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Chapter 4: Problem 39

A \(60.0-\mathrm{kg}\) 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) 447.336 N, (b) 241.326 N

Step by step solution

01

Calculate the Normal Force

The normal force (\(F_N\)) that acts on the crate is equal to the gravitational force acting on it. Since the crate's mass is 60.0 kg and the acceleration due to gravity (\(g\)) is approximately 9.81 \(\text{m/s}^2\), the normal force can be calculated as: \[ F_N = m \, g \]. Thus, \[ F_N = 60.0 \times 9.81 = 588.6 \, \text{N} \].
02

Calculate the Static Frictional Force

The maximum static frictional force (\(f_s\)) can be calculated using the coefficient of static friction (\(\mu_s\)) and the normal force \(F_N\). The formula is \(f_s = \mu_s \, F_N\). Substitute the given values: \[ f_s = 0.760 \times 588.6 = 447.336 \, \text{N} \]. Therefore, a horizontal force of 447.336 N is required to just start moving the crate.
03

Calculate the Kinetic Frictional Force

Once the crate is in motion, the force required to keep it moving at a constant speed is determined by the kinetic frictional force (\(f_k\)). This is calculated using the coefficient of kinetic friction (\(\mu_k\)) and the normal force \(F_N\). The formula is \(f_k = \mu_k \, F_N\). Substitute the given values: \[ f_k = 0.410 \times 588.6 = 241.326 \, \text{N} \]. Therefore, a horizontal force of 241.326 N is needed 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.

Understanding Normal Force
Normal force is the supportive force exerted on an object that is in contact with another stable object. Imagine you place a crate on the floor. The crate pushes down due to gravity, and the floor pushes up with an equal force to support it, which is the normal force. This force is crucial because it balances the weight of the crate, ensuring it doesn't fall through the floor.

In mathematical terms, the normal force (\(F_N\)) for an object resting on a horizontal surface equals the gravitational force on the object. This gravitational force can be calculated by multiplying the mass (\(m\)) of the object by the acceleration due to gravity (\(g\)), usually approximated as 9.81 \(\text{m/s}^2\). Hence, the formula for normal force is:
  • \[ F_N = m \times g \]
To put it into perspective, for a crate that weighs 60 kg, the normal force is 588.6 N. That is how much the floor pushes back against the crate's weight. This helps us understand how other forces, like friction, come into play.
What is Static Friction?
Static friction is the force that keeps an object at rest when a force is applied, keeping it from starting to move. It's like the brakes on a car, preventing it from rolling. This frictional force must be overcome to start moving the object.

The maximum static friction force (\(f_s\)) can be determined using the coefficient of static friction (\(\mu_s\)) and the normal force (\(F_N\)). The coefficient of static friction is a number, usually less than 1, that describes how sticky two surfaces are to each other. Its formula is:
  • \[ f_s = \mu_s \times F_N \]
For the 60 kg crate, with a static friction coefficient of 0.760, the maximum force needed to initiate its movement is 447.336 N. This precise force indicates the threshold where static friction is overcome, allowing the crate to start moving.
Understanding Kinetic Friction
Kinetic friction, unlike static friction, occurs when an object is already in motion. Think of it as the resistance you feel when sliding a book across a table. It's slightly less than the force required to get it moving initially. Once moving, it's the kinetic friction force that you must overcome to keep the object moving at a constant speed.

For a moving object, the kinetic friction force (\(f_k\)) is calculated similarly to static friction, using the coefficient of kinetic friction (\(\mu_k\)) instead. The formula is:
  • \[ f_k = \mu_k \times F_N \]
In the case of the moving crate, where \(\mu_k = 0.410\), this means the force needed to slide it at a constant speed is 241.326 N. Understanding these concepts ensures we can predict and control motion, essential skills in physics.

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Most popular questions from this chapter

A person is trying to judge whether a picture (mass \(=1.10 \mathrm{kg}\) ) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is \(0.660 .\) What is the minimum amount of pressing force that must be used?

Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find (a) the acceleration of the objects and (b) the tension in the string.

On earth, two parts of a space probe weigh 11 000 N and 3400 N. These parts are separated by a center-to-center distance of 12 m and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

Two horizontal forces, \(\overrightarrow{\mathbf{F}}_{1}\) and \(\overrightarrow{\mathbf{F}}_{2},\) are acting on a box, but only \(\overrightarrow{\mathbf{F}}_{1}\) is shown in the drawing. \(\overrightarrow{\mathbf{F}}_{2}\) can point either to the right or to the left. The box moves only along the \(x\) axis. There is no friction between the box and the surface. Suppose that \(\overrightarrow{\mathbf{F}}_{1}=+9.0 \mathrm{N}\) and the mass of the box is \(3.0 \mathrm{kg}\). Find the magnitude and direction of \(\overrightarrow{\mathbf{F}}_{2}\) when the acceleration of the box is (a) \(+5.0 \mathrm{m} / \mathrm{s}^{2},\) (b) \(-5.0 \mathrm{m} / \mathrm{s}^{2},\) and (c) \(0 \mathrm{m} / \mathrm{s}^{2}.\)

A girl is sledding down a slope that is inclined at \(30.0^{\circ}\) with respect to the horizontal. The wind is aiding the motion by providing a steady force of \(105 \mathrm{N}\) that is parallel to the motion of the sled. The combined mass of the girl and the sled is \(65.0 \mathrm{kg}\), and the coefficient of kinetic friction between the snow and the runners of the sled is \(0.150 .\) How much time is required for the sled to travel down a \(175-\mathrm{m}\) slope, starting from rest?
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