The minimum force required to start pushing a body up a rough (frictional coefficient mu) inclined p

Q: Why does the direction of friction change between the two cases?

Friction is a reactive force that always opposes the *impending* relative motion between two surfaces. When you push the block up, it wants to slide up, so friction acts down. When you are just preventing it from sliding down, it wants to slide down, so friction acts up to help you hold it.

Q: Why do we use the maximum static friction formula $f = \mu N$ here?

The problem uses the words 'minimum force required to start pushing' and 'minimum force needed to prevent'. These phrases imply a state of 'limiting equilibrium', where the block is on the absolute verge of moving. At this exact threshold, static friction reaches its maximum possible value, which is $\mu N$.

Q: What would happen if $\tan\theta$ was less than $\mu$?

If $\tan\theta < \mu$, it means the incline is not steep enough for the block to slide down on its own (gravity's pull $mg \sin\theta$ is less than the maximum friction $\mu mg \cos\theta$). In that case, you wouldn't need any force $F_2$ to prevent it from sliding; it would just sit there!

Q: Why did we divide the numerator and denominator by $\cos\theta$?

It's a strategic mathematical manipulation. We were given a condition involving $\tan\theta$ ($\tan\theta = 2\mu$). Our ratio had $\sin\theta$ and $\cos\theta$. By dividing by $\cos\theta$, we force $\tan\theta$ to appear in the equation, allowing us to directly substitute the given condition and simplify the math.

Q: Can the force $F_2$ ever be negative?

Mathematically, if $\tan\theta < \mu$, the formula for $F_2$ yields a negative value. Physically, this just means that instead of pushing *up* the incline to hold it, you would actually have to push *down* the incline to make it slide. But for the context of 'preventing it from sliding down', $F_2$ is zero if the block doesn't slide on its own.

Question

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_1$ while the minimum force needed to prevent it from sliding down is $F_2$. If the inclined plane makes an angle $	heta$ from the horizontal such that $	an	heta = 2\mu$, then the ratio $F_1/F_2$ is

Options:

(A) 4

(B) 1

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Forces

Imagine a block resting peacefully on a rough inclined plane. This classic physics setup is a battleground of forces: gravity trying to pull the block down, the normal force pushing it away from the surface, and friction acting as the ultimate contrarian, always opposing the impending motion.

In this problem, we are asked to analyze two distinct scenarios. In the first scenario, we want to push the block *up* the incline. In the second, we just want to prevent it from sliding *down*. The beauty of this problem lies in how the direction of friction flips between these two cases, completely changing the force equations.

### Case 1: The Uphill Battle

Let's tackle the first scenario. We apply a force $F_1$ to push the block up the incline. Because the block is on the verge of moving upwards, the rough surface fights back. Static friction, $f$, reaches its maximum value and acts *downwards* along the incline, joining forces with the component of gravity that also pulls downwards, $mg \sin	heta$.

To just start moving the block, our applied force $F_1$ must overcome both of these downward forces. We can write the force balance equation as:

$$ F_1 = mg \sin	heta + f $$

We know that the maximum static friction is given by $f = \mu N$, where the normal force $N = mg \cos	heta$. Substituting this in, we get our first master equation:

$$ F_1 = mg(\sin	heta + \mu \cos	heta) $$

### Case 2: Holding the Fort

Now, let's shift gears to the second scenario. The block naturally wants to slide down due to gravity. However, we apply a minimum force $F_2$ to just hold it in place. Because the block is on the verge of sliding *downwards*, friction flips its allegiance! It now acts *upwards* along the incline, helping our force $F_2$ to balance gravity.

In this limiting equilibrium, the upward forces ($F_2$ and friction) must perfectly balance the downward pull of gravity. The equation becomes:

$$ F_2 + f = mg \sin	heta $$

Rearranging this to solve for $F_2$, and again substituting $f = \mu mg \cos	heta$, we get our second master equation:

$$ F_2 = mg(\sin	heta - \mu \cos	heta) $$

### The Grand Unification: Finding the Ratio

The problem asks for the ratio of these two forces, $F_1 / F_2$. Let's divide our two master equations. Notice how the mass $m$ and gravity $g$ beautifully cancel out, showing that this ratio is independent of the block's weight!

$$ \frac{F_1}{F_2} = \frac{mg(\sin	heta + \mu \cos	heta)}{mg(\sin	heta - \mu \cos	heta)} = \frac{\sin	heta + \mu \cos	heta}{\sin	heta - \mu \cos	heta} $$

To simplify this expression, we can use a neat mathematical trick: divide every term in the numerator and the denominator by $\cos	heta$. This transforms our sines and cosines into tangents:

$$ \frac{F_1}{F_2} = \frac{	an	heta + \mu}{	an	heta - \mu} $$

Now, we use the magic key provided in the problem statement: **$	an	heta = 2\mu$**. By substituting this relation into our simplified ratio, everything is expressed purely in terms of the friction coefficient $\mu$:

$$ \frac{F_1}{F_2} = \frac{2\mu + \mu}{2\mu - \mu} $$

$$ \frac{F_1}{F_2} = \frac{3\mu}{\mu} = 3 $$

The $\mu$ cancels out perfectly, leaving us with a clean, elegant integer. The ratio of the forces is exactly 3!

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Visualized Solution

$The Inclined Plane Setup$

$Resolving Gravity$

$Case 1: Pushing the Block Up$

$Equation for F_{1}$

$Case 2: Preventing Sliding Down$

$Equation for F_{2}$

$Ratio of Forces$

$Dividing by cos θ$

$Applying the Given Condition$

$Final Answer$

The Sigma Insight: Static and Kinetic Friction

The Setup

Case 1

Case 2

The Grand Unification

Similar Questions

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.

List-I

List-II

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

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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 Inclined Plane 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; } Case 1: Pushing the Block Up

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for F1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Preventing Sliding Down

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for F2​

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

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

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

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