Dragging problem: Difference between revisions

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.
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The scenario

Suppose a block $m$ rests on a fixed horizontal floor. The limiting coefficient of static friction between $m$ and the floor is $μ_{s}$ and the coefficient of kinetic friction is $μ_{k}$ . A force $F$ is applied on the block in a diagonal direction at an angle $θ$ with the horizontal. The questions are as follows:

For a given angle $θ$ , 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 $F$ ?
For what value of $θ$ 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 $θ$ 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
$F$	external force being applied	given to us that it's being applied	$F$	angle $θ$ with horizontal. The horizontal component is $F \cos θ$ and the vertical component is $F \sin θ$ .
$m g$	gravitational force	unconditional	$m g$	vertical, downward
$N$	normal force	assuming that $F \sin θ \leq m g$ , otherwise the block will lift off the floor.	adjusts so that there is no net force in the vertical direction	vertical, upward
$f_{s}$	static friction	no sliding	adjusts so that there is no net force in the horizontal direction	horizontal, opposite to horizontal component of $F$
$f_{k}$	kinetic friction	sliding	$μ_{k} N$	horizontal, opposite to direction of slipping. Assuming initially at rest, direction of slipping = horizontal component of $F$ , so this is opposite to horizontal component of $F$ .

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:

$m g = N + F \sin θ (1.1)$

Horizontal components and solution assuming no sliding

If there is no sliding, we get:

$F \cos θ = f_{s} (2.1)$

We also have:

$f_{s} \leq μ_{s} N (2.2)$

Plugging in $N$ from (1.1) into (2.2) and plugging $f_{s}$ from this into (2.1) gives:

$F \cos θ \leq μ_{s} m g - μ_{s} F \sin θ (2.3)$

This simplifies to:

$F \leq \frac{μ_{s} m g}{\cos θ + μ_{s} \sin θ} (2.4)$

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 $F$ , denote by $a$ the acceleration along the horizontal component of the direction of $F$ . We get:

$F \cos θ - f_{k} = m a (3.1)$

We also have:

$f_{k} = μ_{k} N (3.2)$

Plugging in $N$ from (1.1) into (3.2) and plugging $f_{k}$ from this into (3.1) gives:

$F \cos θ - μ_{s} m g + μ_{k} F \sin θ = m a (2.3)$

This simplifies to:

$a = \frac{F (\cos θ + μ_{k} \sin θ)}{m} - μ_{s} g (2.4)$

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:

$F = \frac{μ_{s} m g}{\cos θ + μ_{s} \sin θ}$

The numerator is constant, so to minimize this is equvalent to maximizing the denominator, for $0 \leq θ \leq π / 2$ . In other words, we need to maximize:

$\cos θ + μ_{s} \sin θ$

Using either differential calculus or basic trigonometry or the Cauchy-Schwarz inequality, we get that the maximum occurs when $θ = \arctan μ_{s}$ , and the value of the maximum is $\sqrt{1 + μ_{s}^{2}}$ , so that the minimum possible value of the minimum force needed is:

$F = \frac{μ_{s} m g}{\sqrt{1 + μ_{s}^{2}}}$

@@ Line 18: / Line 18: @@
 | <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>