In the figure, the blocks A, B and C have masses 3 kg, 4 kg and 8 kg respectively. The coefficient o

Q: Wait, why is friction $f_2$ acting to the right on Block C?

Friction always opposes *relative* motion. Block C is moving to the left, while Block B is moving to the right. From the perspective of Block C, Block B is scraping across it towards the right. Therefore, Block B drags Block C towards the right via friction. Conversely, from Block B's perspective, Block C is sliding left, so Block C exerts a leftward friction on Block B.

Q: Why do we add the masses to find the normal forces at the lower interfaces?

The normal force is the contact force that prevents objects from passing through each other. At the interface between B and C, the surface of C must support the weight of *both* Block B and Block A resting on top of it. Therefore, $N_2 = (m_A + m_B)g$.

Q: Why is Block A's friction on B acting to the left?

Block A is fixed to the wall and is completely stationary. Block B is being pulled to the right by the string. Since Block B is sliding to the right relative to the stationary Block A, the friction from A will try to oppose this motion, pulling Block B to the left.

Q: Does the tension $T$ act on both blocks B and C?

Yes! The string connects Block B and Block C over a pulley. An ideal massless string transmits the exact same tension $T$ throughout its length. It pulls Block B to the right and pulls Block C to the right (since the string wraps around the pulley and attaches to the right side of C).

Q: Why do we set the net force to zero? Shouldn't a force cause acceleration?

The problem explicitly states that we need to find the force $F$ necessary to drag C at a **constant speed**. According to Newton's First Law, an object moving at a constant velocity has zero acceleration, which means the vector sum of all forces acting on it must be exactly zero.

Question

In the figure, the blocks $A, B$ and $C$ have masses $3	ext{ kg}$, $4	ext{ kg}$ and $8	ext{ kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.25$. $A$ is held at rest by a massless rigid rod fixed to the wall, while $B$ and $C$ are connected by a light flexible cord passing around a fixed frictionless pulley. Find the force $F$ necessary to drag $C$ along the horizontal surface to the left at a constant speed. Assume that the arrangement shown in the figure. i.e. $B$ on $C$ and $A$ on $B$, is maintained throughout.(Take $g = 10	ext{ m/s}^2$).

Sigma Percentile · Accepted Answer

### The Stacked Block Puzzle

Imagine a three-tier cake of physics blocks, stacked neatly on top of each other. We have Block A ($3$ kg) at the top, Block B ($4$ kg) in the middle, and Block C ($8$ kg) at the bottom. Block A is stubbornly anchored to a wall via a rigid rod, refusing to move. Blocks B and C, however, are connected by a light string that loops around a frictionless pulley. When a mysterious force $F$ drags the bottommost Block C to the left at a constant speed, a beautiful dance of relative motion and friction begins. Our mission? To find the exact magnitude of this force $F$.

### Kinematics: Who is moving where?

Before we can even think about forces, we must understand the kinematics—the geometry of motion. When Block C is pulled to the left, its velocity vector points left. Because Block B is tethered to Block C via the pulley, the leftward motion of C pulls the string, which in turn yanks Block B to the right.

Meanwhile, Block A sits at the very top, completely immobilized by the wall. Its velocity is strictly zero. This creates a fascinating scenario where every single interface between the blocks is experiencing relative sliding, meaning kinetic friction is fully engaged everywhere.

### The Weight of the World: Normal Forces

Friction is born from two surfaces pressing against each other. To find the friction at each interface, we first need the normal forces. Think of the normal force as the burden of weight that a specific surface must carry.

At the top interface (between A and B), the surface of B only has to support Block A. Thus, the normal force is:
$$N_1 = m_A g = 3 	imes 10 = 30 	ext{ N}$$

Moving down to the middle interface (between B and C), the surface of C must support the combined weight of both Block A and Block B. Therefore:
$$N_2 = (m_A + m_B) g = (3 + 4) 	imes 10 = 70 	ext{ N}$$

Finally, at the bottom interface (between C and the ground), the floor must bear the weight of the entire three-block tower:
$$N_3 = (m_A + m_B + m_C) g = (3 + 4 + 8) 	imes 10 = 150 	ext{ N}$$

### The Friction Arsenal

Now that we have our normal forces, we can unleash the friction formula, $f = \mu N$, where the coefficient of sliding friction $\mu$ is given as $0.25$ for all surfaces. The tricky part is assigning the correct direction to each friction force. Remember, friction always opposes *relative* motion.

**Friction $f_1$ (Between A and B):**
Block B is sliding to the right relative to the stationary Block A. Therefore, Block A exerts a friction force on Block B towards the left to slow it down.
$$f_1 = \mu N_1 = 0.25 	imes 30 = 7.5 	ext{ N}$$

**Friction $f_2$ (Between B and C):**
This is the danger zone! Block B is moving right, and Block C is moving left. From Block C's perspective, Block B is scraping across it to the right, so C pulls B to the left. Conversely, from Block B's perspective, Block C is sliding away to the left, so B drags C to the right. 
$$f_2 = \mu N_2 = 0.25 	imes 70 = 17.5 	ext{ N}$$

**Friction $f_3$ (Between C and Ground):**
Block C is sliding left across the stationary floor. The floor fights back by exerting a friction force to the right.
$$f_3 = \mu N_3 = 0.25 	imes 150 = 37.5 	ext{ N}$$

### Balancing Act: Block B

The problem states that the blocks are dragged at a **constant speed**. According to Newton's First Law, a constant velocity implies zero acceleration, which means the net force acting on any block must be exactly zero. Let's isolate Block B and look at the horizontal forces.

Block B is being pulled to the right by the tension $T$ in the string. Opposing this motion are two friction forces pulling to the left: $f_1$ (from Block A) and $f_2$ (from Block C). For equilibrium, the rightward forces must perfectly balance the leftward forces:
$$T = f_1 + f_2$$

Substituting our calculated friction values:
$$T = 7.5 + 17.5 = 25 	ext{ N}$$

### The Final Showdown: Block C

Finally, we turn our attention to the star of the show, Block C. It is being pulled to the left by our unknown force $F$. What forces are trying to hold it back?

First, the string pulls it to the right with tension $T$. Second, the friction $f_2$ from Block B drags it to the right. Third, the friction $f_3$ from the ground also drags it to the right. Since Block C is also moving at a constant speed, the leftward force must equal the sum of all rightward forces:
$$F = T + f_2 + f_3$$

We have all the pieces of the puzzle. Let's plug them in:
$$F = 25 + 17.5 + 37.5$$

$$F = 80 	ext{ N}$$

And there we have it! A force of exactly **$80$ N** is required to keep this intricate mechanical system sliding smoothly at a constant speed.

Visualized Solution

System Overview

Kinematic Constraints

Normal Forces at Interfaces

Friction $f_{1}$ on Block B

Friction $f_{2}$ on Blocks B and C

Friction $f_{3}$ on Block C

Equilibrium of Block B

Calculating Tension $T$

Equilibrium of Block C

Final Calculation of Force $F$

The Sigma Insight: Static and Kinetic Friction

The Stacked Block Puzzle

Kinematics

The Weight of the World

The Friction Arsenal

Balancing Act

The Final Showdown

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