/*! 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 83 You must push a crate across a f... [FREE SOLUTION] | 91Ó°ÊÓ

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You must push a crate across a floor to a docking bay. The crate weighs \(165 \mathrm{~N}\). The coefficient of static friction between crate and floor is \(0.510\), and the coefficient of kinetic friction is \(0.32\). Your force on the crate is directed horizontally. (a) What magnitude of your push puts the crate on the verge of sliding? (b) With what magnitude must you then push to keep the crate moving at a constant velocity? (c) If, instead, you then push with the same magnitude as the answer to (a), what is the magnitude of the crate's acceleration?

Short Answer

Expert verified
(a) 84.15 N, (b) 52.8 N, (c) 1.86 m/s².

Step by step solution

01

Understand the Problem for Part (a)

To determine the magnitude of your push to put the crate on the verge of sliding, calculate the maximum static friction. This happens when the force you apply equals the force of static friction.
02

Calculate Maximum Static Friction

The maximum static friction force can be calculated with the formula: \\( f_{s,\max} = \mu_s \times N \)\where \( \mu_s = 0.510 \) and \( N = 165 \text{ N} \) (the weight of the crate).
03

Solve for Maximum Static Friction

\[ f_{s,\max} = 0.510 \times 165 \text{ N} = 84.15 \text{ N} \] \Thus, a horizontal force of 84.15 N will put the crate on the verge of sliding.
04

Recognize the Problem for Part (b)

To keep the crate moving at a constant velocity, you need a force equal to the kinetic friction force, as the velocities are constant.
05

Calculate Kinetic Friction

Kinetic friction force is given by the equation: \\( f_k = \mu_k \times N \) \where \( \mu_k = 0.32 \) and the normal force \( N = 165 \text{ N} \).
06

Solve for Kinetic Friction Force

\[ f_k = 0.32 \times 165 \text{ N} = 52.8 \text{ N} \] \You need to apply a force of 52.8 N to keep the crate moving at constant velocity.
07

Understand the Problem for Part (c)

For this part, calculate the acceleration if you push the crate with the force calculated in part (a), which is now greater than the force needed to overcome kinetic friction.
08

Determine Net Force When Pushing with Part (a) Force

The net force can be found by subtracting the kinetic friction force from your applied force: \\( F_{\text{net}} = 84.15 \text{ N} - 52.8 \text{ N} \)
09

Calculate the Acceleration of the Crate

Using Newton's second law, where the net force equals mass times acceleration: \\( F_{\text{net}} = m \times a \).\First, find the mass \( m \), given by \( m = \frac{165 \text{ N}}{9.8 \text{ m/s}^2} \approx 16.84 \text{ kg} \).\Then, solve for \( a \): \\[ a = \frac{84.15 - 52.8}{16.84} \approx 1.86 \text{ m/s}^2 \]
10

Summary of Solutions

(a) 84.15 N is needed to start sliding, (b) 52.8 N is needed for constant velocity, (c) using 84.15 N gives an acceleration of 1.86 m/s².

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is one of the three fundamental laws laid down by Sir Isaac Newton. It explains how the velocity of an object changes when it is subjected to an external force.
This law is typically expressed with the equation \( F = m \times a \), where:
  • \( F \) represents the net force applied to the object,
  • \( m \) is the mass of the object, and
  • \( a \) is the acceleration.
In this exercise, we're using Newton's Second Law to find the acceleration of a crate when a force is applied. The essence of the law is a direct relationship between the net force acting upon an object and its resulting acceleration. Simply put, if you apply more force to an object, it will accelerate more.
To find the net force, we subtract any opposing forces, like friction, from the force we apply. In the exercise, solving for the crate's acceleration involved using this fundamental relationship by subtracting the frictional force from the applied force to find the net force first. From that net force, kinetic equations helped us find out how quickly the crate speeds up (accelerates). Understanding this law is crucial because it allows us to predict an object's movement in a predictable fashion by knowing the forces at play.
Frictional Force Calculation
Frictional force is the force resisting the relative motion of two surfaces sliding against one another. There are two types of friction relevant to our problem: static friction and kinetic friction.
Static friction acts when an object is not moving, while kinetic friction comes into play once the object is in motion.
To calculate these forces, we use the formulas:
  • For static friction: \( f_{s,\text{max}} = \mu_s \times N \)
  • For kinetic friction: \( f_k = \mu_k \times N \)
Where \( \mu_s \) and \( \mu_k \) are the coefficients of static and kinetic friction, respectively, and \( N \) is the normal force (usually the weight of the object in simple scenarios like this).
In our example, the static friction calculation determined how much force was needed to start moving the crate, while the kinetic friction calculation identified the force necessary to maintain a constant speed. Once the static friction is overcome, only the kinetic friction needs to be countered to keep the object moving smoothly. Understanding these forces helps us realize why different amounts of force are needed in various phases of moving an object.
Acceleration
Acceleration refers to the rate at which an object changes its velocity. In this context, acceleration doesn't just happen on its own; it requires a net force acting on the object.
In our exercise scenario, after the crate is set in motion by overcoming static friction, to further change its state of motion (i.e., to accelerate), the applied force needs to exceed the kinetic friction force.
Using Newton's Second Law, to find the acceleration \( a \), we establish the net force first. We do this by taking the difference between the total force applied and the frictional force:\[ F_\text{net} = F_\text{applied} - f_k \]
With the net force in hand, we can then calculate acceleration using \( a = \frac{F_\text{net}}{m} \). In our situation, applying a force greater than the frictional force resulted in an acceleration of 1.86 m/s². This was calculated by dividing the net force by the mass of the crate.
Knowing how acceleration works and how to calculate it using the precise dynamics of force and mass is essential for predicting how quickly an object will gain speed in practical scenarios. This concept is foundational in physics for understanding the motion of objects when varying forces are applied.

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Most popular questions from this chapter

A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of kinetic friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle \(200 \mathrm{~m}\) slope in the Alps, a skier reduced his top-to-bottom time from \(61 \mathrm{~s}\) with standard skis to \(42 \mathrm{~s}\) with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a \(3.0^{\circ}\) slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

A child places a picnic basket on the outer rim of a merrygo-round that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

A person pushes horizontally with a force of \(220 \mathrm{~N}\) on a \(55 \mathrm{~kg}\) crate to move it across a level floor. The coefficient of kinetic friction between the crate and the floor is \(0.35 .\) What is the magnitude of (a) the frictional force and (b) the acceleration of the crate?

\( \Rightarrow \sqrt{2}\) A sling-thrower puts a stone \((0.250 \mathrm{~kg})\) in the sling's pouch \((0.010 \mathrm{~kg})\) and then begins to make the stone and pouch move in a vertical circle of radius \(0.650 \mathrm{~m}\). The cord between the pouch and the person's hand has negligible mass and will break when the tension in the cord is \(33.0 \mathrm{~N}\) or more. Suppose the slingthrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur?

A locomotive accelerates a 25 -car train along a level track. Every car has a mass of \(5.0 \times 10^{4} \mathrm{~kg}\) and is subject to a frictional force \(f=250 v\), where the speed \(v\) is in meters per second and the force \(f\) is in newtons. At the instant when the speed of the train is \(30 \mathrm{~km} / \mathrm{h}\), the magnitude of its acceleration is \(0.20 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at \(30 \mathrm{~km} / \mathrm{h}\) ?

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