The **question aims** to find the tension in two ropes having mass. In physics, **tension** is defined as the** gravitational force transmitted axially** through a rope, cord, chain, or similar object, or at the end of a rod, truss member, or similar object with three sides; **Tension can also be defined** as **two action-responsive forces acting** on each of the lots of the said element. **Tension** may be the opposite of compression.

At the **atomic level**, when atoms or atoms are separated from one another and receive potentially renewable energy, reciprocal power may create what is also called **tension.**

The **intensity of tension** (such as a transfer force, a dual-action force, or a retrieval force) is measured by **newtons in the International System of Units** (or pound-force in Imperial units). The ends of a bulletproof unit or another object transmitter will exert a force on the wires or rods, which direct the cord to the place of attachment. This force due to the situation’s tension is also called p**assive force**. There are **two basic possibilities** for a system of objects having strings: either the **acceleration is zero**, and the system is equal, or **there is acceleration**, so **total power is present in the system.**

**Expert Answer**

There are **two important things in this question.** The** first is that the length of the rope** is not important in finding tension vectors. Secondly that the **weight of the decoration** is $5kg$. That means a force (in Newtons) $5 \times 9.8 = 49N$ in the negative $j$ direction (straight down). $T_{1}$ is the **tension on the left rope**, and $T_{2}$ is the **tension on the right rope.**

\[T_{1}=-|T_{1}|\cos(52i)+|T_{1}|\sin(52j)\]

\[T_{2}=|T_{2}|\cos(40i)+|T_{2}|\sin(40j)\]

\[\omega=-49j\]

**Since the decoration is not moving,**

\[T_{1}+T_{2}+\omega=0\]

\[=-|T_{1}|\cos(52i)+|T_{1}|\sin(52j)+|T_{2}|\cos(40i)+|T_{2}|\sin(40j)+-49j\]

\[=(-T_{1}\cos(52)+T_{2}\cos(40))i+(T_{1}\sin(52)+T_{2}\sin(40)-49)j\]

**Solve the system of equations**

\[-T_{1}\cos(52)+T_{2}\cos(40)=0\]

\[T_{1}\sin(52)+T_{2}\sin(40)-49=0\]

**Solve equation** for |T_{2}|

\[|T_{2}|=\dfrac{|T_{1}|\cos(52)}{\cos(40)}\]

**Solve equation** for |T_{1}|

\[|T_{1}|=\dfrac{49}{\sin(52)+\cos(52)\tan(40)}\]

\[T_{1}=37.6\]

For $T_{2}$

\[|T_{2}|=\dfrac{|T_{1}|\cos(52)}{\cos(40)}=30.2\]

**Therefore,**

\[T_{1}=-23.1i+29.6j\]

\[T_{2}=23.1i+19.4j\]

**Numerical Result**

**Tension in each wire** is calculated as:

**Tension $T_{1}$, is given as:**

\[T_{1}=-23.1i+29.6j\]

**Tension $T_{2}$, is given as**:

\[T_{2}=23.1i+19.4j\]

**Example**

**3m and 5m long ropes are tied to a holiday decoration hung in the city square. The decoration weighs 5kg. Ropes are tied at different heights, from 52 and 40 degrees horizontally. Find the tension of each wire and the magnitude of each tension.**

**Solution**

There are **two important things here.** The** first is that the length of the rope** is not important in finding tension vectors. Secondly that the **weight of the decoration** is $10kg$. That means a force (in Newtons) $5 \times 9.8 = 49N$ in the negative $j$ direction (straight down). $T_{1}$ is the **tension on the left rope** and $T_{2}$ is the **tension on the right rope.**

\[T_{1}=-|T_{1}|\cos(42i)+|T_{1}|\sin(42j)\]

\[T_{2}=|T_{2}|\cos(30i)+|T_{2}|\sin(30j)\]

\[\omega=-49j\]

**Since the decoration is not moving,**

\[T_{1}+T_{2}+\omega=0\]

\[=-|T_{1}|\cos(42i)+|T_{1}|\sin(42j)+|T_{2}|\cos(30i)+|T_{2}|\sin(30j)+-49j\]

\[=(-T_{1}\cos(42)+T_{2}\cos(30))i+(T_{1}\sin(42)+T_{2}\sin(30)-49)j\]

**Solve the system of equations**

\[-T_{1}\cos(42)+T_{2}\cos(30)=0\]

\[T_{1}\sin(42)+T_{2}\sin(30)-49=0\]

**Solve equation** for |T_{2}|

\[|T_{2}|=\dfrac{|T_{1}|\cos(42)}{\cos(30)}\]

**Solve equation** for |T_{1}|

\[|T_{1}|=\dfrac{49}{\sin(42)+\cos(42)\tan(30)}\]

\[T_{1}=37.6\]

For $T_{2}$

\[|T_{2}|=\dfrac{|T_{1}|\cos(42)}{\cos(30)}=30.2\]

**Therefore,**

\[T_{1}=-23.1i+29.6j\]

\[T_{2}=23.1i+19.4j\]

**Tension in each wire** is calculated as

**Tension $T_{1}$, is given as:**

\[T_{1}=-23.1i+29.6j\]

**Tension $T_{2}$, is given as**:

\[T_{2}=23.1i+19.4j\]