A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the ho

Q: Why does the horizontal force have a component *up* the incline?

Imagine pushing a block horizontally against a wedge. Because the wedge slopes downwards, your horizontal push is partially directed *into* the surface and partially *up* along the surface. Geometrically, the angle between the horizontal force and the incline is $\theta$, so its component along the incline is $F \cos\theta$ directed upwards.

Q: Why don't we need to calculate the normal reaction or the maximum static friction?

The question asks about the *direction* of the frictional force, not its exact magnitude or whether the block will actually slip. As long as the block 'remains stationary' (as stated in the problem), static friction will simply adjust itself to perfectly cancel out the net driving force along the plane.

Q: What happens exactly at $\theta = 45^\circ$?

At $45^\circ$, $\sin(45^\circ)$ equals $\cos(45^\circ)$. Since both the weight and the applied force are $1 \text{ N}$, their components along the plane ($1 \cdot \sin 45^\circ$ and $1 \cdot \cos 45^\circ$) are perfectly equal and opposite. They cancel each other out completely, meaning zero friction is required to hold the block in place!

Q: If $\theta > 45^\circ$, why does friction act towards Q?

When the angle is steeper than $45^\circ$, the sine of the angle is greater than the cosine. This means gravity's pull down the slope ($mg \sin\theta$) is stronger than the horizontal force's push up the slope ($F \cos\theta$). The block wants to slide down towards P, so friction must act upwards towards Q to stop it.

Q: Is it possible for the block to slide up the plane?

Yes! If the incline is very shallow ($\theta < 45^\circ$), the cosine component is larger than the sine component. The horizontal push up the slope overpowers gravity's pull down the slope. If the static friction isn't strong enough, the block would actually accelerate up the incline towards Q!

Question

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

Options:

(A) $	heta = 45^\circ$.

(B) $	heta > 45^\circ$ and a frictional force acts on the block towards $P$

(C) $	heta > 45^\circ$ and a frictional force acts on the block towards $Q$

(D) $	heta < 45^\circ$ and a frictional force acts on the block towards $Q$

Sigma Percentile · Accepted Answer

### The Setup: A Delicate Balance Imagine a block resting on an inclined plane, caught in a tug-of-war between two fundamental forces. On one hand, we have gravity, relentlessly pulling the block downwards with a weight of $1 ext{ N}$. On the other hand, an external horizontal force, also exactly $1 ext{ N}$, is pushing the block directly to the left, into the slope. The question asks us to determine the direction of the static frictional force required to keep this block perfectly stationary at various angles of inclination, $ heta$. To solve this, we must look beyond the obvious vertical and horizontal directions and align our perspective with the inclined plane itself. ### Resolving the Forces The secret to mastering inclined plane problems is to resolve all forces into components that are parallel and perpendicular to the surface of the plane. The perpendicular components will dictate the normal reaction, but it is the parallel components that determine whether the block wants to slide up or down. First, let's analyze the weight, $W = 1 ext{ N}$. Gravity acts straight down, but the incline forces the block to move along a slanted path. The component of the weight acting parallel to the incline, pulling the block downwards towards the bottom point $P$, is given by $W \sin heta$. Since the weight is $1 ext{ N}$, this downward driving force is simply $\sin heta$. Next, we examine the horizontal applied force, $F = 1 ext{ N}$. Because the plane is slanted, pushing horizontally to the left actually helps push the block *up* the incline! The angle between this horizontal force and the inclined plane is exactly $ heta$. Therefore, the component of this force acting parallel to the incline, pushing the block upwards towards the top point $Q$, is $F \cos heta$. Since $F$ is $1 ext{ N}$, this upward driving force is simply $\cos heta$. ### The Battle for Equilibrium For the block to remain stationary, the net force acting along the incline must be zero. We have a downward pull of $\sin heta$ and an upward push of $\cos heta$. Static friction is the ultimate peacekeeper. It is a smart force that only acts when necessary, and it always points in the exact opposite direction of the net driving force to prevent slipping. Let's see how this plays out at different angles. ### The Sweet Spot: $ heta = 45^\circ$ Let's test the specific angle of $45^\circ$. At this angle, a beautiful mathematical symmetry occurs. We know that $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\cos(45^\circ) = \frac{1}{\sqrt{2}}$. This means the downward pull of gravity ($\frac{1}{\sqrt{2}} ext{ N}$) is perfectly matched by the upward push of the horizontal force ($\frac{1}{\sqrt{2}} ext{ N}$). The forces balance each other out completely! Because there is no net driving force, static friction doesn't need to intervene at all. The block remains stationary without the help of friction. This confirms that option (a) is correct. ### The Steep Slope: $ heta > 45^\circ$ What happens if we make the incline steeper, meaning $ heta > 45^\circ$? As the angle increases, the sine function grows larger while the cosine function shrinks. Therefore, $\sin heta > \cos heta$. The downward pull of gravity now overpowers the upward push of the horizontal force. The block has a strong tendency to slide down the plane towards point $P$. To prevent this impending motion, static friction must step in and act *up* the incline, towards point $Q$. This confirms that option (c) is correct. ### The Shallow Slope: $ heta < 45^\circ$ Finally, consider a shallower incline where $ heta < 45^\circ$. In this regime, the cosine function is larger than the sine function. This means $\cos heta > \sin heta$. The horizontal force is now pushing the block so effectively that its upward component overpowers gravity's downward pull. The block actually wants to slide *up* the incline towards point $Q$! To keep the block stationary, static friction must reverse its direction and act *down* the incline, towards point $P$. Option (d) incorrectly states that friction acts towards $Q$ in this scenario, so it is false. By carefully resolving the forces and comparing their magnitudes at different angles, we have completely decoded the mechanics of this system.

Select Answer:

Visualized Solution

The Physical Setup

Component of Weight

Component of Applied Force

Net Force and Friction

Case 1: $θ = 4 5^{\circ}$

Case 2: $θ > 4 5^{\circ}$

Case 3: $θ < 4 5^{\circ}$

Final Conclusion

The Sigma Insight: Static and Kinetic Friction

The Setup

Resolving the Forces

The Battle for Equilibrium

The Sweet Spot: $θ = 4 5^{\circ}$

The Steep Slope: $θ > 4 5^{\circ}$

The Shallow Slope: $θ < 4 5^{\circ}$

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