Notes on Friction and Drag

Friction (Resistive Contact Force)

  • Definition: A resistive contact force that opposes motion, typically denoted by ext{f}.
  • Origin: Arises from intermolecular forces between surfaces and the fact that surfaces are not perfectly smooth.
  • Analogy: Rubbing hands together – more difficult when pressed firmly, due to greater resistance from increased contact (adhesion-like effect, not true adhesion).

Static Friction ( ext{f}_{ ext{s}})

  • Definition: The force that opposes the initiation of sliding or an object that is not moving with respect to another surface.
  • Quantification: Expressed as an inequality: \text{f}{ ext{s}} \le \mu{ ext{s}} \text{F}_{ ext{N}}
    • \mu_{ ext{s}}: Coefficient of static friction. A dimensionless measure of how strong static friction is, relative to the normal force pressing surfaces together.
    • \text{F}_{ ext{N}}: Normal force between the surfaces.
  • Inequality Explanation:
    • Minimum static friction: Can be zero. If no external force is applied, friction doesn't need to resist anything (e.g., a chair sitting still on the floor).
    • Scales with external force: Static friction will increase to match any external force attempting to move the object, up to a maximum value.
      • Example: Pushing a 10 Newton force on a stationary block: static friction exerts 10 Newtons back. Pushing with 100 Newtons: static friction exerts 100 Newtons back, as long as the block remains stationary.
  • Maximum Static Friction: The maximum force static friction can exert is \text{f}{ ext{s,max}} = \mu{ ext{s}} \text{F}_{ ext{N}}.
    • If the applied external force exceeds this maximum, the object will begin to slide or accelerate.
  • Graphical Representation: If external force (F) is plotted against friction force (f), f scales linearly with F until it reaches \mu{ ext{s}} \text{F}{ ext{N}}. At this point, the object breaks free, and the friction force drops and then levels off to a lower value (kinetic friction).

Kinetic Friction ( ext{f}_{ ext{k}})

  • Definition: The friction force experienced when two objects are sliding with respect to one another.
  • Quantification: A constant value, always equal to \text{f}{ ext{k}} = \mu{ ext{k}} \text{F}_{ ext{N}}
    • \mu{ ext{k}}: Coefficient of kinetic friction. A dimensionless measure, similar to \mu{ ext{s}}.
  • Relationship between Static and Kinetic Coefficients: Typically, \mu{ ext{k}} < \mu{ ext{s}}.
    • Significance: It is generally more difficult to get something moving than to keep something moving.
    • Example: Pushing a heavy box (e.g., moving friend's books) – initial push is hard, but once it starts sliding, it becomes slightly easier to maintain motion.
    • Rare Exception: Some materials (e.g., glass on cast iron, as per textbooks) can have nearly identical \mu{ ext{s}} and \mu{ ext{k}} values.

Friction and Motion (Conceptual Points)

  • Static Friction on Moving Objects: An object can be moving while still being acted upon by static friction, provided the contacting surfaces are not moving relative to each other.
    • Example 1 (Tires): Tires rolling on a road are acted upon by static friction because the point of contact between the tire and the road is momentarily stationary (no slipping occurs with respect to the road).
    • Example 2 (Box in Truck Bed): A box in the bed of an accelerating pickup truck is moving with the truck. Static friction between the truck bed and the box causes the box to accelerate with the truck, as there is no relative motion between these two surfaces.
  • Friction Does Not Always Oppose Overall Motion: As seen in the box-in-truck example, static friction can be the cause of an object's motion relative to an external reference frame, not just an opposing force. It always opposes relative motion between the contacting surfaces.

Rolling Friction

  • Description: A force that accounts for energy loss in rolling objects, causing them to stop. It's not primarily due to slipping (as in static friction for rolling without slipping), but rather internal mechanisms and deformation.
  • Origin:
    • Internal Mechanisms: Imperfections in components like ball bearings within a wheel system.
    • Deformation of Solids: Energy is consumed to deform elastic materials. For example, a heavy airplane deforms its rubber tires, leading to energy loss.
  • Practicality: While conceptually important for understanding why rolling objects stop, it's rarely assigned in introductory physics problems due to its complexity, but it is connected to static friction in terms of the contact surfaces.

