To hold the block against the vertical wall, the frictional force (μF) must balance the weight (W). For minimum F, we have μF = W ⇒ F = W/μ. Since μ < 1, F must be greater than W.

The maximum value of static friction that comes into play when a body just starts moving over another surface is called limiting friction.

The minimum force required to move the block is given by \( F = \frac{W \sin \lambda}{\cos(\theta - \lambda)} \), where λ is the angle of friction and θ is the angle of applied force.

For equilibrium, the weight of the hanging part must be balanced by the frictional force on the part lying on the table. If l₁ is the hanging length, then \( l_1 \rho g = \mu (l - l_1) \rho g \), solving gives \( l_1 = \frac{\mu l}{1 + \mu} \).

Rolling friction is actually much smaller than sliding friction. This is why we use wheels and ball bearings to reduce friction in machines.

The frictional force μF must balance the weight of the block (mg = 1 × 9.8 = 9.8 N). So, 0.2 × F = 9.8 ⇒ F = 9.8/0.2 = 49 N.

Using the formula \( \frac{l_1}{l} = \frac{\mu}{1 + \mu} = \frac{0.25}{1.25} = 0.2 \) or 20%.

Work done by friction can be: - Negative when it opposes motion (most common case) - Positive when it helps motion (e.g., feet pushing backward to move forward) - Zero when there's no relative motion (static friction)

For equilibrium, weight of hanging part (l) must be balanced by friction on the part on table (L-l): \( l \rho g = \mu (L - l) \rho g \) ⇒ \( \mu = \frac{l}{L - l} \).

Lubricants reduce friction between surfaces, making them slide more easily upon each other.

Coefficient of static friction μₛ = Fₛ/N = Fₛ/mg = 75/(20×9.8) ≈ 0.38.

Ice has very low friction. Smaller steps reduce the horizontal component of force, making it less likely to exceed the small frictional force available.

At the angle of repose, μ = tanθ = tan60° = √3 ≈ 1.732.

Sand actually increases friction (used on icy roads). Oil, ball bearings, and graphite are all used to reduce friction.

Maximum frictional force = μN = 0.3 × 250 = 75 N (normal reaction equals weight for vertical equilibrium).

Frictional force balances the weight of the body: f = mg = 2 × 9.8 ≈ 19.6 N ≈ 20 N. The maximum possible friction (μN = 0.3×100 = 30 N) is not reached here.

At limiting equilibrium: weight of hanging part (L/3) = friction on part on table (2L/3). So \( \frac{L}{3} \rho g = \mu \frac{2L}{3} \rho g \) ⇒ μ = 1/2.

In a freely falling lift (a = g), the body experiences weightlessness (apparent weight = 0). Friction = μN = μ × 0 = 0.

The angle of friction (λ) is defined as the angle between the resultant reaction (R) and the normal reaction (N), where tanλ = μ (coefficient of friction).

Force to move up: \( F_1 = mg(\sinθ + μ\cosθ) \)
Force to prevent sliding down: \( F_2 = mg(\sinθ - μ\cosθ) \)
Given \( F_1 = 2F_2 \) ⇒ \( \sinθ + μ\cosθ = 2(\sinθ - μ\cosθ) \) ⇒ \( \sinθ = 3μ\cosθ \) ⇒ \( \tanθ = 3μ \).