Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in f

Q: Why does the top block move if no force is applied directly to it?

Friction acts as the driving force! When the bottom block tries to accelerate, the rough surfaces interlock. The static friction from the bottom block drags the top block forward.

Q: Why do we take the normal force as $mg$ and not $(m+M)g$?

The friction we are calculating is between the two blocks. The normal force pressing these two specific surfaces together is only the weight of the top block, which is $mg$.

Q: What happens if $F$ is greater than 21 N?

If $F > 21$ N, the required friction to keep the blocks together exceeds the maximum static friction limit. The grip breaks, kinetic friction takes over, and the top block slips backward relative to the bottom block.

Q: Why did the mass $m$ cancel out in the inequality?

The required friction to accelerate the top block is proportional to its mass $m$ ($f = ma$). The maximum available friction is also proportional to its mass ($f_{\text{max}} = \mu_s mg$). Because both scale with $m$, it cancels out, meaning the slipping condition depends only on the accelerations, not the mass of the top block.

Q: Is there friction acting on the bottom block too?

Yes! By Newton's Third Law, if the bottom block exerts a forward friction force on the top block, the top block exerts an equal and opposite backward friction force on the bottom block. This backward force opposes the applied force $F$.

Question

Two blocks ($m = 0.5$ kg and $M = 4.5$ kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is $3/7$. Then, the maximum horizontal force that can be applied on the larger block so that the blocks move together is ...... N. (Round off to the nearest integer. Take, $g = 9.8 	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

The "Two-Block Problem" is an absolute classic in Newtonian mechanics. It is a beautiful test of your understanding of systems, free-body diagrams, and the subtle nature of static friction.

When you first look at this problem, it might seem intimidating. We have two masses, an applied force, and friction acting between them, but no friction on the floor.

The key to unlocking this problem is to realize that static friction is not just a force that opposes motion; it is often the very force that **causes** motion! Let's dive into the physics and break it down step-by-step.

### Analyzing the Setup

Imagine the scenario: a small block of mass $m$ rests on top of a larger block of mass $M$. The larger block sits on a perfectly smooth, frictionless table.

We apply a horizontal force $F$ to the bottom block $M$.

Because the table is frictionless, any non-zero force $F$ will cause the entire system to accelerate. But here is the critical question: will the top block move *with* the bottom block, or will it slip and fall off?

For the blocks to move together, they must share the exact same acceleration. If their accelerations differ even slightly, relative motion (slipping) occurs.

### The Master Equation

Let's assume the ideal case: the blocks move together perfectly.

If they move together, we can treat them as a single combined object with a total mass of $(m + M)$.

By applying Newton's Second Law to this combined system, we can easily find their common acceleration $a$. The only external horizontal force acting on the system is $F$.

$$a = \frac{F}{m + M}$$

This equation tells us how fast the entire setup is speeding up. But it doesn't tell us *why* the top block is speeding up. For that, we need to zoom in.

### The Invisible Hand

Let's isolate the top block $m$ and draw its Free Body Diagram (FBD).

Look closely at block $m$. Is there any string pulling it? No. Is there any hand pushing it? No. The external force $F$ is applied *only* to the bottom block $M$.

So, what invisible hand is dragging the top block forward? It is **static friction**!

Because the bottom block is trying to accelerate out from under the top block, the rough surfaces interlock. The bottom block drags the top block forward via static friction $f$.

By Newton's Second Law applied only to the top block, this friction must be the force providing its acceleration:

$$f = ma$$

Now, we substitute the common acceleration $a$ we found earlier into this equation:

$$f = m \left( \frac{F}{m + M} ight)$$

This is a profound result. It tells us the *exact* amount of friction required to keep the top block perfectly synced with the bottom block. As we pull harder (increasing $F$), the required friction $f$ increases proportionally.

### The Breaking Point

Here is where reality sets in. Static friction is smart, but it has a limit. It can only grow up to a certain maximum value, known as **limiting friction**.

$$f_{	ext{max}} = \mu_s N$$

For the top block, the normal force $N$ is simply its weight, $mg$. Therefore, the maximum possible static friction is:

$$f_{	ext{max}} = \mu_s mg$$

If the required friction $f$ exceeds this maximum limit $f_{	ext{max}}$, the surfaces will break their grip, and the top block will slip.

To find the maximum force $F$ we can apply without slipping, we set up our critical inequality:

$$f \le f_{	ext{max}}$$

Substituting our expressions:

$$m \left( \frac{F}{m + M} ight) \le \mu_s mg$$

Notice something beautiful? The mass of the top block, $m$, cancels out on both sides!

$$\frac{F}{m + M} \le \mu_s g$$

Rearranging this gives us the master formula for the maximum applied force:

$$F \le \mu_s (m + M) g$$

### Final Calculation

Now, we simply plug in the numbers given in the problem.

We are given $\mu_s = \frac{3}{7}$, $m = 0.5 	ext{ kg}$, $M = 4.5 	ext{ kg}$, and $g = 9.8 	ext{ m/s}^2$.

$$F_{	ext{max}} = \frac{3}{7} 	imes (0.5 + 4.5) 	imes 9.8$$

First, add the masses:

$$F_{	ext{max}} = \frac{3}{7} 	imes 5 	imes 9.8$$

Next, divide $9.8$ by $7$, which gives exactly $1.4$:

$$F_{	ext{max}} = 3 	imes 5 	imes 1.4$$

$$F_{	ext{max}} = 15 	imes 1.4$$

$$F_{	ext{max}} = 21 	ext{ N}$$

The maximum horizontal force we can apply is exactly **$21 	ext{ N}$**.

If we apply $21.1 	ext{ N}$, the required friction will exceed the physical limit of the surfaces, and the top block will begin to slide backward relative to the bottom block.

Physics is all about understanding these limits and the invisible forces that hold our world together. Keep visualizing, keep questioning, and you will master mechanics!

Enter Numerical Value:

Visualized Solution

$System Setup$

$Common Acceleration$

$Free Body Diagram of m$

$Equation of Motion for m$

$Required Friction$

$Limiting Friction$

$No Slipping Condition$

$Maximum Force Expression$

$Substituting Values$

$Final Calculation$

$Conclusion$

$Food for Thought$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Master Equation

The Invisible Hand

The Breaking Point

Final Calculation

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Enter Numerical Value:

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; } Common Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram of m

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } No Slipping Condition

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

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

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

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

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