91Ó°ÊÓ

The pulling force transferred axially by a string, cable, chain, or similar item, or by each end of a rod, truss member, or similar three-dimensional object is known as tension.

Mass of a loudspeaker = $20$kg, Distance of the loudspeaker from the ceiling = $2.0$m and the length of each cable = $3.0$m

The arrangement's free body diagram is seen below.

From the figure, formula to calculate the angle is,

$\cos \mathrm{θ}=\frac{AB}{BC}$

$\mathrm{θ}$= the cable's angle with the vertical.

$AB$= loudspeaker height from the ceiling.

$BC$= rope's length.

Substitute $2.0$m for $AB$and $3.0$m for $BC$to find $\mathrm{θ}$.

$$\begin{gathered} \cos \mathrm{θ}=\frac{2.0m}{3.0m} \\ \mathrm{θ}={\cos }^{-1}\left(0.66\right) \\ =48.19° \end{gathered}$$

Mass of a loudspeaker = $20$kg, Height of the loudspeaker from the ceiling = $2.0$m and the length of each cable = $3.0$m.

For the vertical equilibrium requirement, Newton's second law is used,

$\underset{}{∑}{F}_y=0$

In the y-direction, resolving the components of forces in the ropes,

$-mg+2T\cos \mathrm{θ}=0$

T = Tension in the cable.

$\mathrm{θ}$= Angle made by the cable with the vertical.

$m$= mass of the loudspeaker.

$g$= acceleration due to gravity.

Substitute $20$kg for $m$, localid="1647702114359" $9.81m/s^2$for $g$and $48.19°$for $\mathrm{θ}$to find T.

$$\begin{gathered} -(20kg)(9.81m/s^2)+2T\cos 48.19°=0 \\ T=\frac{196.2kg.m/s^2}{2\cos 48.19°} \\ =147.15N \end{gathered}$$

Hence, the tension in the cable =$147.15N$.