Drag (Resistive Fluid Force)

  • Definition: The force exerted by a fluid (liquid or gas) on an object moving through that fluid.
  • Examples:
    • Reduced gas mileage at high speeds (forcing air molecules out of the way).
    • Turbulence behind moving vehicles (e.g., trucks using fins to reduce wake).
  • Proportionality: Drag force (\text{F}{ ext{D}}) is typically proportional to some factor of the object's speed: \text{F}{ ext{D}} \propto \text{v}^{ ext{n}}
    • Linear Drag (n=1): Often used for speeds below approximately 100 \text{ m/s} or for objects moving in viscous fluids. \text{F}_{ ext{D}} = \text{bv}.
    • Quadratic Drag (n=2): More common for higher speeds or objects moving through air. \text{F}_{ ext{D}} = \text{bv}^{2}.
    • In physics problems, the type of drag (linear or quadratic) will usually be specified.
  • Coefficient 'b': A complex coefficient that depends on:
    • The shape of the object.
    • The properties of the fluid (e.g., density, viscosity).
    • The speed of the object relative to the fluid.
  • Complexity: Drag is largely an empirically driven phenomenon. Predicting it accurately for various shapes and fluid interactions is a complex field (Computational Fluid Dynamics, CFD), and a perfect predictive model for all scales and conditions remains a Nobel Prize-worthy challenge.

Terminal Velocity

  • Definition: The constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration.
  • Condition: Occurs when the gravitational force (\text{F}{ ext{g}}) acting on the object equals the drag force (\text{F}{ ext{D}}) exerted by the fluid. At this point, the net force is zero, meaning acceleration is zero, and thus velocity becomes constant.
    • \text{F}{ ext{g}} = \text{F}{ ext{D}} \implies \text{mg} = \text{bv (or } \text{bv}^2 \text{)}
  • Conceptual Note: It's crucial to understand that terminal velocity means acceleration is zero, not velocity (speed) itself. If velocity were zero, drag would cease, and the object would start accelerating again.
  • Factors Affecting Terminal Velocity: The shape and orientation of the object significantly affect the drag coefficient 'b', and therefore alter the terminal velocity.
    • Example: A skydiver in a pencil dive (minimal surface area) reaches a higher terminal velocity compared to a skydiver splayed out on their belly (maximal surface area).
  • Falling in a Vacuum: In a vacuum, there is no fluid to exert drag. Therefore, an object falling in a vacuum would not reach a terminal velocity due to drag; it would continue to accelerate at gravitational acceleration (g) until it hits a surface.
  • Can Drag Exceed Other Forces?: Yes. If an object is moving significantly faster than its terminal velocity (e.g., a musket ball fired downwards at high velocity), the drag force can initially be greater than the gravitational force. This results in a net upward force, causing the object to slow down until it reaches its terminal velocity.

Practice Problems & Concepts Reviewed

1. Block Pulled at Constant Velocity

  • Problem: A block is pulled horizontally at a constant velocity by a force of 20 \text{ N}.
  • Solution:
    • Constant velocity means zero acceleration (\text{a}=0).
    • According to Newton's Second Law (\Sigma \text{F}_{ ext{x}} = \text{ma}), if \text{a}=0, then the net force is zero.
    • Therefore, the applied pulling force must be equal in magnitude and opposite in direction to the kinetic friction force.
    • \text{f}_{ ext{k}} = 20 \text{ N}.
    • Implicit Calculation: If \mu{ ext{k}} were known (e.g., 0.3), the mass of the block could be found: \text{f}{ ext{k}} = \mu{ ext{k}} \text{F}{ ext{N}} = \mu{ ext{k}} \text{mg}. So, \text{m} = \frac{\text{f}{ ext{k}}}{\mu_{ ext{k}} \text{g}} = \frac{20 \text{ N}}{0.3 \times 9.8 \text{ m/s}^{2}} \approx 6.8 \text{ kg}.

