• Earlier theories proposed that pressure from a skate or foot lowers ice's melting point, creating a water film, as suggested by James Thomson in the 1800s.
• Frictional heating from motion was another idea, where rubbing generates enough warmth to melt the surface slightly. Experiments, like those by Lord Kelvin, supported friction's role by showing higher friction with heat-conducting materials. We also have the theory of pre-existing quasi-liquid layer (QLL) on ice surfaces, formed by weaker bonding of surface molecules exposed to fewer neighbors

Given:

• Mass of box on incline: m
• Incline angle: \theta
• Hanging mass: M
• Pulley: frictionless bearing but with moment of inertia I and radius R (to be included for rotational effects)
• Rope: light and inextensible
• The system is released from rest at t=0
• Incline is frictionless

• Let acceleration of box along the incline be $a$ downwards (positive direction along slope down).
• The hanging mass moves vertically downward with acceleration a.
• Let tensions on the box side of the rope and hanging mass side be T_1 and T_2 respectively.
• Because the pulley has rotational inertia, T_1 \neq T_2.

m a = m g \sin \theta - T_1

M a = M g - T_2

\alpha = \frac{a}{R}

\tau = (T_2 - T_1) R = I \alpha = I \frac{a}{R}

T_2 - T_1 = \frac{I}{R^2} a

1. m a = m g \sin \theta - T_1
2. M a = M g - T_2
3. T_2 - T_1 = \frac{I}{R^2} a

T_1 = m g \sin \theta - m a

T_2 = M g - M a

(M g - M a) - (m g \sin \theta - m a) = \frac{I}{R^2} a

M g - M a - m g \sin \theta + m a = \frac{I}{R^2} a

M g - m g \sin \theta = M a - m a + \frac{I}{R^2} a

M g - m g \sin \theta = a \left( M - m + \frac{I}{R^2} \right)

a = \frac{M g - m g \sin \theta}{M - m + \frac{I}{R^2}}

• If pulley moment of inertia I = 0, this reduces to a simpler formula:

a = \frac{M g - m g \sin \theta}{M - m}

• If the pulley has significant I, the denominator increases, reducing acceleration. :)