A car is moving on a plane inclined at 30 degrees to the horizontal with an acceleration of 10 ms^-2

Q: Why do we use a pseudo force here?

Because we are solving the problem from the perspective of the accelerating car. The car is a non-inertial frame of reference. To make Newton's laws valid in such a frame, we must introduce a pseudo force.

Q: Can we solve this from the ground frame?

Yes, absolutely! In the ground frame, there is no pseudo force. Instead, you would say that the net force on the bob (Tension + Gravity) provides the necessary acceleration $a$ up the incline. The math will lead to the exact same equations.

Q: Why is the pseudo force directed down the incline?

A pseudo force always acts in the exact opposite direction of the frame's acceleration. Since the car accelerates up the incline, the pseudo force pushes the bob down the incline.

Q: Why did we resolve forces horizontally and vertically instead of along and perpendicular to the string?

Resolving horizontally and vertically aligns perfectly with gravity, which is strictly vertical. While you can resolve along the string, it makes the algebra slightly more tedious because both gravity and the pseudo force would need to be resolved into sine and cosine components.

Q: What happens if the car moves with constant velocity?

If the velocity is constant, the acceleration $a$ is zero. The pseudo force vanishes, and the bob simply hangs straight down, meaning $\alpha = 0^\circ$.

Question

A car is moving on a plane inclined at $30^{\circ}$ to the horizontal with an acceleration of $10	ext{ ms}^{-2}$ parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is .........
(Take, $g = 10	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### Introduction to Non-Inertial Frames

Imagine you are sitting inside an accelerating car. Because the car is accelerating, it acts as a **non-inertial frame of reference**. According to Newton's laws, to analyze the motion of any object inside this frame (like our suspended bob), we must introduce a fictitious force known as a **pseudo force**.

This pseudo force always acts in the exact opposite direction of the frame's acceleration. Since the car is accelerating *up* the $30^\circ$ incline with an acceleration $a = 10 	ext{ m/s}^2$, the pseudo force $F_p = ma$ will push the bob *down* the incline.

### Setting up the Free Body Diagram

Let's draw the free body diagram of the bob from the perspective of an observer inside the car. The bob is in equilibrium relative to the car, meaning the net force acting on it must be zero. There are three primary forces at play here:

1. **Tension ($T$)**: Pulling upwards along the string at an angle $\alpha$ with the vertical.
2. **Gravity ($mg$)**: Pulling straight down towards the center of the Earth.
3. **Pseudo Force ($ma$)**: Pushing down the incline at an angle of $30^\circ$ below the horizontal.

### The Art of Resolving Forces

To establish equilibrium, it is mathematically convenient to resolve all forces into standard horizontal ($x$) and vertical ($y$) components.

For the tension $T$, the components are:
- Horizontal: $T_x = T \sin\alpha$ (acting to the right)
- Vertical: $T_y = T \cos\alpha$ (acting upwards)

For the pseudo force $ma$, which acts down the $30^\circ$ incline, the components are:
- Horizontal: $F_{px} = ma \cos 30^\circ$ (acting to the left)
- Vertical: $F_{py} = ma \sin 30^\circ$ (acting downwards)

### Establishing Equilibrium

Since the bob is stationary inside the car, we apply Newton's First Law in both directions.

**Horizontal Equilibrium ($\Sigma F_x = 0$):**
The rightward force must perfectly balance the leftward force.

$$ T \sin\alpha = ma \cos 30^\circ $$

**Vertical Equilibrium ($\Sigma F_y = 0$):**
The upward force must balance the total downward forces (gravity plus the vertical component of the pseudo force).

$$ T \cos\alpha = mg + ma \sin 30^\circ $$

### The Final Mathematical Stroke

We now have a system of two equations. Our goal is to find the angle $\alpha$. The most elegant way to eliminate the unknown tension $T$ is to divide the horizontal equation by the vertical equation.

$$ \frac{T \sin\alpha}{T \cos\alpha} = \frac{ma \cos 30^\circ}{mg + ma \sin 30^\circ} $$

Notice how the mass $m$ beautifully cancels out from every term on the right side, leaving us with a purely kinematic relationship:

$$ 	an\alpha = \frac{a \cos 30^\circ}{g + a \sin 30^\circ} $$

Now, we substitute the given values: $a = 10 	ext{ m/s}^2$, $g = 10 	ext{ m/s}^2$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, and $\sin 30^\circ = \frac{1}{2}$.

$$ 	an\alpha = \frac{10 \left(\frac{\sqrt{3}}{2}ight)}{10 + 10 \left(\frac{1}{2}ight)} $$

$$ 	an\alpha = \frac{5\sqrt{3}}{10 + 5} $$

$$ 	an\alpha = \frac{5\sqrt{3}}{15} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} $$

Since $	an\alpha = \frac{1}{\sqrt{3}}$, we can conclude that the string makes an angle of **$30^\circ$** with the vertical.

Enter Numerical Value:

Visualized Solution

$Non-Inertial Frame$

$Free Body Diagram$

$Resolving Forces$

$Horizontal Equilibrium$

$Vertical Equilibrium$

$Eliminating Tension$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Pseudo Force

Introduction to Non-Inertial Frames

Setting up the Free Body Diagram

The Art of Resolving Forces

Establishing Equilibrium

The Final Mathematical Stroke

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Comprehension Passage

Comprehension Passage

The distance $r$ of the block at time $t$ is :

The net reaction of the disc on the block is :

Enter Numerical Value:

Visualized Solution

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

.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 Forces

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

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

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

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

.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

The Sigma Insight: Pseudo Force

Introduction to Non-Inertial Frames

Setting up the Free Body Diagram

The Art of Resolving Forces

Establishing Equilibrium

The Final Mathematical Stroke

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an acceleration a to keep the block stationary. Then, a is equal to

Comprehension Passage

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance r of the block at time t is :

The net reaction of the disc on the block is :

$Non-Inertial Frame$

$Free Body Diagram$

$Resolving Forces$

$Horizontal Equilibrium$

$Vertical Equilibrium$

$Eliminating Tension$

$Substituting Values$

$Final Calculation$

$The Way Forward$

A block is kept on a frictionless inclined surface with angle of inclination $α$ . The incline is given an acceleration $a$ to keep the block stationary. Then, $a$ is equal to

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