This question aims to find the value of the **smallest angle** theta can make with a rope **without breaking** it by using laws of motion.

Consider a **box of sweets** weighing down the **rope** when people from across the buildings are sending this box. People from one building are sending this box of sweets to the people in the opposite building through a rope. When this box of sweets comes in the **center of the rope**, it makes an** angle** theta with the original position of the rope.

The position of this box of sweets in the center is not determined exactly. Both ends of the rope make an angle theta with the **original position** of the rope. We need to find the** smallest angle** among the two angles by applying **Newton’s second law of motion**.

## Expert Answer

According to Newton’s second law of motion, any **force** acting on the body of **mass m** is equal to the **rate of change** of its velocity.

Applying Newton’s second law of motion:

\[ F = m a \]

Here, gravity is acting on the box of sweets so the **acceleration** will be equal to **gravitational pull**:

\[ F = m g \]

The force is acting along its **vertical component** so it will be written as:

\[ F _ y = 0 \]

\[ {\Sigma} F _ y = 0 \]

\[ 2 T sin \theta – m g = 0 \]

**Tension** in the rope is represented by **T**. It is a force acting on the rope when it is stretched.

\[ 2 T sin \theta = m g \]

To find an angle $ \theta $, we will rearrange the equation:

\[ sin \theta = \frac { m g } { 2 T } \]

Consider the mass of a box is **2 kg** and it produces a tension of **30 N** on the rope then the angle is:

\[ sin \theta = \frac { 2 \times 9 . 8 } { 2 \times 30 } \]

\[ sin \theta = \frac { 19 . 6 } { 60 } \]

\[ sin \theta = 0 . 3 2 6 \]

\[ \theta = sin ^ {-1} ( 0 . 3 2 6 ) \]

\[ \theta = 19 . 0 2 ° \]

## Numerical Solution

**The smallest angle acting on the rope without breaking it is 19.02°**.

## Example

Consider a person in the **circus** doing a **stunt** with the rope by hanging it. Both sides of this **flexible rope** are attached to the opposite cliffs. The mass of the person is **45 kg** and the tension produced in the rope is **4200 N**.

The smallest angle can be found by:

\[ {\Sigma} F _ y = 0 \]

\[ 2 T sin \theta – m g = 0 \]

Tension in the rope is represented by T. It is a force acting on the rope when it is stretched.

\[ 2 T sin \theta = m g \]

To find an angle $ \theta $, we will rearrange the equation:

\[ sin \theta = \frac { m g } { 2 T } \]

\[ sin \theta = \frac { 45 \times 9 . 8 } { 2 \times 4200 } \]

\[ sin \theta = \frac { 441 } { 8400 } \]

\[ sin \theta = 0 . 0 5 2 5 \]

\[ \theta = sin ^ {-1} ( 0 . 0 5 2 5 ) \]

\[ \theta = 3.00 ° \]

*Image/Mathematical drawings are created in Geogebra**.*