Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A box with mass m is dragged across a level floor with coefficient of kinetic friction \(\mu_k\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F\). (a) In terms of \(m, \mu_k, \theta\), and \(g\), obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90-kg patient across a floor at constant speed by pulling on him at an angle of 25\(^\circ\) above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_k\) = 0.35. Use the result of part (a) to answer the instructor's question.

Short Answer

Expert verified
(a) \( F = \frac{\mu_k mg}{\cos\theta + \mu_k \sin\theta} \); (b) 292.8 N.

Step by step solution

01

Identify the Forces Acting on the Box

The forces acting on the box include the gravitational force (weight), the normal force, the applied force (at an angle), and the kinetic frictional force. The gravitational force is given by weight: \( F_g = mg \). The applied force can be broken into horizontal \( F_{ax} = F \cos \theta \) and vertical \( F_{ay} = F \sin \theta \) components. Kinetic friction force is \( f_k = \mu_k N \), where \( N \) is the normal force.
02

Establish Equilibrium Conditions

For the box to move with constant speed, the net horizontal force must be zero, so the applied force \( F_{ax} \) must balance the frictional force \( f_k \). Also, the net vertical force must be zero, which means the normal force \( N \) plus the vertical component of the applied force \( F \sin \theta \) balances the gravitational force \( mg \).
03

Express Normal Force in Terms of Known Quantities

From the vertical force balance, we have: \( N + F \sin \theta = mg \), simplifying to \( N = mg - F \sin \theta \).
04

Solve for the Frictional Force

The frictional force is given by \( f_k = \mu_k N \), substituting for \( N \) yields: \( f_k = \mu_k (mg - F \sin \theta) \).
05

Solve for the Required Force \( F \) Using Horizontal Equilibrium

Horizontal equilibrium implies: \( F \cos \theta = \mu_k (mg - F \sin \theta) \). Rearrange this equation to find \( F \): \[ F = \frac{\mu_k mg}{\cos\theta + \mu_k \sin\theta} \].
06

Calculate the Force for the Given Scenario

Substitute the given values \( m = 90 \text{ kg}, \mu_k = 0.35, \theta = 25^{\circ}, \text{and } g = 9.8 \text{ m/s}^2 \) into the expression for \( F \):\[ F = \frac{0.35 \times 90 \times 9.8}{\cos(25^{\circ}) + 0.35 \sin(25^{\circ})} \].
07

Simplify and Calculate Numeric Solution

Calculate \( \cos(25^{\circ}) \approx 0.9063 \) and \( \sin(25^{\circ}) \approx 0.4226 \). Substitute these to simplify: \[ F = \frac{308.7}{0.9063 + 0.35 \times 0.4226} = \frac{308.7}{1.0544} \approx 292.8 \text{ N} \].

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 Vaia!

Key Concepts

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

Kinetic Friction
Kinetic friction is a type of friction that comes into play when two surfaces slide against each other. It's an opposing force that resists motion. When a box is dragged across the floor, kinetic friction acts opposite to the direction of the box's motion. The strength of this friction is determined by the normal force and the coefficient of kinetic friction, denoted as \(\mu_k\). To calculate kinetic friction, use the formula: \(f_k = \mu_k N\), where \(N\) is the normal force. Kinetic friction does work by dissipating energy, usually in the form of heat, as two surfaces rub against one another. Knowing this, it's clear why understanding kinetic friction is essential in solving problems involving sliding objects.
Force Equilibrium
Force equilibrium occurs when the net force acting on a body is zero. This means all the forces balance each other out, resulting in no acceleration. For the box in the exercise to move at a constant speed, it should be in horizontal force equilibrium.
  • Horizontal equilibrium: The horizontal component of the applied force should equal the kinetic friction.
  • Vertical equilibrium: The normal force plus the vertical component of the applied force should equal the gravitational force.
When these conditions are satisfied, the box moves smoothly without changing its speed. A thorough grasp of force equilibrium helps to ensure systems function in a stable manner without unpredictable movements.
Normal Force
The normal force is the supportive force exerted by a surface perpendicular to the object resting on it. It balances the gravitational force acting on the object. In our case, the box is subject to such a force from the floor. To find the expression for the normal force, consider the vertical forces:\( N + F \sin \theta = mg \), where \(N\) is the normal force, \(F \sin \theta\) is the vertical component of the pulling force, and \(mg\) is the gravitational force on the box.Solving this gives: \N = mg - F \sin \theta\. This adjustment to the normal force due to the vertical component of the pulling force is crucial as it alters the kinetic friction, impacting how easily the box slides.
Horizontal Force Component
When a force acts on an object at an angle, it has both vertical and horizontal components. The horizontal component aids in moving the object across a surface. In our exercise, the horizontal force component is calculated using the expression \(F_{ax} = F \cos \theta\).
  • This force must overcome the kinetic friction for the box to move at a constant speed.
  • To ensure equilibrium, \(F \cos \theta\) must equal the frictional force \(f_k = \mu_k N\).
By calculating this component accurately, we determine the force required to move objects efficiently along surfaces while combating the resisting frictional forces.

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 objects, with masses 5.00 kg and 2.00 kg, hang 0.600 m above the floor from the ends of a cord that is 6.00 m long and passes over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00-kg object.

A flat (unbanked) curve on a highway has a radius of 170.0 m. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum coefficient of static friction that will prevent sliding? (b) Suppose that the highway is icy and the coefficient of static friction between the tires and pavement is only one- third of what you found in part (a). What should be the maximum speed of the car so that it can round the curve safely?

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. The cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor dropped about 0.5 m. The people remained pinned against the wall without touching the floor. (a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the person's mass? (\(Note\): When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

A 5.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force \(F(t)\) is applied to the end of the rope, and the height of the crate above its initial position is given by \(y(t) =\) (2.80 m/s)\(t +\) (0.610 m/s\(^3\))\(t^3\). What is the magnitude of F when \(t =\) 4.00 s?

A bowling ball weighing 71.2 N 116.0 lb2 is attached to the ceiling by a 3.80-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 m/s. At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

See all solutions

Recommended explanations on Physics Textbooks

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