A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is ap

Q: Why does friction act upwards in this case?

Friction always opposes the *tendency* of relative motion. Since we are applying the *minimum* force $F$ to just barely stop the block from sliding down, its tendency is to move downwards. Therefore, static friction acts upwards to help $F$ hold the block in place.

Q: What if the question asked for the maximum force $F$?

If we apply a very large force $F$, the block will tend to slide *up* the incline. In that case, friction would switch direction and act *downwards* to oppose the upward tendency. The equation would become $F = 3 + Mg \sin\theta + \mu Mg \cos\theta$.

Q: Why do we use $Mg \cos\theta$ for the normal reaction?

The block is not moving perpendicular to the inclined plane, meaning the forces in that direction are balanced. The only forces acting perpendicular to the plane are the normal reaction $R$ (outwards) and the component of gravity $Mg \cos\theta$ (inwards). Thus, $R = Mg \cos\theta$.

Q: Why is the $3\text{ N}$ force added to $Mg \sin\theta$?

Both the $3\text{ N}$ force and the gravitational component $Mg \sin\theta$ are pointing downwards along the incline. Since they act in the same direction, their effects add up to pull the block down.

Q: How do we know to use static friction instead of kinetic friction?

The problem asks for the force required such that the block 'does not move'. This implies the block is at rest (in equilibrium), so we use the coefficient of static friction to find the limiting friction force.

Question

A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force F, such that the block does not move downward ? (Take, $g = 10 	ext{ ms}^{-2}$)

Options:

(A) 32 N

(B) 25 N

(C) 23 N

(D) 18 N

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a steep hill, trying to hold a heavy box from sliding down. This is exactly the scenario we are dealing with here. We have a $10 	ext{ kg}$ block resting on a rough inclined plane angled at $45^\circ$.

There is a $3 	ext{ N}$ force actively pulling the block down the slope, and gravity is also doing its part to drag it down. Our mission is to find the **minimum upward force $F$** required to keep the block perfectly stationary.

Because we are looking for the *minimum* force, the block is on the absolute verge of slipping downwards. This is a crucial detail because it tells us the direction of friction. Since friction always opposes the tendency of motion, and the block "wants" to slide down, **static friction will act upwards** along the incline to help our force $F$.

### Resolving the Forces

To solve this, we need to break down the forces into components that are parallel and perpendicular to the inclined plane.

First, let's look at gravity. The weight of the block, $Mg$, acts straight down. We resolve this into two components:
1. A perpendicular component, $Mg \cos	heta$, which presses the block into the surface.
2. A parallel component, $Mg \sin	heta$, which pulls the block down the incline.

Because the block isn't jumping off the plane or sinking into it, the forces perpendicular to the surface must balance. This means the **normal reaction force $R$** provided by the plane is exactly equal to the perpendicular component of gravity:
$$R = Mg \cos	heta$$

### The Master Equation

Now, let's look at the forces acting parallel to the incline. For the block to remain at rest, the total upward forces must perfectly balance the total downward forces.

The forces pulling the block down are the applied $3 	ext{ N}$ force and the parallel component of gravity, $Mg \sin	heta$. The forces pushing the block up are our unknown force $F$ and the static friction $f$.

Setting up our equilibrium equation, we get:
$$F + f = 3 + Mg \sin	heta$$

We know that the maximum static friction $f$ is given by $\mu R$. Substituting $R = Mg \cos	heta$, we get:
$$f = \mu Mg \cos	heta$$

Plugging this back into our master equation, we can isolate $F$:
$$F + \mu Mg \cos	heta = 3 + Mg \sin	heta$$
$$F = 3 + Mg \sin	heta - \mu Mg \cos	heta$$

### Final Calculation

Now comes the satisfying part—plugging in the numbers! We are given $M = 10 	ext{ kg}$, $g = 10 	ext{ m/s}^2$, $	heta = 45^\circ$, and $\mu = 0.6$.

Let's substitute these into our equation:
$$F = 3 + (10)(10) \sin 45^\circ - (0.6)(10)(10) \cos 45^\circ$$

Remember that both $\sin 45^\circ$ and $\cos 45^\circ$ are equal to $\frac{1}{\sqrt{2}}$. 
$$F = 3 + 100 \left(\frac{1}{\sqrt{2}}ight) - 60 \left(\frac{1}{\sqrt{2}}ight)$$

Simplifying the terms, we get:
$$F = 3 + \frac{100 - 60}{\sqrt{2}}$$
$$F = 3 + \frac{40}{\sqrt{2}}$$

To make this easier to calculate, we can rationalize the denominator:
$$F = 3 + 20\sqrt{2}$$

Since $\sqrt{2} \approx 1.414$, we can find the final numerical value:
$$F = 3 + 20(1.414)$$
$$F = 3 + 28.28 = 31.28 	ext{ N}$$

Looking at our options, the closest value is **$32 	ext{ N}$**.

### The Way Forward

This problem beautifully illustrates how friction acts as a "smart" force, adjusting its direction based on the tendency of motion.

A great thought experiment: What if the question asked for the **maximum** force $F$ before the block starts moving *up* the incline? In that scenario, the block's tendency of motion would be upwards, meaning friction would flip and act *downwards*. The equation would then become $F = 3 + Mg \sin	heta + \mu Mg \cos	heta$. Always pay close attention to which way the object "wants" to move!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Free Body Diagram$

$Resolving Gravity$

$Normal Reaction$

$Direction of Friction$

$Balancing Forces Along Incline$

$Substituting Friction$

$Plugging in Values$

$Simplifying the Expression$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

Resolving the Forces

The Master Equation

Final Calculation

The Way Forward

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Select Answer:

Visualized Solution

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

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

.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; } Direction of Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balancing Forces Along Incline

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

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

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

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

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