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Chapter 5: Problem 96

You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

Short Answer

Expert verified
You must apply a force of 105.13 N.

Step by step solution

01

Analyze the Forces

First, identify and analyze the forces acting on the box. As the elevator moves upward, two key forces act on the box: gravity and the normal force. Due to the upward acceleration, the normal force is greater than just the gravitational force.
02

Calculate the Apparent Weight

Calculate the apparent weight of the box as the elevator accelerates upward. The apparent weight is the sum of the gravitational force and the force due to the upward acceleration of the elevator: \[ F_{apparent} = m(g + a) \]where \( m = 28.0 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( a = 1.90 \text{ m/s}^2 \).
03

Determine the Normal Force

The normal force \( N \) is equivalent to the apparent weight calculated in the previous step:\[ N = 28.0 \, \text{kg} \times (9.81 \, \text{m/s}^2 + 1.90 \, \text{m/s}^2) \].
04

Calculate Kinetic Friction Force

Use the coefficient of kinetic friction \( \mu_k = 0.32 \) to find the kinetic friction force \( f_k \):\[ f_k = \mu_k \times N \].
05

Apply Newton's First Law

Since the box is moving at constant speed, the horizontal forces must be balanced. Hence, the applied force \( F \) equals the frictional force:\[ F = f_k \].
06

Calculate the Applied Force

Substitute the values into the equation to find the applied force:1. Calculate the apparent weight: \[ F_{apparent} = 28.0 \, \text{kg} \times (9.81 \, \text{m/s}^2 + 1.90 \, \text{m/s}^2) = 328.52 \, \text{N} \]2. Calculate the friction force: \[ f_k = 0.32 \times 328.52 \, \text{N} = 105.1264 \, \text{N} \]The magnitude of the force you must apply is 105.13 N.

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Key Concepts

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

Kinetic Friction
When you push a box across a surface, it doesn't glide effortlessly; this resistance is due to kinetic friction. Kinetic friction comes into play when there is relative motion between two objects in contact. In this case, it's the box sliding against the elevator's floor.
  • Kinetic friction is always opposite to the direction of motion.
  • Its magnitude depends on two factors: the coefficient of kinetic friction (\(\mu_k\)) and the normal force (\(N\)).
  • This relationship is expressed by the formula: \(f_k = \mu_k \times N\).
Here, with a kinetic friction coefficient of 0.32, the force resisting the box's movement can be determined by multiplying this coefficient by the normal force. This frictional force must be balanced by your applied force to maintain a constant speed, as explained by Newton's First Law.
Apparent Weight
Apparent weight isn't the actual weight of an object. Instead, it's the weight you feel when other forces are acting on you. In an elevator, as it accelerates upwards, you feel heavier than when it's at rest. Why? Because your apparent weight increases.
  • In an upward-accelerating elevator, apparent weight is the sum of the gravitational force and the force due to the elevator's acceleration.
  • The formula for apparent weight \( F_{apparent} \) is: \( F_{apparent} = m(g + a) \).
With an upward acceleration of 1.90 m/s虏, your apparent weight increases the force pressing against the floor, simulating being heavier. This change impacts the normal force, which is crucial for calculating kinetic friction.
Normal Force
The normal force is a supporting force that acts perpendicular to the contact surface, counterbalancing other forces like gravity. When not in an elevator, this force typically equals the gravitational force. However, when acceleration comes into play鈥攕uch as an elevator moving鈥攊t alters the situation slightly.
  • The normal force is crucial for calculating kinetic friction, as it serves as the basis for the friction calculation.
  • In an accelerating elevator, the normal force increases because it not only balances gravity but also counters the force from acceleration.
The normal force in our scenario can be calculated as \( N = 28.0 \, \text{kg} \times (9.81 \, \text{m/s}^2 + 1.90 \, \text{m/s}^2) \). This increase in normal force means more kinetic friction to overcome, hence requiring a stronger push to maintain the box's constant speed.

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Most popular questions from this chapter

An airplane flies in a loop (a circular path in a vertical plane) of radius 150 \(\mathrm{m} .\) The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom. (a) At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point? (b) At the bottom of the loop, the speed of the airplane is 280 \(\mathrm{km} / \mathrm{h}\) . What is the apparent weight of the pilot at this point? His true weight is 700 \(\mathrm{N}\) .

A 6.00 -kg box sits on a ramp that is inclined at \(37.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the ramp is \(\mu_{k}=0.30 .\) What horizontal force is required to move the box up the incline with a constant acceleration of 4.20 \(\mathrm{m} / \mathrm{s}^{2}\) ?

A flat (unbanked) curve on a highway has a radius of 220.0 \(\mathrm{m} .\) A car rounds the curve at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is the minimum coefficient of friction that will prevent sliding? (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely?

While a person is walking, his arms swing through approximately a \(45^{\circ}\) angle in \(\frac{1}{2}\) s. As a reasonable approximation, we can assume that the arm moves with constant speed during each swing. A typical arm is 70.0 \(\mathrm{cm}\) long, measured from the shoulder joint. (a) What is the acceleration of a \(1.0-\mathrm{g}\) drop of blood in the fingertips at the bottom of the swing? (b) Draw a free-body diagram of the drop of blood in part (a). ( c)Find the force that the blood vessel must exert on the drop of blood in part (a). Which way does this force point? (d) What force would the blood vessel exert if the arm were not swinging?

A stockroom worker pushes a box with mass 11.2 \(\mathrm{kg}\) on a horizontal surface with a constant speed of 3.50 \(\mathrm{m} / \mathrm{s}\) . The coefficient of kinetic friction between the box and the surface is 0.20 (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?
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