# Difference between revisions of "Dragging problem"

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| <math>mg</math> || [[gravitational force]] || unconditional || <math>mg</math> || vertical, downward | | <math>mg</math> || [[gravitational force]] || unconditional || <math>mg</math> || vertical, downward | ||

|- | |- | ||

− | | <math>N</math> || [[normal force]] || assuming that <math>F \sin \theta \le mg</math>, otherwise the block will lift off the floor. || vertical, upward | + | | <math>N</math> || [[normal force]] || assuming that <math>F \sin \theta \le mg</math>, otherwise the block will lift off the floor. || adjusts so that there is no net force in the vertical direction || vertical, upward |

|- | |- | ||

− | | <math>f_s</math> || [[static friction]] || no sliding || horizontal, opposite to horizontal component of <math>F</math> | + | | <math>f_s</math> || [[static friction]] || no sliding || adjusts so that there is no net force in the horizontal direction || horizontal, opposite to horizontal component of <math>F</math> |

|- | |- | ||

− | | <math>f_k</math> || [[kinetic friction]] || sliding || horizontal, opposite to direction of slipping. Assuming initially at rest, direction of slipping = horizontal component of <math>F</math>, so this is opposite to horizontal component of <math>F</math>. | + | | <math>f_k</math> || [[kinetic friction]] || sliding || <math>\mu_kN</math> || horizontal, opposite to direction of slipping. Assuming initially at rest, direction of slipping = horizontal component of <math>F</math>, so this is opposite to horizontal component of <math>F</math>. |

|} | |} | ||

+ | |||

+ | ===Vertical components=== | ||

+ | |||

+ | Taking components in the vertical direction, and assuming no acceleration in the vertical direction (because the block does not lift off), we get: | ||

+ | |||

+ | <math>mg = N + F \sin \theta \qquad (1.1)</math> | ||

+ | |||

+ | ===Horizontal components and solution assuming no sliding=== | ||

+ | |||

+ | If there is no sliding, we get: | ||

+ | |||

+ | <math>F \cos \theta = f_s \qquad (2.1)</math> | ||

+ | |||

+ | We also have: | ||

+ | |||

+ | <math>f_s \le \mu_sN \qquad (2.2)</math> | ||

+ | |||

+ | Plugging in <math>N</math> from (1.1) into (2.2) and plugging <math>f_s</math> from this into (2.1) gives: | ||

+ | |||

+ | <math>F \cos \theta \le \mu_smg - \mu_sF\sin \theta \qquad (2.3)</math> | ||

+ | |||

+ | This simplifies to: | ||

+ | |||

+ | <math>F \le \frac{\mu_smg}{\cos \theta + \mu_s \sin \theta} \qquad (2.4)</math> | ||

+ | |||

+ | This is a necessary and sufficient condition for there to be no sliding ''assuming'' the block was initially at rest. | ||

+ | |||

+ | ===Horizontal components and solution assuming sliding=== | ||

+ | |||

+ | If there is sliding along the horizontal component of the direction of <math>F</math>, denote by <math>a</math> the acceleration along the horizontal component of the direction of <math>F</math>. We get: | ||

+ | |||

+ | <math>F \cos \theta - f_k = ma \qquad (3.1)</math> | ||

+ | |||

+ | We also have: | ||

+ | |||

+ | <math>f_k = \mu_kN \qquad (3.2)</math> | ||

+ | |||

+ | Plugging in <math>N</math> from (1.1) into (3.2) and plugging <math>f_k</math> from this into (3.1) gives: | ||

+ | |||

+ | <math>F \cos \theta - \mu_smg + \mu_kF\sin \theta = ma\qquad (2.3)</math> | ||

+ | |||

+ | This simplifies to: | ||

+ | |||

+ | <math>a = \frac{F(\cos \theta + \mu_k \sin \theta)}{m} - \mu_sg \qquad (2.4)</math> | ||

+ | |||

+ | ==Angle needed to minimize the minimum force needed to start sliding== | ||

+ | |||

+ | As noted in equation (2.4), the minimum force needed to ''start'' sliding is: | ||

+ | |||

+ | <math>F = \frac{\mu_smg}{\cos \theta + \mu_s \sin \theta} \qquad</math> | ||

+ | |||

+ | The numerator is constant, so to minimize this is equvalent to maximizing the denominator, for <math>0 \le \theta \le \pi/2</math>. In other words, we need to maximize: | ||

+ | |||

+ | <math>\cos \theta + \mu_s \sin \theta</math> | ||

+ | |||

+ | Using either differential calculus or basic trigonometry or the Cauchy-Schwarz inequality, we get that the maximum occurs when <math>\theta = \arctan \mu_s</math>, and the value of the maximum is <math>\sqrt{1 + \mu_s^2}</math>, so that the minimum possible value of the minimum force needed is: | ||

+ | |||

+ | <math>F = \frac{\mu_smg}{\sqrt{1 + \mu_s^2}}</math> |

## Latest revision as of 02:02, 14 August 2011

This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.

View other mechanics scenarios

## Contents

## The scenario

Suppose a block rests on a fixed horizontal floor. The limiting coefficient of static friction between and the floor is and the coefficient of kinetic friction is . A force is applied on the block in a diagonal direction at an angle with the horizontal. The questions are as follows:

- For a given angle , what is the minimum magnitude of force needed to get the block to start sliding, and how is the acceleration of the block given as a function of ?
- For what value of is the minimum magnitude of force needed to get the block to start sliding as low as possible, and what is this minimum magnitude of force?
- For what value of is the minimum magnitude of force needed to get the block to
*keep*sliding as low as possible, and what is this minimum magnitude of force?

## Basic components of force diagram

Force (letter) | Nature of force | Condition for existence | Magnitude | Direction |
---|---|---|---|---|

external force being applied | given to us that it's being applied | angle with horizontal. The horizontal component is and the vertical component is . | ||

gravitational force | unconditional | vertical, downward | ||

normal force | assuming that , otherwise the block will lift off the floor. | adjusts so that there is no net force in the vertical direction | vertical, upward | |

static friction | no sliding | adjusts so that there is no net force in the horizontal direction | horizontal, opposite to horizontal component of | |

kinetic friction | sliding | horizontal, opposite to direction of slipping. Assuming initially at rest, direction of slipping = horizontal component of , so this is opposite to horizontal component of . |

### Vertical components

Taking components in the vertical direction, and assuming no acceleration in the vertical direction (because the block does not lift off), we get:

### Horizontal components and solution assuming no sliding

If there is no sliding, we get:

We also have:

Plugging in from (1.1) into (2.2) and plugging from this into (2.1) gives:

This simplifies to:

This is a necessary and sufficient condition for there to be no sliding *assuming* the block was initially at rest.

### Horizontal components and solution assuming sliding

If there is sliding along the horizontal component of the direction of , denote by the acceleration along the horizontal component of the direction of . We get:

We also have:

Plugging in from (1.1) into (3.2) and plugging from this into (3.1) gives:

This simplifies to:

## Angle needed to minimize the minimum force needed to start sliding

As noted in equation (2.4), the minimum force needed to *start* sliding is:

The numerator is constant, so to minimize this is equvalent to maximizing the denominator, for . In other words, we need to maximize:

Using either differential calculus or basic trigonometry or the Cauchy-Schwarz inequality, we get that the maximum occurs when , and the value of the maximum is , so that the minimum possible value of the minimum force needed is: