Question

As shown in the following figures, block of mass 2 kg is in equilibrium in the vertical plane.

At a certain instant right spring in case-1 and right string in case-2 are cut. Then the ratio of instantaneous acceleration of block in case-1 to case-2 just after the cutting is (strings and springs are ideal)

• A. 0

Answer 1

1

Answer 2

Let $m$ be the mass of the block and $\theta = 37^\circ$ be the angle the springs/strings make with the vertical.

Case 1: Right spring is cut Initially, the block is in equilibrium. Let $F_s$ be the force in each spring. Vertical equilibrium: $2 F_s \cos(\theta) = mg$ Horizontal equilibrium: $F_s \sin(\theta) - F_s \sin(\theta) = 0$ (already satisfied) So, $F_s = \frac{mg}{2 \cos(\theta)}$.

Just after the right spring is cut, the forces acting on the block are:

1. The force from the left spring, $\vec{F}_{s1}$, which is unchanged immediately after cutting. Its magnitude is $F_s$.
2. The weight of the block, $m\vec{g}$, acting vertically downwards.

Let's set up a coordinate system with the y-axis vertically upwards and the x-axis horizontally. Assuming the left spring is to the left of the vertical: $\vec{F}_{s1} = -F_s \sin(\theta) \hat{i} + F_s \cos(\theta) \hat{j}$ $m\vec{g} = -mg \hat{j}$

The net force in Case 1 is: $\vec{F}_{net,1} = \vec{F}_{s1} + m\vec{g} = -F_s \sin(\theta) \hat{i} + (F_s \cos(\theta) - mg) \hat{j}$ Substitute $F_s \cos(\theta) = \frac{mg}{2}$: $\vec{F}_{net,1} = -F_s \sin(\theta) \hat{i} + (\frac{mg}{2} - mg) \hat{j} = -F_s \sin(\theta) \hat{i} - \frac{mg}{2} \hat{j}$

The instantaneous acceleration in Case 1 is: $\vec{a}_1 = \frac{\vec{F}_{net,1}}{m} = -\frac{F_s \sin(\theta)}{m} \hat{i} - \frac{g}{2} \hat{j}$

Case 2: Right string is cut Initially, the block is in equilibrium. Let $T$ be the tension in each string. Vertical equilibrium: $2 T \cos(\theta) = mg$ Horizontal equilibrium: $T \sin(\theta) - T \sin(\theta) = 0$ (already satisfied) So, $T = \frac{mg}{2 \cos(\theta)}$.

Just after the right string is cut, the forces acting on the block are:

1. The tension from the left string, $\vec{T}_1$, which is unchanged immediately after cutting. Its magnitude is $T$.
2. The weight of the block, $m\vec{g}$, acting vertically downwards.

Assuming the left string is to the left of the vertical: $\vec{T}_1 = -T \sin(\theta) \hat{i} + T \cos(\theta) \hat{j}$ $m\vec{g} = -mg \hat{j}$

The net force in Case 2 is: $\vec{F}_{net,2} = \vec{T}_1 + m\vec{g} = -T \sin(\theta) \hat{i} + (T \cos(\theta) - mg) \hat{j}$ Substitute $T \cos(\theta) = \frac{mg}{2}$: $\vec{F}_{net,2} = -T \sin(\theta) \hat{i} + (\frac{mg}{2} - mg) \hat{j} = -T \sin(\theta) \hat{i} - \frac{mg}{2} \hat{j}$

The instantaneous acceleration in Case 2 is: $\vec{a}_2 = \frac{\vec{F}_{net,2}}{m} = -\frac{T \sin(\theta)}{m} \hat{i} - \frac{g}{2} \hat{j}$

Comparison of Accelerations: Since the springs and strings are ideal, and the geometry (mass, angle) is the same, the force in the left spring ($F_s$) and the tension in the left string ($T$) are equal: $F_s = T = \frac{mg}{2 \cos(\theta)}$

Substituting this into the acceleration expressions: $\vec{a}_1 = -\frac{\frac{mg}{2 \cos(\theta)} \sin(\theta)}{m} \hat{i} - \frac{g}{2} \hat{j} = -\frac{g \tan(\theta)}{2} \hat{i} - \frac{g}{2} \hat{j}$ $\vec{a}_2 = -\frac{\frac{mg}{2 \cos(\theta)} \sin(\theta)}{m} \hat{i} - \frac{g}{2} \hat{j} = -\frac{g \tan(\theta)}{2} \hat{i} - \frac{g}{2} \hat{j}$

Thus, $\vec{a}_1 = \vec{a}_2$. The instantaneous accelerations in both cases are identical vectors. The ratio of the magnitudes of the instantaneous accelerations is: $\frac{|\vec{a}_1|}{|\vec{a}_2|} = \frac{|\vec{a}_1|}{|\vec{a}_1|} = 1$

Using $\theta = 37^\circ$, $\tan(37^\circ) \approx 0.75$: $\vec{a}_1 = \vec{a}_2 = -\frac{g \times 0.75}{2} \hat{i} - \frac{g}{2} \hat{j} = -0.375g \hat{i} - 0.5g \hat{j}$ $|\vec{a}_1| = |\vec{a}_2| = \sqrt{(-0.375g)^2 + (-0.5g)^2} = g \sqrt{0.140625 + 0.25} = g \sqrt{0.390625} = 0.625g = \frac{5}{8}g$. The ratio is $\frac{5g/8}{5g/8} = 1$.

The user's provided answer of 0 is incorrect. The ratio of the instantaneous accelerations is 1.