Question: System is shown in figure and wedge is moving towards left with speed 2 m/s. Then velocity of the bl...

Question

System is shown in figure and wedge is moving towards left with speed 2 m/s. Then velocity of the block B will be:

Answer 1

Solution

To determine the velocity of block B, we will use the principle of constant string length and vector addition of velocities.

1. Analyze the String Constraint:

Let $L$ be the total length of the string. The string consists of several segments:

Segment 1: From block B to pulley P1 (on the wedge A). Let its length be $l_1$ .
Segment 2: From pulley P1 to pulley P2 (fixed on the right wall). Let its length be $l_2$ .
Segment 3: From pulley P2 to pulley P3 (fixed on the right wall). Let its length be $l_3$ .
Segment 4: From pulley P3 to the left wall (fixed). Let its length be $l_4$ .

The total length of the string is constant: $L = l_1 + l_2 + l_3 + l_4$ . Since pulleys P2 and P3 are fixed, and the string ends are fixed to the walls, the lengths $l_3$ and $l_4$ are constant. Therefore, differentiating the total length with respect to time:

$\frac{dL}{dt} = \frac{dl_1}{dt} + \frac{dl_2}{dt} + \frac{dl_3}{dt} + \frac{dl_4}{dt} = 0$ .

Since $\frac{dl_3}{dt} = 0$ and $\frac{dl_4}{dt} = 0$ , we have:

$\frac{dl_1}{dt} + \frac{dl_2}{dt} = 0$ .

Let $x_{P1}$ be the horizontal position of pulley P1 and $x_{P2}$ be the horizontal position of pulley P2. The segment $l_2$ is horizontal, so $l_2 = x_{P2} - x_{P1}$ . Differentiating $l_2$ : $\frac{dl_2}{dt} = \frac{dx_{P2}}{dt} - \frac{dx_{P1}}{dt}$ . Since P2 is fixed, $\frac{dx_{P2}}{dt} = 0$ . So, $\frac{dl_2}{dt} = - \frac{dx_{P1}}{dt}$ .

Substituting this into the earlier equation:

$\frac{dl_1}{dt} - \frac{dx_{P1}}{dt} = 0$ .

$\frac{dl_1}{dt} = \frac{dx_{P1}}{dt}$ .

The wedge A is moving towards the left with a speed of 2 m/s. Pulley P1 is on the wedge A, so its horizontal velocity is the same as the wedge's velocity. Let's define the positive x-direction to the right. The velocity of the wedge A is $\vec{v}_A = -2 \hat{i}$ m/s. So, $\frac{dx_{P1}}{dt} = -2$ m/s. Therefore, $\frac{dl_1}{dt} = -2$ m/s.

This means the length of the string segment $l_1$ (from block B to pulley P1) is decreasing at a rate of 2 m/s. This implies that block B is moving towards pulley P1 along the incline with a speed of 2 m/s. In other words, the magnitude of the velocity of block B relative to wedge A, along the incline, is $v_{B/A} = 2$ m/s. Since $l_1$ is decreasing, block B is moving up the incline.

2. Vector Addition of Velocities:

The absolute velocity of block B, $\vec{v}_B$ , is the vector sum of its velocity relative to the wedge A, $\vec{v}_{B/A}$ , and the velocity of the wedge A, $\vec{v}_A$ .

$\vec{v}_B = \vec{v}_{B/A} + \vec{v}_A$ .

We have $\vec{v}_A = -2 \hat{i}$ m/s.

Let $\theta$ be the angle of inclination of the wedge with the horizontal. From the diagram, the incline appears to be at $60^\circ$ to the horizontal. This is a common angle in physics problems when not explicitly stated but implied by the visual representation (e.g., an equilateral triangle-like shape for the wedge's incline). Let's assume $\theta = 60^\circ$ .

Block B is moving up the incline relative to A with a speed of 2 m/s. The incline goes up and to the right. So, the x-component of $\vec{v}_{B/A}$ is $v_{B/A} \cos\theta = 2 \cos\theta$ . The y-component of $\vec{v}_{B/A}$ is $v_{B/A} \sin\theta = 2 \sin\theta$ . Thus, $\vec{v}_{B/A} = (2 \cos\theta) \hat{i} + (2 \sin\theta) \hat{j}$ .

Now, substitute the values into the vector sum:

$\vec{v}_B = (2 \cos\theta) \hat{i} + (2 \sin\theta) \hat{j} + (-2 \hat{i})$

$\vec{v}_B = (2 \cos\theta - 2) \hat{i} + (2 \sin\theta) \hat{j}$ .

Now, substitute $\theta = 60^\circ$ :

$\cos 60^\circ = \frac{1}{2}$

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\vec{v}_B = (2 \cdot \frac{1}{2} - 2) \hat{i} + (2 \cdot \frac{\sqrt{3}}{2}) \hat{j}$

$\vec{v}_B = (1 - 2) \hat{i} + (\sqrt{3}) \hat{j}$

$\vec{v}_B = -1 \hat{i} + \sqrt{3} \hat{j}$ .

The magnitude of the velocity of block B is:

$|\vec{v}_B| = \sqrt{(-1)^2 + (\sqrt{3})^2}$

$|\vec{v}_B| = \sqrt{1 + 3}$

$|\vec{v}_B| = \sqrt{4}$

$|\vec{v}_B| = 2$ m/s.

This matches option (C).

Explanation of the solution:

The problem involves a system with a wedge and a block connected by a string over pulleys. By applying the constraint of constant string length, it's found that the block B moves up the incline relative to the wedge A with a speed equal to the speed of the wedge (2 m/s). The absolute velocity of block B is then determined by vectorially adding its relative velocity to the velocity of the wedge. Assuming the incline angle is $60^\circ$ (a common implied angle in such diagrams), the components of relative velocity are calculated. Summing these with the wedge's velocity yields the total velocity of block B, whose magnitude is 2 m/s.