In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force F$\vec{\mathrm{F}}$ required to slide the block C with constant speed is ____ N. (Used g = 10 m/s²$g = 10 \text{ m/s}^{2}$)
Three stacked blocks A, B, C with an external pulling force F applied to C.
Three stacked blocks A, B, C with an external pulling force F applied to C.
Since C moves with constant speed, a=0$a=0$, meaning net force is zero.
For block A (4 kg$4\text{ kg}$) over B, sliding friction fA = μ (mA)g = 0.5 × 4 × 10 = 20 N$f_A = \mu (m_A)g = 0.5 \times 4 \times 10 = 20\text{ N}$.
For block B (6 kg$6\text{ kg}$) over C, the normal force includes block A. NB = (mA + mB)g = 100 N$N_B = (m_A + m_B)g = 100\text{ N}$.
Friction on B from C is fB = 0.5 × 100 = 50 N$f_B = 0.5 \times 100 = 50\text{ N}$.
For block C (8 kg$8\text{ kg}$) over ground, normal force is total weight NC = (mA + mB + mC)g = 180 N$N_C = (m_A + m_B + m_C)g = 180\text{ N}$.
Friction on C from ground is fC = 0.5 × 180 = 90 N$f_C = 0.5 \times 180 = 90\text{ N}$.
Step 1: Force Balance Equations
For block B to move relative to A and C (assuming standard string configuration where pulling C left causes B to be dragged right relative to C via pulley):
The tension T$T$ on B balances friction from A and C:
For block C moving left, the force F$F$ must overcome friction from ground, friction from B, and the tension T$T$ attached to it (depending on pulley setup. If pulley is on wall, string pulls C right):
Assuming standard setup where string connects B and C around wall pulley:
In multi-block pulley systems, pulling the bottom block accumulates all nested friction forces. The bottom block fights the ground, the block above it, AND the tension of the string anchoring the block above it.
Keywords:#stacked blocks friction force#JEE Main 2026 Morning Q49#Laws of Motion JEE Main 2026#Friction JEE Main 2026#stacked blocks#pulley string#sliding friction#external force
More Laws of Motion Previous-Year Questions — Page 7
Q31jee_main_2024_31_jan_eveningPulley Systems and Tension
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
The image displays two masses suspended over a single fixed smooth pulley.
For standard Atwood machines, a = g × Difference in massSum of mass$a = g \times \frac{\text{Difference in mass}}{\text{Sum of mass}}$. If a/g = 1/8$a/g = 1/8$, then (m₁-m₂)/(m₁+m₂) = 1/8$(m_1-m_2)/(m_1+m_2) = 1/8$, which can be solved using componendo and dividendo: m₁/m₂ = (8+1)/(8-1) = 9/7$m_1/m_2 = (8+1)/(8-1) = 9/7$.
Chapter Mix
Class 11 Physics: Laws of Motion
Q42jee_main_2024_31_jan_eveningFriction on an Inclined Plane
A block of mass 5 kg$5 \text{ kg}$ is placed on a rough inclined surface as shown in the figure.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
If F₁$\vec{F}_1$ is the force required to just move the block up the inclined plane and F₂$\vec{F}_2$ is the force required to just prevent the block from sliding down, then the value of | F₁| - | F₂|$|\vec{F}_1| - |\vec{F}_2|$ is: [Use g = 10 m/s²$g = 10 \text{ m/s}^2$]
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.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
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
Qjee_main_2024_31_jan_morningPulley And Incline Friction
In the given arrangement of a doubly inclined plane two blocks of masses M$M$ and m$m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25$0.25$. The value of m$m$, for which M = 10 kg$M = 10\mathrm{\ kg}$ will move down with an acceleration of 2 m/s²$2\mathrm{\ m/s^2}$ is : (take g = 10 m/s²$g = 10\mathrm{\ m/s^2}$ and 37° = 3 / 4$\tan 37^{\circ} = 3 / 4$)
Two blocks M and m on opposite sides of a double inclined plane linked by a rope over a top pulley. M is on the 53-degree slope and moving downwards, m is on the 37-degree slope.
A.9 kg$9\mathrm{\ kg}$
B.4.5 kg$4.5\mathrm{\ kg}$
C.6.5 kg$6.5\mathrm{\ kg}$
D.2.25 kg$2.25\mathrm{\ kg}$
Solution
Related Formula
Σ F = ma$$\sum F = ma$$fk = μk N = μk mg θ$$f_k = \mu_k N = \mu_k mg \cos\theta$$
Core Logic
Two blocks M and m on opposite sides of a double inclined plane linked by a rope over a top pulley. M is on the 53-degree slope and moving downwards, m is on the 37-degree slope.
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).
A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ$\mu$. If the distance of the coin from the center of the disc is r$r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
