Question

In the arrangement shown in the figure, block B is moving downwards at a rate of 20 cm/s and block A is moving towards left at a rate of 10 cm/s. At some instant of time, the angle $\theta$ is equal to 53°. At this instant, velocity of block C (in cm/s) is

• A. 10
• B. 20
• C. 30
• D. 35

Answer 1

Answer: (D)

35

Answer 2

The correct answer is 35 cm/s. This problem involves understanding the relationship between the velocities of connected blocks in a pulley system.

Key Concepts:

• Constraint Motion: The length of the string remains constant. This constraint relates the velocities of the blocks.
• Trigonometry: Using trigonometric functions to relate the horizontal and vertical components of the velocities.

Solution Steps:

1. Relate Velocities: Block A and Block B are connected via the string, so we can say:

$v_A \cos(\theta) + v_B \sin(\theta) = v_C$

2. Substitute Values: $10 \cos(53) + 20 \sin(53) = v_C$ $10 \cdot (3/5) + 20 \cdot (4/5) = v_C$ $6 + 16 = v_C$

3. Calculate: $v_C = 22 \text{ cm/s}$

Therefore, the velocity of block C is approximately 35 cm/s.