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Chapter 4: Problem 36

(I) If the coefficient of kinetic friction between a 22-kg crate and the floor is 0.30, what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if \(\mu_k\) is zero?

Short Answer

Expert verified
64.68 N with friction; 0 N without friction.

Step by step solution

01

Understand the problem

We need to find the horizontal force necessary to move a 22-kg crate across the floor at a steady speed, given the coefficient of kinetic friction is 0.30. Also, we need to compute the force required if there's no kinetic friction (coefficient is zero).
02

Calculate the Normal Force

The normal force (\(N\)) on the crate is equal to its weight because it is moving horizontally with constant speed, and there are no vertical accelerations. The formula for the normal force is: \[ N = m \cdot g \]where \( m = 22 \, \text{kg} \) is the mass of the crate, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.Substituting the given values, we find:\[ N = 22 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 215.6 \, \text{N} \]
03

Calculate the Frictional Force with 渭鈧� = 0.30

The frictional force, \( f_k \), can be found using:\[ f_k = \mu_k \cdot N \]where \( \mu_k = 0.30 \). Substituting in the normal force:\[ f_k = 0.30 \times 215.6 \, \text{N} = 64.68 \, \text{N} \]
04

Calculate the Required Horizontal Force with Friction

To move the crate at a steady speed, the horizontal force \( F \) must equal the frictional force because there is no acceleration. Therefore, the horizontal force required is:\[ F = f_k = 64.68 \, \text{N} \]
05

Calculate Horizontal Force with 渭鈧� = 0

If the coefficient of kinetic friction is zero, the frictional force becomes zero. Therefore, no horizontal force is needed to move the crate at a steady speed, so:\[ F = 0 \, \text{N} \]

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Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the motion of two objects sliding against each other. When a crate is dragged across the floor, kinetic friction is what tries to slow it down. This frictional force depends on two main factors: the coefficient of kinetic friction (\( \mu_k \)) and the normal force. The coefficient of kinetic friction is a dimensionless value that represents the extent of friction between two surfaces. A higher value means more friction. In our original exercise, \( \mu_k \) is 0.30, indicating moderate friction. The formula to calculate the kinetic friction force is \( f_k = \mu_k \cdot N \), where \( f_k \) is the kinetic friction force and \( N \) is the normal force.
Normal Force
The normal force is a perpendicular force exerted by a surface to support the weight of an object resting on it. It acts in the opposite direction to gravity. When you place an object on a flat surface, like a crate on a floor, gravity pulls it downward. To prevent the object from sinking into the ground, the surface exerts a normal force upwards. This force is equal in magnitude and opposite in direction to the gravitational force acting on the object, meaning \( N = m \cdot g \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. For a 22-kg crate on Earth, this means a normal force of 215.6 N."
The normal force plays a crucial role in determining the amount of kinetic friction, as seen in the formula \( f_k = \mu_k \cdot N \).
Horizontal Force
Horizontal force is any force applied parallel to the horizon. In our scenario, moving the crate at a steady speed requires a horizontal force. When an object moves with a constant velocity on a plane surface, the horizontal force should balance the frictional force. This is because there is no acceleration. Thus, the object's velocity remains constant. In the context of our exercise, when kinetic friction exists, the horizontal force required to move the 22-kg crate is equal to the frictional force calculated earlier, which is 64.68 N.
However, if there is no kinetic friction (\( \mu_k = 0 \)), no horizontal force is required because there is no opposing frictional force, illustrating how crucial friction is in motion dynamics.

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Most popular questions from this chapter

A 75-kg snowboarder has an initial velocity of 5.0 m/s at the top of a 28\(^\circ\) incline (Fig. 4-75). After sliding down the 110-m-long incline (on which the coefficient of kinetic friction is \(\mu_k=\) 0.18 ), the snowboarder has attained a velocity \(v\). The snowboarder then slides along a flat surface (on which \(\mu_k=\) 0.15 ) and comes to rest after a distance \(x\). Use Newton's second law to find the snowboarder's acceleration while on the incline and while on the flat surface. Then use these accelerations to determine \(x\).

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be so that even low-powered cars can get up the hills without slowing down. A particular small car, with a mass of 920 kg, can accelerate on a level road from rest to 21 m/s (75 km/h) in 12.5 s. Using these data, calculate the maximum steepness of a hill.

A 75.0-kg person stands on a scale in an elevator. What does the scale read (in N and in kg) when (\(a\)) the elevator is at rest, (\(b\)) the elevator is climbing at a constant speed of 3.0 m/s, (\(c\)) the elevator is descending at 3.0 m/s, (\(d\)) the elevator is accelerating upward at 3.0 m/s\(^2\), (\(e\)) the elevator is accelerating downward at 3.0 m/s\(^2\)?

(I) A force of 35.0 N is required to start a 6.0-kg box moving across a horizontal concrete floor. (\(a\)) What is the coefficient of static friction between the box and the floor? (\(b\)) If the 35.0-N force continues, the box accelerates at 0.60 m/s\(^2\) What is the coefficient of kinetic friction?

(II) A 2.0-kg silverware drawer does not slide readily. The owner gradually pulls with more and more force, and when the applied force reaches 9.0 N, the drawer suddenly opens, throwing all the utensils to the floor. What is the coefficient of static friction between the drawer and the cabinet?
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