This question aims to find the **tension in the string** when a **body of mass** with **weight $w$** is suspended from it. Figure 1 shows the two formations of suspension.

The question is based on the concept of **tension. Tension** can be defined by the **force** exerted by the **string or cord** when a body of the **weight** is **suspended** by it. Simple **trigonometric ratios** of a right-angle triangle and basic **triangle geometry** are also needed to solve this question. Let’s suppose a body of weight **$W$** is attached to a string, and the other end of the string is attached to a fixed point. The **tension $T$** in the string is given as :

\[ T = W \]

Here, the weight of the body will be downward, and tension in the string will be in the upward direction.

## Expert Answer

a) In the first part of the question, we can see that the **$T_1$** makes an angle of **$30^{\circ}$** and **$T_2$** makes an angle of **$45^{\circ}$.** As the weight and cord are **balanced,** the **tension in left cord** must be **equal** to **tension in the right cord**. This can be written as:

\[ T_1 \cos (30^{\circ}) = T_2 \cos(45^{\circ}) \hspace{0.4in} (1) \]

According to the definition of the tension, the **forces** pointing **upwards** are equal to the **forces** pointing **downwards.** This means that the **tension** in both cords pointing **upward** is equal to the **weight** **of the object** pointing **downward.** The equation can be written as:

\[ T_1 \cos (60^{\circ}) + T_2 \cos (45^{\circ}) = W \]

Calculated in equation $(1)$, the **tension** in the** right cord** is equal to the **tension** in the **left cord**. We can replace the value $T_2$ with $T_1$.

\[ T_1 \cos (60^{\circ}) + T_1 \cos (30^{\circ}) = W \]

\[ T_1 = \dfrac{2W}{1 + \sqrt{3}} \]

Putting the value of **$T_1$** in equation $(1)$ to find the tension in the cord on the right side:

\[ (\dfrac{2W}{1 + \sqrt{3}}) \cos(30^{\circ}) = T_2 \cos(45^{\circ}) \]

Solving for $T_2$, we get:

\[ T_2 = \dfrac{\sqrt{6} W}{1 + \sqrt{3}} \]

b) In the second part of the question, the **cord** on the **left side** also has **tension** pointing **downwards,** same as the **weight**. We can write this equation in this manner:

\[ T_1 \cos(60^{\circ}) + W = T_2 \cos(45^{\circ}) \]

Here, the tension on the right side will be equal to the horizontal component of the cord on the left side.

\[ T_1 \cos(30^{\circ}) = T_2 \cos(45^{\circ}) \hspace{0.4in} (2) \]

Substituting this value of **$T_1$** in the above equation to find its value, we get:

\[ T_1 \cos(60^{\circ}) + W = T_1 \cos(30^{\circ}) \]

\[ T_1 = \dfrac{2 W}{1 – \sqrt{3}} \]

Substituting this value in equation $(2)$ to get the value of $T_2$:

\[ (\dfrac{2W}{1 – \sqrt{3}}) \cos(30^{\circ}) = T_2 \cos(45^{\circ}) \]

Solving for **$T_2$**, we get:

\[ T_2 = \dfrac{\sqrt{6}W}{1 – \sqrt{3}} \]

## Numerical Results

a) The **tension in the cords** in the first part of the question are given as:

\[ [T_1, T_2] = \Bigg{[}\dfrac{2W}{1 + \sqrt{3}}, \dfrac{\sqrt{6}W}{1 + \sqrt{3}}\Bigg{]} \]

b) The **tension in the cords** in the second part of the question are given as:

\[ [T_1, T_2] = \Bigg{[}\dfrac{2W}{1 – \sqrt{3}}, \dfrac{\sqrt{6}W}{1 – \sqrt{3}}\Bigg{]} \]

## Example

Find the **weight of the body** if it is suspended with two strings with **tension** amounting to **$5N$** and **$10N$**.

According to the definition of **tension,** the **weight** is equal to the **tension** in the **cords.** We can write this problem as:

\[ T_1 + T_2 = W \]

Substituting the values, we get:

\[ W = 5N + 10N \]

\[ W = 15N \]

The **weight of the body** suspended by the cords is **$15N$**.