Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The declaration has a mass of 5kg. The ropes, fastened at different heights, make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.

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 passive 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.

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:

$T_{2}=23.1i+19.4j$