Question

A block is dragged on smooth curved plane with the help of a rope which moves with a speed v as shown in figure. The speed of block at this instant will be

• A. $\frac{V}{\sin(\theta+\alpha)}$
• B. $\frac{V}{\cos(\theta+\alpha)}$
• C. $\frac{V}{\sin(\theta-\alpha)}$
• D. none of the above

Answer 1

Answer: (C)

$\frac{V}{\sin(\theta-\alpha)}$

Answer 2

To determine the speed of the block, we use the principle of constraint motion, specifically the inextensibility of the rope. The speed of the rope segment being pulled horizontally is given as v. Due to the inextensibility of the rope, the component of the block's velocity along the rope must be equal to this speed v.

Let V_block be the speed of the block. The block moves along the tangent to the curved plane at its instantaneous position.

Let's define angles:

• Let $\theta$ be the angle between the tangent to the curved plane (direction of $V_{block}$) and the vertical. The tangent is inclined downwards from the horizontal. So, the angle of the tangent with the horizontal is $(90^\circ - \theta)$.
• Let $\alpha$ be the angle between the rope and the horizontal. The rope is inclined upwards from the horizontal.

The tangent (direction of $V_{block}$) is inclined downwards from the horizontal by $(90^\circ - \theta)$. The rope is inclined upwards from the horizontal by $\alpha$. Since they are on opposite sides of the horizontal, the angle $\phi$ between the velocity vector $V_{block}$ and the rope is the sum of these two angles:

$\phi = (90^\circ - \theta) + \alpha$

Apply the constraint equation:

$v = V_{block} \cdot \cos(\phi)$

Substitute $\phi$:

$v = V_{block} \cdot \cos(90^\circ - \theta + \alpha)$

Using the trigonometric identity $\cos(90^\circ - x) = \sin(x)$:

$v = V_{block} \cdot \sin(\theta - \alpha)$

Solve for $V_{block}$:

$V_{block} = \frac{v}{\sin(\theta - \alpha)}$