Dragging problem

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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 \mu_s and the coefficient of kinetic friction is \mu_k. A force F is applied on the block in a diagonal direction at an angle \theta with the horizontal. The questions are as follows:

  • For a given angle \theta, 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 \theta 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 \theta 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 \theta with horizontal. The horizontal component is F \cos \theta and the vertical component is F \sin \theta.
mg gravitational force unconditional mg vertical, downward
N normal force assuming that F \sin \theta \le mg, 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 \mu_kN 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:

mg = N + F \sin \theta \qquad (1.1)

Horizontal components and solution assuming no sliding

If there is no sliding, we get:

F \cos \theta = f_s \qquad (2.1)

We also have:

f_s \le \mu_sN \qquad (2.2)

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

F \cos \theta \le \mu_smg - \mu_sF\sin \theta \qquad (2.3)

This simplifies to:

F \le \frac{\mu_smg}{\cos \theta + \mu_s \sin \theta} \qquad (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 \theta - f_k = ma \qquad (3.1)

We also have:

f_k = \mu_kN \qquad (3.2)

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

F \cos \theta - \mu_smg + \mu_kF\sin \theta = ma\qquad (2.3)

This simplifies to:

a = \frac{F(\cos \theta + \mu_k \sin \theta)}{m} - \mu_sg \qquad (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{\mu_smg}{\cos \theta + \mu_s \sin \theta} \qquad

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

\cos \theta + \mu_s \sin \theta

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

F = \frac{\mu_smg}{\sqrt{1 + \mu_s^2}}