A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is $\mu$ and the acceleration due to gravity is $g$, calculate the minimum force required to be applied by the finger to hold the block against the wall ?

> [!TLDR] > To hold the block against the wall, the upward static friction must balance the downward gravitational force. The minimum applied force ensures that this required friction is exactly the maximum possible static friction, which depends on the normal force. The minimum force required is $\boxed{\frac{Mg}{\mu}}$. When you press a block against a rough vertical wall, several forces act on it. For the block to remain stationary, these forces must be balanced. The key idea here is to understand how the force you apply horizontally influences the friction force that can act vertically to counteract gravity. The block has a tendency to slide down due to its weight. To prevent this, an upward force is needed. This upward force is provided by static friction between the block and the wall. The amount of static friction available depends directly on how hard you press the block against the wall. 1. **Identify and Draw Forces:** Let's consider the forces acting on the block: * **Gravitational Force ($Mg$):** Acts vertically downwards, due to the mass of the block and acceleration due to gravity. * **Applied Force ($F$):** This is the force exerted by your finger, acting horizontally into the wall. This is what we need to find. * **Normal Force ($N$):** The wall exerts an equal and opposite force on the block, acting horizontally outwards, perpendicular to the wall's surface. * **Static Frictional Force ($f_s$):** Since the block tends to slide downwards, the wall exerts an upward static frictional force, parallel to the wall's surface, opposing this tendency. Imagine a free-body diagram: * $Mg$ points down. * $F$ points left (into the wall). * $N$ points right (out from the wall). * $f_s$ points up. 2. **Apply Newton's First Law (Equilibrium Conditions):** For the block to be held stationary, it must be in equilibrium. This means the net force in both the horizontal and vertical directions must be ze

A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of fr

NCERT Exemplar · Q30

\mu

Punjab PsebShort· 3mImportance★★★★★est

84% · 65/77 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Static Friction Limit

The Static Friction Limit: Why a Heavy Box Won't Move Until You Really Push

Imagine you're trying to push a heavy wooden crate across a rough floor. You lean into it gently — nothing happens. You push a little harder — still nothing. The crate stays perfectly still, as if glued to the spot. Then, at some point, you push just a bit more, and suddenly the crate lurches forward.

That invisible "sticking point" — the exact moment the crate finally gives way — is the static friction limit.

The Intuition: Friction as a "Smart" Force

Friction between two surfaces that aren't sliding is called static friction. What makes it special is that it's self-adjusting. It doesn't have a fixed value. Instead, it automatically grows to match whatever force you apply — up to a point.

Think of it like a tug-of-war where your opponent (static friction) matches your pull exactly, but only until you exceed their maximum strength. As long as you pull less than their limit, you both stay in equilibrium and nothing moves. The moment you exceed that limit, you win — and the crate starts sliding.

Note

Static friction only exists when there is no relative motion between the surfaces. Once sliding begins, it's replaced by kinetic friction, which is usually weaker.

The Precise Statement

The static friction limit (also called limiting friction) is the maximum possible value of static friction that can act between two surfaces in contact before they start sliding relative to each other.

Mathematically:

f_s \leq \mu_s N

Where:

$f_s$ = static friction force (the actual value, which can be anything from 0 up to the limit)
$\mu_s$ = coefficient of static friction (a constant that depends on the two materials — rubber on concrete is high, ice on steel is low)
$N$ = normal reaction force (the force pressing the surfaces together, usually equal to weight on a horizontal surface)

The static friction limit is the equality case:

f_{s,\text{max}} = \mu_s N

This is the maximum static friction the surfaces can provide. Apply a force less than this, and the object stays put. Apply a force equal to this, and the object is on the verge of moving (impending motion). Apply a force greater than this, and the object accelerates.

A Concrete Example

A 10 kg block rests on a horizontal floor. $\mu_s = 0.4$ between the block and floor. Take $g = 10 \text{ m/s}^2$ .

Normal reaction: $N = mg = 10 \times 10 = 100 \text{ N}$

Static friction limit: $f_{s,\text{max}} = 0.4 \times 100 = 40 \text{ N}$

Now, what happens as you push?

Applied Force	Static Friction	Result
10 N	10 N (matches)	Block stays still
25 N	25 N (matches)	Block stays still
40 N	40 N (matches)	Block is just about to move
45 N	40 N (cannot exceed limit)	Block accelerates forward

Start your 14-day free trial to unlock the full solution →

To hold the block against the wall, the upward static friction must balance the downward gravitational force. The minimum applied force ensures that this required friction is exactly the maximum possible static friction, which depends on the normal force. The minimum force required is $\boxed{\frac{Mg}{\mu}}$ .

When you press a block against a rough vertical wall, several forces act on it. For the block to remain stationary, these forces must be balanced. The key idea here is to understand how the force you apply horizontally influences the friction force that can act vertically to counteract gravity.

The block has a tendency to slide down due to its weight. To prevent this, an upward force is needed. This upward force is provided by static friction between the block and the wall. The amount of static friction available depends directly on how hard you press the block against the wall.

Identify and Draw Forces:

Let's consider the forces acting on the block:
- Gravitational Force ( $Mg$ ): Acts vertically downwards, due to the mass of the block and acceleration due to gravity.
- Applied Force ( $F$ ): This is the force exerted by your finger, acting horizontally into the wall. This is what we need to find.
- Normal Force ( $N$ ): The wall exerts an equal and opposite force on the block, acting horizontally outwards, perpendicular to the wall's surface.
- Static Frictional Force ( $f_s$ ): Since the block tends to slide downwards, the wall exerts an upward static frictional force, parallel to the wall's surface, opposing this tendency.
Imagine a free-body diagram:
- $Mg$ points down.
- $F$ points left (into the wall).
- $N$ points right (out from the wall).
- $f_s$ points up.
Apply Newton's First Law (Equilibrium Conditions):

For the block to be held stationary, it must be in equilibrium. This means the net force in both the horizontal and vertical directions must be zero.
- Horizontal Equilibrium: The applied force $F$ must be balanced by the normal force $N$ .

\Sigma F_x = 0 \implies F - N = 0

N = F \quad \text{(Equation 1)}

    This tells us that the harder you press, the greater the normal force exerted by the wall on the block.

*   **Vertical Equilibrium:** The upward static frictional force $f_s$ must balance the downward gravitational force $Mg$.

\Sigma F_y = 0 \implies f_s - Mg = 0

f_s = Mg \quad \text{(Equation 2)}

    This means that to prevent the block from falling, the static friction must be exactly equal to the block's weight.

3. Relate Friction to Normal Force:

The maximum possible static frictional force ( $f_{s,max}$ ) that can be exerted by a surface is proportional to the normal force pressing the surfaces together.

> [!FORMULA]

> The maximum static frictional force is given by:

> $f_{s,max} = \mu N$

> where $\mu$ is the coefficient of static friction. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ29 — Block A, of weight 100N, rests on a frictionless inclined plane whose …→Next questionQ31 — A 100 kg gun fires a ball of 1kg horizontally from a cliff of height 5…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days