Animated Solution for Physics - Laws of Motion: The boxes of masses 2 kg and 8 kg are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass 8 kg to strike the ground starting from rest. (Use, $g=10{\text{ m/s}}^2$)

Mastering Constraint Relations

Imagine a system where two masses, $2\text{ kg}$ and $8\text{ kg}$, are connected by a single continuous string. This string passes over a fixed pulley and a movable pulley. At first glance, it might seem like a standard Atwood machine, but the presence of the movable pulley introduces a fascinating twist known as a constraint relation.

The key to unlocking this problem lies in understanding how the string's length constrains the motion of the two masses. Look closely at the movable pulley holding the $8\text{ kg}$ mass. It is supported by two segments of the string.

This means that if the $2\text{ kg}$ mass moves up by a distance $x$, the string must lengthen by $x$ on that side. This extra length $x$ is distributed equally between the two segments supporting the movable pulley. Consequently, the movable pulley and the $8\text{ kg}$ mass will only move down by a distance of $\frac{x}{2}$.

Therefore, if the $2\text{ kg}$ mass has an upward acceleration $a$, the $8\text{ kg}$ mass will have a downward acceleration of exactly half that amount, or $\frac{a}{2}$.

Free Body Diagrams

Now, let's draw the free body diagrams for both masses to set up our equations of motion.

For the $2\text{ kg}$ mass ($m_1$), the string pulls it upwards with a tension $T$, while gravity pulls it downwards with a force $m_{1}g$. Applying Newton's second law for its upward motion, we get: Substituting $m_1=2\text{ kg}$, our first equation becomes:

Moving to the $8\text{ kg}$ mass ($m_2$), the movable pulley is pulled upwards by two segments of the string, so the total upward force is $2T$. Downwards, we have its weight, $m_{2}g$. For its downward motion with acceleration $\frac{a}{2}$, the equation is: Substituting $m_2=8\text{ kg}$: Dividing the entire equation by 2 simplifies it to:

Let's solve these two equations simultaneously. By adding equation (i) and equation (ii), the tension $T$ cancels out beautifully:

Using $g=10{\text{ m/s}}^2$, we find the acceleration $a$ of the $2\text{ kg}$ mass is $5{\text{ m/s}}^2$.

But remember, we need the time taken by the $8\text{ kg}$ mass to hit the ground. Its acceleration is $a_2=\frac{a}{2}$, which gives us:

Now, we use kinematics. The $8\text{ kg}$ mass starts from rest ($u=0$) and needs to cover a distance of $20\text{ cm}$, which must be converted to standard SI units as $0.2\text{ m}$. We use the second equation of motion:

Substituting our known values:

Taking the square root, we find the time $t$ is exactly $0.4\text{ s}$. The $8\text{ kg}$ block will strike the ground in a mere fraction of a second, perfectly dictated by the laws of physics!