This question aims to develop a **practical understanding of Newton’s laws of motion**. It uses the concepts of **tension in a string**, the **weight of a body**, and the **centripetal/centrifugal force**.

Any force acting along a string is called the **tension in the string**. It is denoted by **T**. The **weight of a body** with mass **m** is given by the following formula:

**w = mg**

Where **g = 9.8 m/s^2** is the **gravitational acceleration**. The **centripetal force** is the force acting toward the center of a circle whenever **a body is moving in the circular path**. It is mathematically given by the following formula:

\[ F = \dfrac{ m v^2 }{ r } \]

Where $ v $ is the **speed of the body** while $ r $ is the **radius of the circle** in which the body is moving.

## Expert Answer

During the **part of motion** where the **speed of the van is uniform** (constant), the block is **hanging vertically downward**. In this case, the **weight** $ w \ = \ m g $ is acting **vertically downward**. According to **Newton’s third law** of motion, there is an equal and opposite **tension force** $ T \ = \ w \ = m g $ must be acting **vertically upward** to balance the force exerted by the weight. We can say that the **system is in equilibrium** under such circumstances.

During the **part of motion** where the **van is moving along a circular path** of radius $ r \ = \ 175 \ m $ with a speed of $ v \ = \ 24 \ m/s $, this equilibrium is disturbed and the **block has moved horizontally** towards the outer edge of the curve due to the **centrifugal force** acting in the horizontal direction.

In this case, the **weight** $ w \ = \ m g $ acting downward is **balanced by** the **vertical component of tension force** $ T cos( \theta ) \ = \ w \ = m g $ and the **centrifugal force** $ F \ = \ \dfrac{ m v^{ 2 } }{ r } $ is **balanced by** the horizontal component **horizontal component of tension force** $ T sin( \theta ) \ = \ F \ = \ \dfrac{ m v^{ 2 } }{ r } $.

So we have **two equations**:

\[ T cos( \theta ) \ = \ m g \ … \ … \ … \ ( 1 ) \]

\[ T sin( \theta ) \ = \ \dfrac{ m v^{ 2 } }{ r } \ … \ … \ … \ ( 2 ) \]

**Dividing** equation (1) by equation (2):

\[ \dfrac{ T sin( \theta ) }{ T cos( \theta ) } \ = \ \dfrac{ \dfrac{ m v^{ 2 } }{ r } }{ m g } \]

\[ \Rightarrow \dfrac{ sin( \theta ) }{ cos( \theta ) } \ = \ \dfrac{ v^{ 2 } }{ g r } \]

\[ \Rightarrow tan( \theta ) \ = \ \dfrac{ v^{ 2 } }{ g r } \ … \ … \ … \ ( 3 ) \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } \bigg ( \dfrac{ v^{ 2 } }{ g r } \bigg ) \]

**Substituting numerical values:**

\[ \theta \ = \ tan^{ -1 } \bigg ( \dfrac{ ( 24 \ m/s )^{ 2 } }{ ( 9.8 \ m/s^2 ) ( 175 \ m ) } \bigg ) \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } ( 0.336 ) \]

\[ \Rightarrow \theta \ = \ 18.55^{ \circ } \]

## Numerical Result

\[ \theta \ = \ 18.55^{ \circ } \]

## Example

Find the angle theta in the **same scenario** given above if the **speed was 12 m/s**.

Recall **equation no. (3):**

\[ tan( \theta ) \ = \ \dfrac{ v^{ 2 } }{ g r } \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } \bigg ( \dfrac{ ( 12 \ m/s )^{ 2 } }{ ( 9.8 \ m/s^2 ) ( 175 \ m ) } \bigg ) \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } ( 0.084 ) \]

\[ \Rightarrow \theta \ = \ 4.8^{ \circ } \]