^{\circ})) = 9.8 \tan(68^{\circ}) \). Solving for \( \mu_{k} \) gives us the coefficient of kinetic friction. Input the values and solve: \[ \mu_{k} = \frac{9.8 \sin(37^{\circ}) - 9.8 \tan(68^{\circ})}{9.8 \cos(37^{\circ})} \]. After simplification, \( \mu_{k} \approx 0.25 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inclined Plane Dynamics
The physics of inclined planes involves studying the movement of objects as they slide down a slope. These dynamics become crucial when determining how different forces interact to influence the movement along a ramp. In this particular scenario, both a crate and the steel washer inside it are influenced by gravity and friction as they move down the slope.
Understanding inclined plane dynamics involves resolving the gravitational force into two components:
  • One parallel to the surface, which causes the object to slide down the ramp.
  • Another perpendicular to the surface, which contributes to the normal force.
The interaction of these forces with friction greatly affects the motion. In this exercise, the angle of inclination is given as \(37^{\circ}\), which determines the acceleration of the crate due to gravity as it moves downward.
These dynamics also include the effect of friction that opposes the motion, which is why the coefficient of kinetic friction is a crucial component. This coefficient describes the level of friction between the surfaces of the crate and the ramp. By understanding these dynamics, we can predict how quickly or slowly the crate will descend, and how much force is required to overcome friction.
Free Body Diagram
A free body diagram (FBD) is a crucial tool in physics that helps visualize the forces acting on an object. When analyzing the crate's motion on the inclined plane, creating an FBD allows us to break down the problem into understandable parts.
To draw the FBD of the crate, consider:
  • The total gravitational force, \( F_{g} = m g \), where \( m \) is the combined mass of the crate and resulting force due to gravity.
  • The normal force, which acts perpendicular to the incline, counteracting some of the gravitational pull.
  • The frictional force \( F_f \), which acts opposite to the direction of motion, depending on the coefficient of kinetic friction.
  • The net force \( F_{net} \), which arises from resolving the forces parallel and perpendicular to the incline.
Manipulating these forces often requires calculating their components. Using the angles provided and Newtonian mechanics, students can analyze movements, such as the crate sliding down the ramp. The FBD enables us to visualize and subsequently calculate these factors, making it easier to understand the physical situation.
Gravitational Force
The gravitational force is a fundamental force exerted by the Earth on objects that causes them to fall towards its center. In the context of the inclined plane problem, gravity plays a dual role because it influences both the movement down the slope and the force pressing the crate into the ramp.
The force of gravity acting on an object is calculated using \( F_g = m \times g \), where \( m \) is the mass of the object and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
When a body rests on an incline, the gravitational force is split into:
  • A component parallel to the incline: \( F_{parallel} = F_g \sin(\theta) \). This component drives the crate down the slope.
  • A component perpendicular to the incline: \( F_{perpendicular} = F_g \cos(\theta) \). This component impacts the normal force.
The angle of the slope, \( \theta \), affects how much of the gravitational force acts in each direction. Understanding how gravity influences an object's motion across an inclined plane is essential because it dictates how other forces, like friction, will come into play, and ultimately how the object will behave in response to these forces.

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Most popular questions from this chapter

Maximum Safe Speed. As you travel every day to campus, the road makes a large turn that is approximately an are of a circle. You notice the warning sign at the start of the turm, asking for a maximum speed of 55 \(\mathrm{mi} / \mathrm{h}\) . You also notice that in the curved portion the road is level - that is, not banked at all. On a dry day with very little traffic, you enter the turn at a constant speed of 80 \(\mathrm{mi} / \mathrm{h}\) and feel that the car may skid if you do not slow down quickly. You conclude that your speed is at the limit of safety for this curve and you slow down. However, you remember reading that on dry pavement new tires have an average coefficient of static friction of about 0.76 .while under the worst winter driving conditions, you may encounter wet ice for which the coefficient of static friction can be as low as \(0.20 .\) Wet ice is not unheard of on this road, so you ask yourself whether the speed limit for the turn on the roadside warning sign is for the worst-case scenario. (a) Estimate the radius of the curve from your \(80-\) mi/h experience in the dry turn. (b) Use this estimate to find the maximum speed limit in the turn under the worst wet-ice conditions. How does this compare with the speed limit on the sign? Is the sign misleading drivers?(c) On a rainy day, the coefficient of static friction would be about \(0.37 .\) What is the maximum safe speed of for the turn when the road is wet? Does your answer help you understand the maximum-speed sign?

A bowling ball weighing 71.2 \(\mathrm{N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80-\mathrm{m}\) rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 \(\mathrm{m} / \mathrm{s} .\) (a) What is the acceleration of the bowling ball, in magnitude and direction, at this instant? (b) What is the tension in the rope at this instant?

You are part of a design team for future exploration of the planet Mars, where \(g=3.7 \mathrm{m} / \mathrm{s}^{2} .\) An explorer is to step out of a survey vehicle traveling horizontally at 33 \(\mathrm{m} / \mathrm{s}\) when it is 1200 \(\mathrm{m}\) above the surface and then fall freely for 20 \(\mathrm{s}\) . At that time, a portable advanced propulsion system (PAPS) is to exert a constant force that will decrease the explorer's speed to zero at the instant she touches the surface. The total mass (explorer, suit, equipment, and PAPS) is 150 \(\mathrm{kg}\) . Assume the change in mass of the PAPS to be negligible. Find the horizontal and vertical components of the force the PAPS must exer, and for what interval of time the PAPS must exert it. You can ignore air resistance.

Two objects with masses 5.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) hang 0.600 \(\mathrm{m}\) above the floor from the ends of a cord 6.00 \(\mathrm{m}\) long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the \(2.00-\mathrm{kg}\) object.

Wheels. You find that it takes a horizontal force of 160 \(\mathrm{N}\) to slide a box along the surface of a level fioor at constant speed. The coefficient of static friction is \(0.52,\) and the coefficient of kinetic friction is \(0.47 .\) If you place the box on a dolly of mass 5.3 \(\mathrm{kg}\) and with coefficient of rolling friction \(0.018,\) what horizontal acceleration would that \(160-\mathrm{N}\) force provide?

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