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    Friction

    Flashcards for topic Friction

    Intermediate44 cardsGeneral

    Preview Cards

    Card 1

    Front

    Explain the key distinction between static and kinetic friction at the fundamental level.

    Back

    • Static friction: Force that operates between bodies not slipping on each other

      • Self-adjustable in magnitude (adjusts to maintain relative rest)
      • Limited by maximum value: fs≤μsNf_s \leq \mu_s Nfs​≤μs​N
      • Direction: configured to prevent relative motion
    • Kinetic friction: Force between bodies slipping over each other

      • Constant magnitude: fk=μkNf_k = \mu_k Nfk​=μk​N (for given normal force)
      • Direction: always opposes relative motion
      • Generally μk<μs\mu_k < \mu_sμk​<μs​ for same surfaces
    • Transition occurs when applied force exceeds maximum static friction

    Card 2

    Front

    Compare the microscopic mechanisms of friction and static friction at the atomic level. Why do these forces exist?

    Back

    Microscopic mechanisms of friction:

    • Surface reality: Apparently smooth surfaces have microscopic irregularities
    • Contact points: True contact occurs only at scattered high points (much smaller than apparent area)
    • Molecular bonds: At contact points, intermolecular forces form temporary bonds between surfaces

    For static friction:

    • Molecular bonds must be overcome to initiate motion
    • These bonds resist deformation until a threshold force is reached
    • Explains why static friction has a maximum value

    For kinetic friction:

    • Continuous breaking and forming of new bonds as surfaces move
    • Energy dissipation through:
      • Breaking molecular bonds
      • Local deformation at contact points
      • Generation of vibration waves that convert to heat

    This atomic perspective explains why friction is independent of area but dependent on normal force, which determines the strength and number of microscopic bonds.

    Card 3

    Front

    A block on a horizontal surface is pulled by an angle force FFF at angle θ above horizontal. How does increasing θ affect the friction force? Analyze all effects.

    Back

    Increasing the angle θ creates multiple competing effects:

    Direct effects:

    1. Horizontal component decreases: Fh=Fcos⁡θF_h = F\cos\thetaFh​=Fcosθ decreases as θ increases

      • Reduces tendency to move horizontally
      • Decreases required friction to maintain equilibrium
    2. Vertical component increases: Fv=Fsin⁡θF_v = F\sin\thetaFv​=Fsinθ increases as θ increases

      • Reduces normal force: N=mg−Fsin⁡θN = mg - F\sin\thetaN=mg−Fsinθ
      • Decreases maximum possible friction: fmax=μsN=μs(mg−Fsin⁡θ)f_{max} = \mu_s N = \mu_s(mg - F\sin\theta)fmax​=μs​N=μs​(mg−Fsinθ)

    Combined effect on static case:

    • Two opposing influences on friction required for equilibrium
    • At small angles: friction decreases as θ increases (horizontal component effect dominates)
    • At large angles: maximum available friction decreases more rapidly (vertical component effect dominates)
    • There exists an optimal angle that minimizes the required force magnitude

    For kinetic friction:

    • As θ increases, normal force decreases
    • Results in lower kinetic friction: fk=μk(mg−Fsin⁡θ)f_k = \mu_k(mg - F\sin\theta)fk​=μk​(mg−Fsinθ)
    • Makes it easier to maintain motion once started
    Card 4

    Front

    For a system with multiple contact interfaces each with different friction coefficients, how do you determine which surface will experience slipping first?

    Back

    To determine which surface slips first in a multi-interface system:

    1. Calculate the friction threshold at each interface:

      • For each surface: fmax=μsNf_{max} = \mu_s Nfmax​=μs​N (where N is normal force at that interface)
    2. Calculate the actual friction required at each interface:

      • Apply Newton's laws to determine required friction force at each interface to maintain static equilibrium
      • This depends on the entire system's configuration and external forces
    3. Compare ratios of required friction to maximum friction:

      • For each interface, calculate: R=frequiredfmax=frequiredμsNR = \frac{f_{required}}{f_{max}} = \frac{f_{required}}{\mu_s N}R=fmax​frequired​​=μs​Nfrequired​​
      • Surface with highest R value will reach its limit first
      • Slipping occurs when any R ≥ 1
    4. After first slip:

      • Recalculate forces with kinetic friction at the slipping interface
      • Determine if this causes other interfaces to reach their limits

    This analysis applies to systems like stacked blocks, wheeled vehicles, or objects with multiple contact points where forces are distributed across different surfaces.

    Card 5

    Front

    What is the maximum value of static friction that can exist between two surfaces, and what happens when an applied force exceeds this value?

    Back

    Maximum static friction (limiting friction):

    • Value: fₘₐₓ = μₛN (coefficient of static friction × normal force)
    • This represents the maximum resistance before motion begins
    • When applied force exceeds this threshold:
      • Static friction fails to increase further
      • Surfaces begin to slip relative to each other
      • Static friction is replaced by kinetic friction (typically lower)
      • The object accelerates in the direction of the net force

    Example: A 10kg box with μₛ = 0.5 on a flat surface has maximum static friction of 49N. Applying 50N will cause the box to start moving, with friction dropping to the kinetic value.

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    Card 6

    Front

    How can you experimentally verify that friction is independent of contact area using the horizontal table apparatus?

