A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block

Q: Wait, why does friction change its sign from positive to negative?

Static friction is a self-adjusting force that always opposes the *tendency* of relative motion. When you apply a weak force ($P_1$), the block wants to slide down, so friction acts up (positive). When you apply a very strong force ($P_2$), the block wants to slide up, so friction flips and acts down (negative) to stop it.

Q: How do we know the graph is a straight line and not a curve?

By writing the equilibrium equation: $P + f - mg \sin\theta = 0$. When we isolate friction, we get $f = -P + mg \sin\theta$. This equation is in the standard linear form $y = mx + c$, where the variable is $P$, the slope $m$ is $-1$, and the y-intercept $c$ is $mg \sin\theta$. A constant slope means it's a straight line.

Q: Why is the normal force $N$ equal to $mg \cos\theta$ and not just $mg$?

The normal force only balances the forces perpendicular to the surface. Gravity acts straight down, so only a component of it ($mg \cos\theta$) presses into the inclined plane. Since the block doesn't accelerate into or off the plane, $N$ must exactly balance this perpendicular component.

Q: What does the condition $\tan\theta > \mu$ mean in the problem?

This condition tells us that the angle of inclination $\theta$ is greater than the angle of repose. Physically, it means that if you didn't apply any force $P$, the block would definitely slide down because gravity ($mg \sin\theta$) is stronger than the maximum possible static friction ($\mu mg \cos\theta$). That's why a positive force $P_1$ is required just to hold it in place.

Q: What would happen to the graph if we kept increasing $P$ beyond $P_2$?

If $P$ exceeds $P_2$, the net force upwards overcomes the maximum static friction. The block will break equilibrium and start accelerating up the plane. At that point, static friction transitions into kinetic friction, which has a constant magnitude ($\mu_k N$) and would remain a flat horizontal line on the graph for any $P > P_2$.

Question

A block of mass $m$ is on an inclined plane of angle $	heta$. The coefficient of friction between the block and the plane is $\mu$ and $	an 	heta > \mu$. The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_1 = mg(\sin 	heta - \mu \cos 	heta)$ to $P_2 = mg(\sin 	heta + \mu \cos 	heta)$, the frictional force $f$ versus $P$ graph will look like (2010)

Sigma Percentile · Accepted Answer

### The Physics of Self-Adjusting Friction

Friction is often misunderstood as a simple, constant force that always opposes motion. However, static friction is much more intelligent than that; it is a **self-adjusting force**. It only exerts exactly as much force as is necessary to maintain equilibrium, up to a certain limit. This problem is a beautiful exploration of how static friction adapts its magnitude and even flips its direction based on the external forces applied to a system.

Imagine you are standing on a rough inclined plane, trying to hold a heavy block of mass $m$ stationary. The plane is inclined at an angle $	heta$. Gravity is relentless; it pulls the block straight down with a force $mg$. However, because the block is constrained to the surface of the plane, we must resolve this gravitational force into two perpendicular components.

First, there is the component perpendicular to the plane, $mg \cos	heta$. This force presses the block into the surface. The surface responds by pushing back with an equal and opposite Normal reaction force, $N = mg \cos	heta$.

Second, there is the component parallel to the plane, $mg \sin	heta$. This is the driving force that desperately wants to slide the block down the ramp.

### The Master Equation of Equilibrium

To prevent the block from sliding, you apply a force $P$ parallel to the plane, pointing upwards. The problem explicitly dictates a crucial sign convention: **the direction pointing up the plane is positive**.

Since the block is held stationary, it is in a state of static equilibrium. According to Newton's First Law, the net force acting along the plane must be exactly zero. Let's sum the forces:

$$P + f - mg \sin	heta = 0$$

Here, $f$ represents the static frictional force. By rearranging this equation, we can express friction as a function of our applied force $P$:

$$f = mg \sin	heta - P$$

Take a moment to appreciate this equation. It is a linear equation of the form $y = mx + c$. The variable is $P$, the y-intercept is $mg \sin	heta$, and the slope is $-1$. This immediately tells us that the graph of $f$ versus $P$ must be a straight line sloping downwards.

### Testing the Boundaries

Let's analyze how friction behaves at the extreme limits of our applied force.

**Case 1: The Lower Limit ($P_1$)**
Suppose you are tired and apply the absolute minimum force required to just keep the block from sliding down. The problem defines this minimum force as $P_1 = mg(\sin	heta - \mu \cos	heta)$. Let's substitute this into our friction equation:

$$f = mg \sin	heta - [mg(\sin	heta - \mu \cos	heta)]$$

Expanding the brackets, the $mg \sin	heta$ terms cancel out perfectly, leaving:

$$f = +\mu mg \cos	heta$$

Because the result is positive, it means friction is acting in the positive direction (up the plane). It is working together with your weak force $P_1$ to fight against gravity. This is the maximum static friction acting upwards.

**Case 2: The Zero Friction Point**
What if you apply a force exactly equal to the downward pull of gravity? That is, $P = mg \sin	heta$.

$$f = mg \sin	heta - (mg \sin	heta) = 0$$

In this scenario, you are perfectly balancing the block yourself. The block has no tendency to slide either up or down. Consequently, the "smart" static friction realizes it isn't needed and drops to zero. This corresponds to the x-intercept on our graph.

**Case 3: The Upper Limit ($P_2$)**
Now, suppose you push the block very hard up the incline, applying the maximum possible force before the block actually starts moving upwards. This force is $P_2 = mg(\sin	heta + \mu \cos	heta)$. Substituting this into our equation:

$$f = mg \sin	heta - [mg(\sin	heta + \mu \cos	heta)]$$

Again, the $mg \sin	heta$ terms cancel, leaving:

$$f = -\mu mg \cos	heta$$

Notice the negative sign! Because you are pushing so hard up the plane, the block now has a tendency to slide *up*. To oppose this impending upward motion, friction flips its direction and acts *down* the plane (the negative direction). This is the maximum static friction acting downwards.

### The Final Verdict

By tracking the behavior of friction, we see a fascinating journey. As the applied force $P$ increases from $P_1$ to $P_2$, the frictional force $f$ starts at a positive maximum ($+\mu mg \cos	heta$), decreases linearly to zero, and continues decreasing until it reaches a negative maximum ($-\mu mg \cos	heta$).

This linear, downward-sloping relationship perfectly matches the visual representation in **Graph (a)**. The beauty of this problem lies in trusting the math and the sign convention to reveal the physical reality of a self-adjusting force.

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Visualized Solution

$The Physical Setup$

$Resolving Gravity$

$Normal Reaction$

$Sign Convention$

$The Equilibrium Equation$

$Isolating Friction$

$Analyzing the Lower Limit (P_{1})$

$Friction at P_{1}$

$The Zero Friction Point$

$Analyzing the Upper Limit (P_{2})$

$Friction at P_{2}$

$Conclusion: The Graph Shape$

The Sigma Insight: Static and Kinetic Friction

The Physics of Self-Adjusting Friction

The Master Equation of Equilibrium

Testing the Boundaries

The Final Verdict

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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 block of mass $m$ rests on a horizontal floor with which it has a coefficient of static friction $μ$ . It is desired to make the body move by applying the minimum possible force $F$ . Find the magnitude of $F$ and the direction in which it has to be applied.

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Visualized Solution

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Lower Limit (P1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Friction at P1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Upper Limit (P2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Friction at P2​

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