A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coeffici

Q: Why is the normal force horizontal instead of vertical?

The normal force is always perpendicular to the contact surface. Since the wall is vertical, the normal force pushing away from it must be horizontal.

Q: Why does friction act upwards?

Friction always opposes the relative motion or the tendency of relative motion. Gravity tries to pull the block downwards, so friction acts upwards to oppose this tendency.

Q: What does 'just hold' mean in physics problems?

It means the system is in limiting equilibrium. The static friction has reached its maximum possible value ($f = \mu_s R$), and any slight decrease in the applied force would cause the block to slip.

Q: What would happen if we applied a force of 20 N instead of 10 N?

The normal force would increase to 20 N, which means the maximum possible friction would increase to 4 N. However, the actual friction acting on the block would still only be 2 N to balance the weight. The block would remain stationary.

Q: Does the weight of the block affect the normal force?

No. In this specific setup, the horizontal forces (applied force and normal force) are completely independent of the vertical forces (weight and friction).

Question

A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Options:

(A) 20 N

(B) 50 N

(C) 100 N

(D) 2 N

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are trying to hold a heavy book against a vertical wall by pressing it horizontally with your hand. If you don't press hard enough, the book slips and falls. If you press just hard enough, it stays perfectly still. This is exactly the scenario we are dealing with here.

We have a block held stationary against a vertical wall by a horizontal force of $10	ext{ N}$. Our goal is to find the weight of this block, given that the coefficient of static friction between the block and the wall is $0.2$.

### The Horizontal Equilibrium

Let's break down the forces acting on the block. Since the block is not moving horizontally (it's neither sinking into the wall nor flying away from it), the net horizontal force must be zero.

The applied force $F = 10	ext{ N}$ pushes the block into the wall. According to Newton's Third Law, the wall pushes back with an equal and opposite force. This is our normal reaction, $R$.

Therefore, we can confidently state:
$$R = F = 10	ext{ N}$$

### The Vertical Equilibrium

Now, let's look at the vertical direction. Gravity is relentlessly pulling the block downwards with a force equal to its weight, $w$. But the block isn't falling! This means there must be an upward force perfectly balancing the weight.

This upward force is the static friction, $f$, acting parallel to the surface of the wall. For the block to remain stationary, the upward friction must equal the downward weight:
$$f = w$$

### The Limiting Friction

The problem uses a very specific phrase: the force is necessary to **just hold** the block. This is a massive clue! It tells us that the block is on the absolute verge of slipping. In this critical state, the static friction has reached its maximum possible value, known as limiting friction.

The formula for limiting friction is:
$$f_{	ext{max}} = \mu R$$

Since our friction $f$ is at this maximum limit, we can substitute $f_{	ext{max}}$ for $w$:
$$w = \mu R$$

### Final Calculation

We have all the pieces of the puzzle. We know the coefficient of friction $\mu = 0.2$, and we found the normal reaction $R = 10	ext{ N}$. Let's plug these values into our equation:

$$w = 0.2 	imes 10$$
$$w = 2	ext{ N}$$

The weight of the block is exactly $2	ext{ N}$. It's a beautifully simple application of Newton's laws and the concept of limiting friction!

Animated Solution for Physics - Laws of Motion: A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Select Answer:

Visualized Solution

$F_{net} = 0$

$R = 10 N$

$f = w$

$f_{max} = μ R$

$w = μ R$

$w = 2 N$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Horizontal Equilibrium

The Vertical Equilibrium

The Limiting Friction

Final Calculation

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