Two blocks connected by a massless string slides down an inclined plane having an angle of inclinati

Q: Wait, why does tension act downwards on $M_1$?

Tension is a pulling force; it always points away from the mass you are analyzing, along the string. Since $M_1$ is connected to $M_2$ which is physically below it on the incline, the string pulls $M_1$ down the slope.

Q: What would happen if $\mu_1$ was less than $\mu_2$?

If $M_1$ had less friction, it would naturally want to slide faster than $M_2$. In that scenario, $M_1$ would slide down and bump into $M_2$, causing the string to go slack. The tension $T$ would become zero, and they would push against each other instead of pulling.

Q: Why do we use $g = 9.8$ instead of $10$?

The problem explicitly states at the end: `(Take g = 9.8 m/s^2)`. In JEE/NEET, always scan the question for specific constants. Using $10$ here would lead to incorrect decimal answers and might cost you marks.

Q: Can we treat both blocks as a single system to find acceleration faster?

Yes, absolutely! Since they move with the same acceleration, you can treat them as a single $6\text{ kg}$ block. The net external force down the incline is $(M_1+M_2)g \sin 37^\circ - f_1 - f_2$. Setting this equal to $(M_1+M_2)a$ gives you the acceleration directly without needing to solve for tension first.

Q: Why did the gravity component exactly cancel the friction for $M_1$?

It's a beautiful numerical coincidence designed by the examiner! The angle is $37^\circ$, and $\tan 37^\circ = 0.75$. The friction coefficient $\mu_1$ is also $0.75$. Since $\mu = \tan \theta$, block $M_1$ is exactly at its angle of repose. Without the string, it would slide with constant velocity or stay at rest.

Question

Two blocks connected by a massless string slides down an inclined plane having an angle of inclination of $37^\circ$. The masses of the two blocks are $M_1 = 4 	ext{ kg}$ and $M_2 = 2 	ext{ kg}$ respectively and the coefficients of friction of $M_1$ and $M_2$ with the inclined plane are $0.75$ and $0.25$ respectively. Assuming the string to be taut, find (a) the common acceleration of two masses and (b) the tension in the string. ($\sin 37^\circ = 0.6, \cos 37^\circ = 0.8$). (Take $g = 9.8 	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Blocks

Imagine a classic physics scenario: a smooth, yet slightly rough inclined plane angled precisely at $37^\circ$ to the horizontal. Resting on this slope are two blocks, $M_1$ at the top with a mass of $4	ext{ kg}$, and $M_2$ at the bottom with a mass of $2	ext{ kg}$. They are tethered together by a light, inextensible string.

Before we dive into the mathematics, we must ask a crucial physical question: **Will the string remain taut as they slide?**

### The Intuition Check: Will the String Slack?

To answer this, we must look at the coefficients of kinetic friction. Block $M_1$ has a high friction coefficient of $\mu_1 = 0.75$, while $M_2$ has a much lower coefficient of $\mu_2 = 0.25$.

Because $M_1$ experiences significantly more relative friction, its natural tendency is to slide down the incline much slower than $M_2$. However, since $M_2$ is positioned in front of $M_1$ and wants to rush ahead, it will continuously pull on $M_1$. This forward pull guarantees that the string remains perfectly taut throughout the motion. Consequently, both blocks will descend with a shared, common acceleration $a$.

### Analyzing Block $M_1$: The Magic of Cancellation

Let's isolate block $M_1$ and draw its Free Body Diagram. The forces acting parallel to the incline are:
1. The component of gravity pulling it down: $M_1 g \sin 37^\circ$.
2. The tension $T$ from the string, which also pulls it down (since $M_2$ is below it).
3. The kinetic friction $f_1$ opposing the motion, acting up the incline.

First, we calculate the kinetic friction $f_1$. The normal force is $N_1 = M_1 g \cos 37^\circ$. Therefore, the friction is:

$$f_1 = \mu_1 M_1 g \cos 37^\circ = 0.75 	imes 4 	imes 9.8 	imes 0.8 = 23.52	ext{ N}$$

Next, we calculate the driving force due to gravity:

$$M_1 g \sin 37^\circ = 4 	imes 9.8 	imes 0.6 = 23.52	ext{ N}$$

Notice something beautiful here? The downward pull of gravity exactly equals the upward resistance of friction! This is a numerical coincidence because $	an 37^\circ = 0.75$, which exactly matches $\mu_1$.

Applying Newton's Second Law ($F_{	ext{net}} = ma$) for $M_1$:

$$23.52 + T - 23.52 = 4a$$

This simplifies elegantly to our first master equation:

$$T = 4a$$

### Analyzing Block $M_2$: The Driving Force

Now, let's shift our focus to block $M_2$. The forces acting parallel to the incline are:
1. The component of gravity pulling it down: $M_2 g \sin 37^\circ$.
2. The tension $T$ from the string, which now pulls it *up* the incline (holding it back).
3. The kinetic friction $f_2$ opposing the motion, also acting up the incline.

Let's calculate the kinetic friction $f_2$:

$$f_2 = \mu_2 M_2 g \cos 37^\circ = 0.25 	imes 2 	imes 9.8 	imes 0.8 = 3.92	ext{ N}$$

And the driving force due to gravity:

$$M_2 g \sin 37^\circ = 2 	imes 9.8 	imes 0.6 = 11.76	ext{ N}$$

Applying Newton's Second Law for $M_2$:

$$11.76 - T - 3.92 = 2a$$

Simplifying this gives us our second master equation:

$$7.84 - T = 2a$$

### The Grand Finale: Solving the System

We now possess a clean system of two linear equations:
1. $T = 4a$
2. $7.84 - T = 2a$

To find the common acceleration, we simply add the two equations together. The internal tension $T$ perfectly cancels out:

$$7.84 = 6a$$

Solving for $a$:

$$a = \frac{7.84}{6} \approx 1.3	ext{ m/s}^2$$

Finally, to find the tension in the string, we substitute the exact value of $a$ back into our simplest equation, $T = 4a$:

$$T = 4 	imes \left(\frac{7.84}{6}ight) = \frac{15.68}{3} \approx 5.2	ext{ N}$$

And there we have it! By carefully breaking down the system into atomic Free Body Diagrams and trusting Newton's Laws, we have completely unraveled the dynamics of these connected blocks.

Visualized Solution

$System Setup$

$Will the string be taut?$

$FBD of M_{1}$

$Calculating f_{1}$

$Equation for M_{1}$

$FBD of M_{2}$

$Calculating f_{2}$

$Equation for M_{2}$

$Common Acceleration$

$Tension in String$

The Sigma Insight: Static and Kinetic Friction

The Setup

The Intuition Check

Analyzing Block $M_{1}$

Analyzing Block $M_{2}$

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Will the string be taut?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FBD of M1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating f1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for M1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FBD of M2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating f2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for M2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Common Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tension in String