2. Block Pulled by a String at an Angle

  • Problem: A block of mass 'm' is pulled at constant velocity across a horizontal surface by a string with tension 'T' at an angle \theta to the horizontal.
  • Analysis:
    • Since velocity is constant, acceleration is \text{a}=0. Net forces in both x and y directions are zero.
    • Forces in y-direction: Normal force (\text{F}_{ ext{N}}) up, gravitational force (\text{mg}) down, and the vertical component of tension (\text{T} \sin\theta) up.
      • \Sigma \text{F}{ ext{y}} = \text{F}{ ext{N}} + \text{T} \sin\theta - \text{mg} = 0 \implies \text{F}_{ ext{N}} = \text{mg} - \text{T} \sin\theta
    • Friction force: \text{f}{ ext{k}} = \mu{ ext{k}} \text{F}{ ext{N}} = \mu{ ext{k}} (\text{mg} - \text{T} \sin\theta).
    • Forces in x-direction: Horizontal component of tension (\text{T} \cos\theta) in the direction of motion, friction force (\text{f}_{ ext{k}}) opposing motion.
      • \Sigma \text{F}{ ext{x}} = \text{T} \cos\theta - \text{f}{ ext{k}} = 0 \implies \text{f}_{ ext{k}} = \text{T} \cos\theta
  • Conclusion: Both \text{f}{ ext{k}} = \text{T} \cos\theta and \text{f}{ ext{k}} = \mu_{ ext{k}} (\text{mg} - \text{T} \sin\theta) are valid expressions for the friction force, depending on what values are known. In an actual exam, only one would typically be presented as a choice.

3. Atwood Machine with Inclined Plane and Friction

  • Problem: Mass m1 = 4 \text{ kg} on a ext{30}^ ext{o} inclined plane, connected via a cord over a pulley to a hanging mass m2 = 5 \text{ kg}. Coefficient of kinetic friction between m1 and the plane is \mu{ ext{k}} = 0.24. Find acceleration 'a' and tension 'T'.
  • Approach: Apply Newton's Second Law (\Sigma \text{F} = \text{ma}) to each mass separately, considering the forces acting on each, and assuming a direction for acceleration (e.g., m2 falls, m1 slides up the incline).
  • Forces on m_2 (hanging mass):
    • Downward: Gravitational force (\text{m}_2 \text{g})
    • Upward: Tension (\text{T})
    • Equation of motion: \text{m}2 \text{g} - \text{T} = \text{m}2 \text{a} (Assuming m_2 accelerates downward)
  • Forces on m_1 (on incline):
    • Parallel to incline:
      • Downward (due to gravity): \text{m}_1 \text{g} \sin(30^ ext{o})
      • Upward: Tension (\text{T})
      • Downward (due to friction): \text{f}{ ext{k}} = \mu{ ext{k}} \text{F}_{ ext{N}}
    • Perpendicular to incline:
      • Normal force (\text{F}{ ext{N}}): equals the perpendicular component of gravity, \text{m}1 \text{g} \cos(30^ ext{o})
      • Equation for friction: \text{f}{ ext{k}} = \mu{ ext{k}} \text{m}_1 \text{g} \cos(30^ ext{o})
    • Equation of motion parallel to incline: \text{T} - \text{m}1 \text{g} \sin(30^ ext{o}) - \text{f}{ ext{k}} = \text{m}1 \text{a} (Assuming m1 accelerates up the incline)
  • Solving the system:
    1. Substitute \text{f}{ ext{k}} into the second equation: \text{T} - \text{m}1 \text{g} \sin(30^ ext{o}) - \mu{ ext{k}} \text{m}1 \text{g} \cos(30^ ext{o}) = \text{m}_1 \text{a}
    2. Add the two equations:
      (m2 g - T) + (T - m1 g \sin(30^ ext{o}) - \muk m1 g \cos(30^ ext{o})) = (m2 a) + (m1 a)
      \text{m}2 \text{g} - \text{m}1 \text{g} \sin(30^ ext{o}) - \mu{ ext{k}} \text{m}1 \text{g} \cos(30^ ext{o}) = (\text{m}1 + \text{m}2) \text{a}
    3. Solve for acceleration 'a':
      \text{a} = \frac{\text{g} (\text{m}2 - \text{m}1 \sin(30^ ext{o}) - \mu{ ext{k}} \text{m}1 \cos(30^ ext{o}))}{(\text{m}1 + \text{m}2)}
      Substituting values: \text{a} = \frac{9.8 \text{ m/s}^{2} (5 \text{ kg} - 4 \text{ kg} \times 0.5 - 0.24 \times 4 \text{ kg} \times 0.866)}{(4 \text{ kg} + 5 \text{ kg})} \approx \text{2.36 m/s}^{2}
    4. Solve for tension 'T' using the equation for m2: \text{T} = \text{m}2 \text{g} - \text{m}2 \text{a} = \text{m}2 (\text{g} - \text{a})
      Substituting values: \text{T} = 5 \text{ kg} (9.8 \text{ m/s}^{2} - 2.36 \text{ m/s}^{2}) \approx \text{37.2 N}