Friction

• 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: $f_s \leq \mu_s N$
  • Direction: configured to prevent relative motion

• Kinetic friction: Force between bodies slipping over each other

  • Constant magnitude: $f_k = \mu_k N$ (for given normal force)
  • Direction: always opposes relative motion
  • Generally $\mu_k < \mu_s$ for same surfaces

• Transition occurs when applied force exceeds maximum static friction

The kinetic friction on the block acts forward (in the direction of the cart's motion).

Key principle: Kinetic friction on a body A slipping against body B is opposite to the velocity of A with respect to B.

Since the block is sliding backward relative to the cart:

• The block's velocity relative to the cart is backward
• Therefore, friction from the cart acts on the block in the forward direction
• This counterintuitive case demonstrates that friction doesn't always oppose absolute motion, but rather opposes relative motion between contacting surfaces

Three key forces act on the rider:

1. Weight (Mg): Gravitational force acting downward
2. Normal force (N): Contact force from the horse acting upward, perpendicular to the contact surface
3. Static friction (fs): Horizontal force acting in the direction of acceleration

The static friction between the rider and horse prevents the rider from sliding backward relative to the horse during acceleration. This static friction adjusts automatically (up to a maximum value) to maintain the relative position between rider and horse.

Formula: The maximum static friction possible is fs ≤ μs·N, where μs is the coefficient of static friction and N is the normal force.

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.

The horizontal table method for measuring the coefficient of static friction involves:

1. A block placed on a horizontal plank with a frictionless pulley at one end
2. A string connecting the block to a hanger that holds weights
3. Gradually increasing weights until the block just begins to slide
4. Calculating using the formula: μs = W₂/W₁, where:
   • W₁ = weight of the block
   • W₂ = weight of the hanger plus weights when sliding begins
   • μs = coefficient of static friction

The maximum static friction equals the tension in the string (W₂) at the point of impending motion.

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.

The inclined plane method demonstrates area independence of friction through:

1. Testing the same block on different faces with varying surface areas
2. Observing that the critical angle θ remains constant regardless of which face contacts the plane

This occurs because:

• The maximum static friction force is proportional to the normal force (fs = μs·N)
• The normal force equals mg·cos θ, which depends only on mass and angle
• As contact area changes, the pressure redistributes, but the total normal force remains constant

This validates Amontons' Law that friction depends on normal force but not apparent contact area, contradicting the intuitive notion that more contact area would create more friction.

Ball bearings reduce friction by substituting rolling motion for sliding motion:

• Steel balls roll between fixed and rotating parts, minimizing surface contact
• Rolling friction is significantly lower than sliding friction (often 10-20 times less)
• The balls maintain a small point of contact rather than broad surface contact
• As parts move relative to each other, the balls simply roll rather than slide
• This mechanism transforms potential energy loss into rotational motion of the balls

Example: In industrial machinery, ball bearings often reduce energy consumption by 15-25% compared to simple bushings or sliding surfaces.

When a block is at rest on an inclined plane in limiting equilibrium:

• The static friction force equals: fs = μs·N = μs·mg·cos(θ)
• This friction force exactly balances the parallel component of weight: mg·sin(θ)
• Setting these equal: μs·mg·cos(θ) = mg·sin(θ)
• Simplifying: μs = tan(θ)

Therefore, the maximum angle before sliding is θmax = tan⁻¹(μs)

This is a fundamental relationship in mechanics: an object at rest on an incline will begin to slide when the angle of inclination exceeds the angle whose tangent equals the coefficient of static friction.

To determine if a block will remain stationary:

1. Compare the angle of inclination (θ) with the critical angle (θcrit = tan⁻¹(μs)):

   • If θ < θcrit (or equivalently, if μs > tan(θ)): Block remains stationary
   • If θ > θcrit (or equivalently, if μs < tan(θ)): Block will slide

2. When the block remains stationary, the friction force adjusts to balance the parallel component of weight:

   • fs = mg·sin(θ)
   • This is less than the maximum possible static friction (fs,max = μs·mg·cos(θ))
   • The friction force is self-adjusting up to this maximum value

3. The static friction force is always directed up the incline, opposite to the component of weight trying to pull the block down.

Example: For a block with mass 2 kg on a 15° incline with μs = 0.5, we have tan(15°) ≈ 0.27 < 0.5, so the block remains stationary with a friction force of fs = (2 kg)(9.8 m/s²)sin(15°) ≈ 5.1 N.