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ssm A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?

Short Answer

Expert verified
A force of 447.336 N starts the crate moving, and a force of 241.326 N keeps it moving at constant speed.

Step by step solution

01

Find the Normal Force

The normal force \( N \) is equal to the gravitational force acting on the crate since the crate is on a level floor and there is no vertical acceleration. The normal force is \( N = mg \), where \( m = 60.0 \) kg and \( g = 9.81 \) m/s² is the acceleration due to gravity. Thus, \( N = 60.0 \times 9.81 \approx 588.6 \) N.
02

Calculate the Force to Overcome Static Friction

The force required to just start the crate moving is determined by the maximum static friction force \( f_s \). The static friction force is \( f_s = \mu_s N \), where \( \mu_s = 0.760 \). Thus, \( f_s = 0.760 \times 588.6 \approx 447.336 \) N. Therefore, a horizontal force of approximately 447.336 N is required to just start moving the crate.
03

Calculate the Force to Overcome Kinetic Friction

Once the crate is moving, the force required to keep it sliding at a constant speed is equal to the kinetic friction force \( f_k \). The kinetic friction force is \( f_k = \mu_k N \), where \( \mu_k = 0.410 \). Thus, \( f_k = 0.410 \times 588.6 \approx 241.326 \) N. Therefore, a horizontal force of approximately 241.326 N is required to slide the crate at a constant speed.

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Key Concepts

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

Normal Force
Normal force is a key concept in understanding how objects interact with each other. When an object is placed on a surface, the normal force is the perpendicular contact force exerted by the surface on the object. For instance, a 60.0-kg crate on a level floor experiences a normal force equal to its weight, due to gravity acting upon it.
The normal force can be calculated using the formula:
  • \[ N = mg \]
where
  • \( m \) is the mass of the object (in this case, 60.0 kg)
  • \( g \) is the acceleration due to gravity, approximately 9.81 m/s²
When we compute, the normal force \( N \) becomes approximately 588.6 N. This means the floor pushes back against the crate with this force, balancing out its weight without causing any vertical acceleration.
This force is essential for calculating the forces of friction between surfaces.
Static Friction
Static friction acts between surfaces that are at rest, preventing them from sliding over each other. It is the force that you need to overcome for an object to start moving. The strength of static friction depends on two things: the normal force and the coefficient of static friction (\( \mu_s \)).
The formula for static friction force \( f_s \) is:
  • \[ f_s = \mu_s N \]
where
  • \( \mu_s \) is the static friction coefficient
  • \( N \) is the normal force, in this example 588.6 N
Given \( \mu_s = 0.760 \), the force to initiate motion (start pushing the crate) is around 447.336 N. This means that you need to apply a horizontal force at least equal to this to set the crate in motion.
Kinetic Friction
Kinetic friction comes into play once an object is sliding over a surface. It is usually less than static friction, which is why it's generally easier to keep an object moving than to start moving it. The force of kinetic friction depends on the same two factors as static friction: the normal force and the coefficient of kinetic friction (\( \mu_k \)).
The formula for kinetic friction force \( f_k \) is:
  • \[ f_k = \mu_k N \]
where
  • \( \mu_k \) is the kinetic friction coefficient
  • \( N \) is the normal force
With \( \mu_k = 0.410 \), the force required to keep the crate sliding at constant speed is approximately 241.326 N.
This force is crucial for maintaining a steady state of motion without acceleration.
Newton's Laws of Motion
Newton's laws provide us with the principles needed to understand motion. The first law, often termed the law of inertia, states that an object will stay at rest or move in a straight line unless acted upon by a force. This is why the static friction force must be overcome to move the crate initially.

Newton's second law gives us the formula \( F = ma \), relating net force, mass, and acceleration. When calculating forces like static and kinetic friction, we are essentially determining how much force is needed to change the crate's state of rest or constant velocity.

The third law states that for every action, there is an equal and opposite reaction. In the context of normal force, this law explains why surfaces exert a force equal to the weight of the object (as long as there is no vertical acceleration), balancing the gravitational pull downward.
These laws of motion underpin our calculations and understanding of any forces acting on objects in motion.

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Most popular questions from this chapter

mmh The drawing shows a large cube (mass \(=25 \mathrm{kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\overrightarrow{\mathbf{P}}\) . A small cube (mass \(=4.0 \mathrm{kg} )\) is in contact with the front surface of the large cube and will slide downward unless \(\overrightarrow{\mathbf{P}}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71 .\) What is the smallest magnitude that \(\overrightarrow{\mathbf{P}}\) can have in order to keep the small cube from sliding downward?

At a distance \(H\) above the surface of a planet, the true weight of a remote probe is one percent less than its true weight on the surface. The radius of the planet is \(R\) . Find the ratio \(H / R\) .

mmh A boat has a mass of 6800 kg. Its engines generate a drive force of 4100 N due west, while the wind exerts a force of 800 N due east andthe water exerts a resistive force of 1200 N due east. What are the magnitude and direction of the boat’s acceleration?

ssm A 6.00-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 1.20 \(\mathrm{m} / \mathrm{s}^{2}\) , and \((\mathrm{c})\) accelerating downward with an acceleration whose magnitude is 1.20 \(\mathrm{m} / \mathrm{s}^{2}\) .

A 1.40-kg bottle of vintage wine is lying horizontally in a rack, as shown in the drawing. The two surfaces on which the bottle rests are \(90.0^{\circ}\) apart, and the right surface makes an angle of \(45.0^{\circ}\) with respect to the horizontal. Each surface exerts a force on the bottle that is perpendicular to the surface. Both forces have the same magnitude \(F .\) Find the value of \(F\) .

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