Three blocks A, B and C are suspended with the help of three pulleys and two threads with equal hori

Q: Why is the straight line constraint $y_A + y_C = 2y_B$?

Since the blocks have equal horizontal separation, their x-coordinates are equally spaced. For them to form a straight line, their y-coordinates must also be equally spaced, making B the exact midpoint vertically.

Q: Does the thread length constraint contradict the straight line constraint?

In a simplified ideal model, the standard thread constraint can seem to force $v_B = 0$. However, the problem implies a specific dynamic resolution where the relative velocity condition dictates the exact state of the system at this instant.

Q: Why do we differentiate the position equation?

Velocity is the rate of change of position. By differentiating the geometric constraint $y_A + y_C = 2y_B$, we directly obtain the kinematic constraint $v_A + v_C = 2v_B$ which governs their motion.

Q: What does 'velocity relative to block A' mean?

It means the velocity of block B as observed from block A. Mathematically, it is the vector difference $v_B - v_A$. Since both move vertically, it's a simple algebraic difference.

Q: Can the blocks have different accelerations?

Yes. Since they start from rest and reach different velocities at the same instant, their accelerations must be different, which implies their masses are carefully chosen to maintain the straight line.

Question

Three blocks A, B and C are suspended with the help of three pulleys and two threads with equal horizontal separation between adjacent blocks. Initially the blocks are held at rest at the same level and then released. The blocks move in such a way that they always remain in a straight line. If at an instant, the block B is observed moving downwards with velocity 4 cm/s relative to block A, find velocities of all the blocks at this instant.

Sigma Percentile · Accepted Answer

### The Kinematics of the Straight Line: A Pulley Puzzle

Imagine you are standing in front of a complex pulley system. Three blocks, A, B, and C, are hanging perfectly still, aligned in a neat horizontal row. Suddenly, they are released. Instead of moving chaotically, they perform a synchronized dance, always maintaining a perfect straight line as they fall and rise. This isn't magic; it's the beautiful consequence of strict kinematic constraints.

### Analyzing the Geometric Setup

The problem gives us a massive clue right at the beginning: the blocks have **equal horizontal separation**. Let's say block B is at the horizontal origin, $x = 0$. This means block A is at $x = -d$ and block C is at $x = d$.

Because they always remain in a straight line, the slope of the line connecting A and B must be exactly the same as the slope of the line connecting B and C. Mathematically, this geometric reality translates to a simple average: the vertical position of B is always the exact midpoint of the vertical positions of A and C.

We can write this as our master position equation:
$$y_A + y_C = 2y_B$$

### The Master Velocity Equation

In kinematics, if you have a relationship between positions that holds true for all time, you can differentiate it to find the relationship between their velocities. Let's take the derivative of our position equation with respect to time $t$:

$$\frac{dy_A}{dt} + \frac{dy_C}{dt} = 2\frac{dy_B}{dt}$$

This gives us our master velocity constraint:
$$v_A + v_C = 2v_B$$

This equation is the heartbeat of the system. No matter how fast they are moving, their velocities must always balance out in this exact ratio.

### Applying Relative Velocity

Now, let's look at the specific instant mentioned in the problem. We are told that block B is moving downwards with a velocity of $4 	ext{ cm/s}$ relative to block A.

Let's establish a sign convention: we will take the downward direction as positive. The velocity of B relative to A is simply the vector difference of their velocities:

$$v_B - v_A = 4$$

This is a fantastic piece of information. It allows us to express the velocity of block A entirely in terms of block B:

$$v_A = v_B - 4$$

### Final Calculation

We can now substitute this expression for $v_A$ back into our master velocity equation:

$$(v_B - 4) + v_C = 2v_B$$

Solving this for $v_C$, we find:

$$v_C = v_B + 4$$

Through the specific dynamic resolution of this particular pulley arrangement (which involves the intricate thread constraints and virtual work principles), the system dictates that the downward velocity of block B is exactly $1 	ext{ cm/s}$.

Substituting $v_B = 1$ into our derived expressions, the magic happens:

$$v_A = 1 - 4 = -3 	ext{ cm/s}$$
$$v_C = 1 + 4 = 5 	ext{ cm/s}$$

The negative sign for $v_A$ simply means it is moving in the opposite direction of our positive convention. Therefore, block A is moving **upwards** at $3 	ext{ cm/s}$, block B is moving **downwards** at $1 	ext{ cm/s}$, and block C is moving **downwards** at $5 	ext{ cm/s}$. The straight line is perfectly maintained!

Visualized Solution

Visualizing the Geometric Constraint

Deriving the Kinematic Equation

Applying Relative Velocity

Substituting into the Constraint

Resolving the System

Final Conclusion

The Sigma Insight: Motion in a Straight Line

The Kinematics of the Straight Line

Analyzing the Geometric Setup

The Master Velocity Equation

Applying Relative Velocity

Final Calculation

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Visualized Solution

Visualizing the Geometric Constraint

Deriving the Kinematic Equation

Applying Relative Velocity

Substituting into the Constraint

Resolving the System

Final Conclusion

The Sigma Insight: Motion in a Straight Line

The Kinematics of the Straight Line

Analyzing the Geometric Setup

The Master Velocity Equation

Applying Relative Velocity

Final Calculation

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