A light string passing over a smooth light fixed pulley connects two blocks of masses m₁$m_1$ and m₂$m_2$. If the acceleration of the system is g/8$g/8$, then the ratio of masses is

Related Formula

Core Logic

To move the block up, the applied force F₁$F_1$ must overcome both the downward gravitational component and the downward frictional force. To prevent it from sliding down, the applied force F₂$F_2$ acts upwards and is aided by friction which acts upwards to oppose impending downward slip.

Step 1: Calculate Friction

Step 2: Force Equations

Moving up:

Preventing slip down:

Step 3: Difference calculation

Note: The official options had an anomaly where 5√(3) N$5\sqrt{3} \text{ N}$ was missing or evaluated as a bonus. Option 2 was listed as 50√(3)$50\sqrt{3}$ in the primary text. We track the closest logic path indicating Bonus.

Pattern Recognition

The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely 2 fk$2 f_k$ (2 μ mg θ$2 \mu mg \cos \theta$). Bypass calculating the mg θ$mg \sin \theta$ terms entirely.

Chapter Mix

Class 11 Physics: Laws of Motion

Related Formula

Core Logic

Since block M$M$ moves down the incline, kinetic friction opposes its motion (acts upwards). Block m$m$ is pulled up the incline, so kinetic friction opposes its motion (acts downwards).

For M$M$ block (53^°$53^\circ$ slope):

Step 1: Tension Calculation

Given M = 10 kg$M = 10\mathrm{\,kg}$, a = 2 m/s²$a = 2\mathrm{\,m/s^2}$, μ = 0.25$\mu = 0.25$, g = 10 m/s²$g = 10\mathrm{\,m/s^2}$. 53^° = 4/5 = 0.8$\sin 53^\circ = 4/5 = 0.8$, 53^° = 3/5 = 0.6$\cos 53^\circ = 3/5 = 0.6$.

Step 2: Evaluate mass m

For m$m$ block (37^°$37^\circ$ slope) moving upward:

37^° = 3/5 = 0.6$\sin 37^\circ = 3/5 = 0.6$, 37^° = 4/5 = 0.8$\cos 37^\circ = 4/5 = 0.8$.

45 - 8m = 2m$45 - 8m = 2m$

Chapter Mix

Class 11 Physics: Laws Of Motion

Related Formula

Core Logic

To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion. f = mω² r$f = m\omega^2 r$

The normal force on the flat disc is N = mg$N = mg$. The maximum static friction is fmax = μ N = μ mg$f_{\text{max}} = \mu N = \mu mg$.

For no slipping:

Chapter Mix

Class 11 Physics: Laws Of Motion