/*! 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 5 A \(2.5 \mathrm{~kg}\) block is ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 5

A \(2.5 \mathrm{~kg}\) block is initially at rest on a horizontal surface. A horizontal force \(\vec{F}\) of magnitude \(6.0 \mathrm{~N}\) and a vertical force \(\vec{P}\) are then applied to the block (Fig. \(6-17) .\) The coefficients of friction for the block and surface are \(\mu_{s}=0.40\) and \(\mu_{k}=0.25 .\) Determine the magnitude of the frictional force acting on the block if the magnitude of \(\vec{P}\) is (a) \(8.0 \mathrm{~N},\) (b) \(10 \mathrm{~N}\), and (c) \(12 \mathrm{~N}\).

Short Answer

Expert verified
(a) 6.0 N, (b) 3.625 N, (c) 3.125 N

Step by step solution

01

Determine the normal force for case (a)

Calculate the normal force for the case where \( P = 8.0 \text{ N} \). The gravitational force on the block is \( F_g = mg \), where \( m = 2.5 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Thus, \( F_g = 2.5 \cdot 9.8 = 24.5 \text{ N} \). Since \( P \) is the vertical force acting upwards, the normal force \( N = F_g - P = 24.5 - 8.0 = 16.5 \text{ N} \).
02

Calculate static friction for case (a)

The maximum static friction is \( f_s = \mu_s N \). With \( \mu_s = 0.40 \) and \( N = 16.5 \text{ N} \), \( f_s = 0.40 \times 16.5 = 6.6 \text{ N} \).
03

Compare static friction and applied force for case (a)

The applied horizontal force is \( F = 6.0 \text{ N} \). Since \( F < f_s \) (6.0 < 6.6), the block does not move, and the frictional force is equal to the applied force: \( f = 6.0 \text{ N} \).
04

Determine normal force for case (b)

For \( P = 10 \text{ N} \), the normal force \( N \) is calculated as \( N = F_g - P = 24.5 - 10 = 14.5 \text{ N} \).
05

Calculate static friction for case (b)

With \( N = 14.5 \text{ N} \), the maximum static friction is \( f_s = \mu_s N = 0.40 \times 14.5 = 5.8 \text{ N} \).
06

Compare static friction and applied force for case (b)

Since the applied force \( F = 6.0 \text{ N} \) is greater than the maximum static friction \( f_s = 5.8 \text{ N} \), the block begins to move. Hence, the kinetic friction is \( f_k = \mu_k N = 0.25 \times 14.5 = 3.625 \text{ N} \).
07

Determine normal force for case (c)

For \( P = 12 \text{ N} \), the normal force \( N = F_g - P = 24.5 - 12 = 12.5 \text{ N} \).
08

Calculate static friction for case (c)

With \( N = 12.5 \text{ N} \), the maximum static friction is \( f_s = \mu_s N = 0.40 \times 12.5 = 5.0 \text{ N} \).
09

Compare static friction and applied force for case (c)

Since the applied force \( F = 6.0 \text{ N} \) is greater than the maximum static friction \( f_s = 5.0 \text{ N} \), the block moves. The kinetic friction here is \( f_k = \mu_k N = 0.25 \times 12.5 = 3.125 \text{ N} \).

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Key Concepts

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

Static Friction
Static friction is the force that prevents an object at rest from moving when a force is applied. It acts opposite to the direction of the applied force. This force is determined by the coefficient of static friction (\( \mu_s \)), which is a ratio between the maximum static friction force and the normal force acting on an object. Here's what you need to know about static friction:
  • If the applied force is less than the maximum static friction, the object will not move.
  • The formula for static friction is: \[ f_s = \mu_s \times N \]where \( N \) is the normal force.
For example, in case (a) of our exercise, the applied force of \( 6.0 \text{ N} \) was less than the maximum static friction of \( 6.6 \text{ N} \), meaning the block stayed at rest. The key takeaway is that static friction varies and can adjust to match the applied force until it reaches the maximum value, beyond which movement occurs.
Kinetic Friction
Kinetic friction takes over once an object starts moving. Unlike static friction that varies, kinetic friction is constant for the motion of an object at a given speed and surface condition. The coefficient of kinetic friction (\( \mu_k \)) is typically lower than static friction, leading to less force required to keep an object moving than to start the movement.
  • The formula for kinetic friction is: \[ f_k = \mu_k \times N \]where \( N \) is the normal force.
  • Kinetic friction always acts opposite to the direction of motion, working to slow down the moving object.
In parts (b) and (c) of the exercise, once the applied force surpassed static friction, the block began to slide. The kinetic friction then was calculated to be \( 3.625 \text{ N} \) and \( 3.125 \text{ N} \) for \( P = 10 \text{ N} \) and \( P = 12 \text{ N} \) respectively.
Normal Force
The normal force (\( N \)) is one of the key elements in understanding frictional forces. It is the perpendicular force exerted by a surface against an object resting on it. The magnitude of the normal force changes based on other forces acting on the object, such as vertical pushes or pulls.
  • In essence, it balances the gravitational force to prevent the object from sinking into the surface.
  • In calculations, normal force can be found using: \[ N = F_g - P \]where \( F_g \) is the gravitational force and \( P \) is any other vertical force acting on the object.
In our example, different magnitudes of \( P \) (8.0 N, 10.0 N, and 12.0 N) adjusted the normal force, influencing the frictional forces the block experienced. Knowing the normal force is crucial because it directly affects both static and kinetic friction calculations.

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Most popular questions from this chapter

An airplane is flying in a horizontal circle at a speed of \(480 \mathrm{~km} / \mathrm{h}\) (Fig. 6-41). If its wings are tilted at angle \(\theta=40^{\circ}\) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

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?

A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68,\) and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) \(222 \mathrm{~N}\), (b) \(334 \mathrm{~N},\) (c) \(445 \mathrm{~N},\) and (d) \(556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

Two blocks, of weights \(3.6 \mathrm{~N}\) and \(7.2 \mathrm{~N},\) are connected by a massless string and slide down a \(30^{\circ}\) inclined plane. The coefficient of kinetic friction between the lighter block and the plane is \(0.10,\) and the coefficient between the heavier block and the plane is \(0.20 .\) Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string.

A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod (Fig. \(6-42\) ). The strobe rate is 2000 flashes per second; the bolt has mass \(30 \mathrm{~g}\) and is at radius \(3.5 \mathrm{~cm} .\) What is the magnitude of the force on the bolt from the rod?
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