    Back

    To verify friction's independence from contact area:

    1. Measure the coefficient of friction with the block resting on its largest face
    2. Repeat the measurement with the block positioned on smaller faces
    3. Keep the normal force constant by:
      • Either using the same block weight in all trials
      • Or adding weights to maintain identical normal force when necessary
    4. Compare the resulting friction coefficients (μ = W₂/W₁)

    If friction is truly independent of area, the measured coefficients will be the same regardless of which face contacts the surface, provided the normal force remains constant and the surface materials are identical.

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    Card 7

    Front

    Why does a person pushing a box with wheels require less force than pushing the same box directly on the floor?

    Back

    • Wheels convert sliding friction into rolling friction, which has a much lower coefficient of friction
    • In rolling motion, surfaces don't rub against each other, eliminating the shearing forces present in sliding
    • The molecular bonds between surfaces don't need to be continuously broken and reformed during rolling
    • While sliding friction dissipates energy through heat from surface deformation, rolling dissipates much less energy
    • The normal force remains the same in both cases, but the friction coefficient is substantially lower for rolling
    • Example: Moving a 100 kg crate might require 500 N of force when sliding but only 50 N when rolling
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    Card 8

    Front

    In a block-pulley system where a 2 kg block on a horizontal surface is connected via string to a hanging mass, what is the maximum value of the hanging mass that will keep the system in static equilibrium if the coefficient of static friction is 0.2?

    Back

    To find the maximum hanging mass:

    1. Analyze forces on the 2 kg block:

      • Weight = 20 N downward
      • Normal force N = 20 N upward
      • Tension T pulling right
      • Static friction f pulling left (opposing motion)
    2. Maximum static friction possible: f_max = μ_s × N = 0.2 × 20 N = 4 N

    3. For equilibrium, tension must equal maximum friction: T = f_max = 4 N

    4. The hanging mass experiences:

      • Weight = mg downward
      • Tension T upward
    5. At equilibrium: T = mg Therefore: 4 N = m × 9.8 m/s² m = 4 N ÷ 9.8 m/s² ≈ 0.41 kg

    The maximum hanging mass is approximately 0.41 kg before the system begins to move.

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    Card 9

    Front

    How does the direction of static friction change in a block-pulley system compared to most typical friction scenarios, and why is this counterintuitive?

    Back

    In a block-pulley system:

    1. Static friction acts in the direction that maintains equilibrium (prevents motion), which can be counterintuitive because:

      • In typical scenarios, we think of friction as opposing applied forces
      • Here, friction acts to prevent the block from being pulled by the string tension
    2. Direction determination:

      • Friction acts opposite to the direction the object would move if friction were absent
      • With the pulley system, the hanging mass creates tension that would pull the block right
      • Therefore, static friction acts left to prevent this motion
    3. This illustrates a key principle: static friction is not always opposed to the applied force, but rather to the impending motion.

    4. The magnitude adjusts as needed (up to μ_s×N) to maintain equilibrium, making static friction a self-adjusting force rather than a constant one.

    This understanding helps solve pulley problems by focusing on the direction of potential motion rather than applied forces.

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    Card 10

    Front

    Derive the relationship between the coefficient of static friction (μs\mu_sμs​) and the critical angle of incline (θ\thetaθ) for a body at limiting equilibrium, and explain its physical significance.

    Back

    Force Analysis at Limiting Equilibrium:

    For an object on an inclined plane at the verge of sliding:

    1. Forces acting on the object:

      • Weight (mgmgmg) acting downward
      • Normal force (N=mgcos⁡θN = mg\cos\thetaN=mgcosθ) perpendicular to incline
      • Static friction force (fs=μsNf_s = \mu_s Nfs​=μs​N) opposing motion up the incline
      • Weight component parallel to incline (mgsin⁡θmg\sin\thetamgsinθ) pulling object down
    2. Mathematical derivation:

      • At limiting equilibrium: fs=mgsin⁡θf_s = mg\sin\thetafs​=mgsinθ
      • Substituting fs=μsN=μsmgcos⁡θf_s = \mu_s N = \mu_s mg\cos\thetafs​=μs​N=μs​mgcosθ:
      • μsmgcos⁡θ=mgsin⁡θ\mu_s mg\cos\theta = mg\sin\thetaμs​mgcosθ=mgsinθ
      • Simplifying: μs=sin⁡θcos⁡θ=tan⁡θ\mu_s = \frac{\sin\theta}{\cos\theta} = \tan\thetaμs​=cosθsinθ​=tanθ
    3. Key relationship: μs=tan⁡θ\boxed{\mu_s = \tan\theta}μs​=tanθ​ or θ=tan⁡−1(μs)\boxed{\theta = \tan^{-1}(\mu_s)}θ=tan−1(μs​)​

    Physical significance:

    • This relationship is independent of the object's mass and surface area
    • The angle θ\thetaθ where sliding begins is called the "angle of repose"
    • Practical applications include:
      • Determining safe slopes for roads in different weather conditions
      • Designing ramps and inclined conveyor systems
      • Civil engineering for hillside construction

    Example: If μs=0.577\mu_s = 0.577μs​=0.577, the object will begin to slip when θ=tan⁡−1(0.577)=30°\theta = \tan^{-1}(0.577) = 30°θ=tan−1(0.577)=30°

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