As shown in the figure, a block of mass √3 kg is kept on a horizontal rough surface of coefficient o

Q: Why does the applied force increase the normal reaction?

Because the force is applied at a downward angle, it has a vertical component pushing the block into the floor. The floor must push back with an equal and opposite force to prevent the block from sinking, thus increasing the normal reaction.

Q: Why do we use static friction instead of kinetic friction?

The problem asks for the 'critical force' such that the block 'does not move'. This implies the block is on the verge of slipping but is still stationary. Therefore, we use the maximum limit of static friction.

Q: What if the force was applied horizontally?

If the force was purely horizontal, it would have no vertical component. The normal reaction would simply be equal to the weight of the block ($N = mg$), and the required force would be $F = \mu mg$.

Q: Why did we equate the driving force exactly to the maximum static friction?

At the critical point just before motion begins, the driving force has increased to the exact point where it matches the maximum possible resistance the surface can offer. Any tiny increase in force beyond this point will cause acceleration.

Q: Can the block move vertically?

No, the block is constrained by the solid floor. Unless the upward forces exceed the downward forces (which isn't happening here), the block will remain in vertical equilibrium.

Question

As shown in the figure, a block of mass $\sqrt{3}	ext{ kg}$ is kept on a horizontal rough surface of coefficient of friction $1/3\sqrt{3}$. The critical force to be applied on the vertical surface as shown at an angle $60^\circ$ with horizontal such that it does not move, will be $3x$. The value of $x$ will be ......... [$g = 10	ext{ ms}^{-2}$; $\sin 60^\circ = \frac{\sqrt{3}}{2}$; $\cos 60^\circ = \frac{1}{2}$]

Sigma Percentile · Accepted Answer

The study of mechanics often brings us face-to-face with the delicate balance of forces. In this problem, we are tasked with finding the exact tipping point—the critical force—just before a block yields to an applied push and begins to slide.

Imagine a block resting peacefully on a rough horizontal surface. We apply a force, but not just any force; we push it at a downward angle. This seemingly simple act sets off a fascinating interplay between gravity, the surface, and friction. Let's unravel this physical puzzle step by step.

### Analyzing the Setup

We are given a block of mass $m = \sqrt{3} 	ext{ kg}$. It sits on a surface with a coefficient of static friction $\mu = \frac{1}{3\sqrt{3}}$.

A force $F$ is applied at the top right corner of the block, pushing inwards at an angle of $60^\circ$ with the horizontal. Because we are looking for the **critical force** such that the block *does not move*, we are dealing with a state of **limiting equilibrium**.

In this state, the block is on the absolute verge of slipping. The static friction has reached its maximum possible value, and the net force in every direction is exactly zero.

### Breaking Down the Forces

To understand what's happening, we must resolve the applied force $F$ into its horizontal and vertical components.

The force pushes into the block at a downward angle. This means it has a horizontal component pushing to the left:

$$
F_x = F \cos 60^\circ
$$

And a vertical component pushing downwards:

$$
F_y = F \sin 60^\circ
$$

This downward push is crucial. It doesn't just try to move the block; it actively presses the block harder against the floor!

### The Vertical Balance

Let's look at the vertical direction. The block isn't flying into the air, nor is it sinking into the ground. The vertical forces must perfectly cancel each other out.

What is pushing down? We have the intrinsic weight of the block, $mg$, and the downward component of our applied force, $F \sin 60^\circ$.

What is pushing up? The floor responds with a normal reaction, $N$. Therefore, the normal reaction must support both of these downward forces:

$$
N = mg + F \sin 60^\circ
$$

This equation reveals a beautiful physical truth: **pushing down on an object increases the normal reaction**, which in turn will increase the maximum possible friction!

### The Horizontal Balance

Now, let's turn our attention to the horizontal plane. The horizontal component of our applied force, $F \cos 60^\circ$, is the driving force trying to slide the block to the left.

The rough surface fights back with static friction, $f$, acting to the right. Since we are at the critical point of slipping, this friction is at its absolute limit:

$$
f = \mu N
$$

For the block to remain stationary, the driving force must exactly equal this limiting friction:

$$
F \cos 60^\circ = \mu N
$$

### The Master Equation

We now have our two fundamental equations. By substituting the expression for the normal reaction $N$ into our horizontal balance equation, we create a single master equation:

$$
F \cos 60^\circ = \mu (mg + F \sin 60^\circ)
$$

This equation perfectly encapsulates the entire physical situation. It balances the desire to move against the resistance to motion, accounting for the extra grip caused by the downward push.

### Executing the Mathematics

It is time to bring in our numerical values. We know $m = \sqrt{3}$, $g = 10$, and $\mu = \frac{1}{3\sqrt{3}}$. We also know the trigonometric values: $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

Substituting these into our master equation:

$$
F \left(\frac{1}{2}ight) = \frac{1}{3\sqrt{3}} \left(\sqrt{3} \cdot 10 + F \frac{\sqrt{3}}{2}ight)
$$

Let's simplify the right side. Notice how every term inside the parenthesis contains a $\sqrt{3}$. We can factor this out and elegantly cancel it with the $\sqrt{3}$ in the denominator outside:

$$
\frac{F}{2} = \frac{1}{3} \left(10 + \frac{F}{2}ight)
$$

Expanding the right side gives:

$$
\frac{F}{2} = \frac{10}{3} + \frac{F}{6}
$$

To isolate $F$, we bring all terms containing $F$ to the left side:

$$
\frac{F}{2} - \frac{F}{6} = \frac{10}{3}
$$

Finding a common denominator of $6$:

$$
\frac{3F - F}{6} = \frac{10}{3}
$$

$$
\frac{2F}{6} = \frac{10}{3}
$$

Simplifying the fraction:

$$
\frac{F}{3} = \frac{10}{3}
$$

The denominators perfectly cancel out, leaving us with a beautifully clean result:

$$
F = 10 	ext{ N}
$$

### The Final Conclusion

We have successfully determined that the critical force required is $10 	ext{ N}$.

However, the problem states that this critical force is equal to $3x$. To find our final answer, we simply set up the equivalence:

$$
3x = 10
$$

Solving for $x$:

$$
x = \frac{10}{3} \approx 3.33
$$

And there we have it! By carefully dissecting the forces and respecting the physical constraints of limiting equilibrium, we navigated through the algebra to arrive at the precise value.

Enter Numerical Value:

Visualized Solution

$System Setup$

$Limiting Equilibrium$

$Vertical Equilibrium$

$Horizontal Equilibrium$

$Substituting Normal Reaction$

$Substituting Values$

$Simplifying the Equation$

$Solving for F$

$Finding x$

$Food for Thought$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

Breaking Down the Forces

The Vertical Balance

The Horizontal Balance

The Master Equation

Executing the Mathematics

The Final Conclusion

Similar Questions

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A block of mass $2 kg$ rests on a rough inclined plane making an angle of $3 0^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$ . The frictional force on the block is

A small block of mass of $0.1 kg$ lies on a fixed inclined plane $P Q$ which makes an angle $θ$ with the horizontal. A horizontal force of $1 N$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if (Take $g = 10 m/s^{2}$ )

A block of mass $m$ is placed on a surface with a vertical cross-section given by $y = \frac{x ^{3}}{6}$ . If the coefficient of friction is $0.5$ , the maximum height above the ground at which the block can be placed without slipping is

What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move? (2003)

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; } Limiting Equilibrium

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

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

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

.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; } Simplifying the Equation

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

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

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