/*! 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 38 A box with mass \(m\) is dragged... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A box with mass \(m\) is dragged across a level floor with coefficient of kinetic friction \(\mu_{\mathrm{k}}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F\). (a) In terms of \(m, \mu_{\mathrm{k}}, \theta,\) 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 \(90 \mathrm{~kg}\) 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 \(\mu_{\mathrm{k}}=0.35 .\) Use the result of part (a) to answer the instructor's question.

Short Answer

Expert verified
A force of approximately 385 N is required.

Step by step solution

01

Analysis of forces

Identify all the forces on the box: There is a downwards force due to gravity equal to \(mg\). There is a normal force upwards from the ground. The kinetic friction force \(F_{k}\) opposes the motion of the box and there is an applied force \(F\) at an angle \(\theta\) above the horizontal.
02

Newton’s second law in the vertical direction

According to Newton’s second law, the sum of the forces in the vertical direction is zero, because the box is not moving vertically. So, we write \(N - mg + F \sin \theta = 0\). From this equation, we find the normal force \(N = mg - F\sin\theta\)
03

Newton’s second law in the horizontal direction

Again, the sum of the forces in the horizontal direction is zero, because the box is moving with a constant speed (no acceleration). The friction force is given by \(F_{k} = \mu_k N\). So we write \(- F_{k} + F \cos \theta = 0\), and replace \(F_{k}\) and \(N\) with their expressions from above. This gives \(F = \frac{ mg \mu_k}{\cos\theta + \mu_k \sin\theta}\)
04

Substitute numbers from part (b)

To estimate the necessary force to move the patient, plug in the given values \(m = 90\, \mathrm{kg}\), \(\mu_{\mathrm{k}} = 0.35\), \(\theta = 25^{\circ}, g = 9.8\, \mathrm{m/s^2}\) in the formula from step 3. Note that you need to convert the angle from degree to radian before substituting it.
05

Calculation

After substituting the given values, the formula gives \(F \approx 385 \mathrm{N}\). Therefore, it would take a force of approximately 385 N to drag the patient across the floor at a constant speed.

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 91Ó°ÊÓ!

Key Concepts

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

Newton's Second Law
Newton's Second Law plays a crucial role in understanding how forces affect motion. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be expressed through the equation: \[ F_{ ext{net}} = ma \] where \( F_{\text{net}} \) is the net force, \( m \) is the mass, and \( a \) is the acceleration of the object. In this exercise, however, we focused on the scenario where the net force, particularly in the vertical and horizontal directions, is zero. This is because the box is moving at a constant speed and not accelerating. For the vertical direction, we applied Newton's Second Law by setting the sum of forces to zero: \[ N - mg + F \sin \theta = 0 \] This ensures that the vertical forces are balanced since the box doesn’t move up or down. Similarly, in the horizontal direction, we balance forces to achieve constant speed by:\[ - F_{k} + F \cos \theta = 0 \] This states that the horizontal applied force is perfectly countered by the opposing friction force.
Normal Force
The normal force, denoted as \( N \), is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In the context of this problem, the normal force counteracts gravity and the component of the applied force, which tries to lift the box off the ground.The normal force is crucial because it affects the frictional force, which relies on the normal force's magnitude. From the vertical forces balance, as per Newton's Second Law:\[ N = mg - F \sin \theta \] Here, \( mg \) is the gravitational force, and \( F \sin \theta \) is the upward force component. This formula shows that as the angle \( \theta \) increases, resulting in a larger upward component, the normal force decreases, reducing the frictional force.Normal force plays an integral role when calculating the kinetic friction, as friction depends on this adaptive force.
Constant Speed Motion
Constant speed motion implies there is no acceleration. In practical terms, this means the forces acting on the object must be perfectly balanced out.Since the object's speed remains unchanged, according to Newton's Second Law, the net horizontal force is zero. This is because:\[ F_{\text{applied}} = F_{k} \]Where \( F_{\text{applied}} \) is the part of the force that aims to pull the box horizontally, represented by \( F \cos \theta \). The friction force \( F_{k} \) opposes this motion. Hence:\[ F \cos \theta = mg \mu_k \]When moving at constant speed, forces in both the horizontal and vertical directions must sum to zero. Thus, the applied force must exactly match the kinetic friction force to maintain that speed. This principle helped derive the formula necessary to exert just the right force to move the box without accelerating.

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

When a crate with mass \(25.0 \mathrm{~kg}\) is placed on a ramp that is inclined at an angle \(\alpha\) below the horizontal, it slides down the ramp with an acceleration of \(4.9 \mathrm{~m} / \mathrm{s}^{2} .\) The ramp is not friction less. To increase the acceleration of the crate, a downward vertical force \(\overrightarrow{\boldsymbol{F}}\) is applied to the top of the crate. What must \(F\) be in order to increase the acceleration of the crate so that it is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) ? How does the value of \(F\) that you calculate compare to the weight of the crate?

A man pushes on a piano with mass \(180 \mathrm{~kg} ;\) it slides at constant velocity down a ramp that is inclined at \(19.0^{\circ}\) above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.

A box of bananas weighing \(40.0 \mathrm{~N}\) rests on a horizontal surface. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is \(0.20 .\) (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of \(6.0 \mathrm{~N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of \(18.0 \mathrm{~N},\) what is the magnitude of the friction force and what is the box's acceleration?

BIO Force During a Jump. When jumping straight up from a crouched position, an average person can reach a maximum height of about \(60 \mathrm{~cm} .\) During the jump, the person's body from the knees up typically rises a distance of around \(50 \mathrm{~cm} .\) To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of \(60 \mathrm{~cm} ?\) (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(w\), what force does the ground exert on him or her during the jump?

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 \(2.5 \mathrm{~m} .\) The cylinder started to rotate, and when it reached a constant rotation rate of \(0.60 \mathrm{rev} / \mathrm{s},\) the floor dropped about \(0.5 \mathrm{~m}\). 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 wall to the floor.

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.