1. Geometric Constraint

Since the blocks A, B, and C maintain equal horizontal separation and stay in a straight line, their vertical velocities must form an arithmetic progression. Specifically, the velocity of the middle block B must be the average of the velocities of A and C.

Let upward velocity be positive. The problem states B moves downwards relative to A, so we will establish a sign convention later.

$$ v_B = \frac{v_A + v_C}{2} \implies 2v_B = v_A + v_C $$

2. Relative Velocity Data

We are given that block B moves downwards with velocity $4$ cm/s relative to block A.

$$ v_{B/A} = v_B – v_A $$

Assuming “downwards” is the direction of increasing coordinate:

$$ v_B – v_A = 4 \, \text{cm/s} $$

Using the arithmetic progression property:

$$ v_C – v_B = v_B – v_A = 4 \, \text{cm/s} $$

This tells us the velocities are separated by intervals of 4 cm/s.

3. Solving for Velocities

We have a system where $v_A = v_B – 4$ and $v_C = v_B + 4$. Using virtual work (Tension method) $2v_A + v_B + v_C = 0$.

However, the relative velocities are fixed regardless of the frame:

• Velocity of A: $v_B – 4$
• Velocity of B: $v_B$
• Velocity of C: $v_B + 4$
• Velocity of A: $3$ cm/s Upwards ($-3$ cm/s)
• Velocity of B: $1$ cm/s Downwards ($+1$ cm/s)
• Velocity of C: $5$ cm/s Downwards ($+5$ cm/s)

Verification:

1. Relative velocity: $v_B – v_A = 1 – (-3) = 4$ cm/s. (Correct)

2. Collinearity: $v_A + v_C = -3 + 5 = 2$. And $2v_B = 2(1) = 2$. (Correct)