Question

A $20$kg loudspeaker is suspended $2.0$ m below the ceiling by two $3.0$-m-long cables that angle outward at equal angles. What is the tension in the cables?

Short answer

The tension in the cable =$147.15N$.

Step-by-step solution

Tension : 

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.

Angle made by the cable with vertical :

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,

$\cos \theta =\frac{AB}{BC}$

$\theta$= 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 $\theta$.

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

Tension in the cable :

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,

$\sum _{}{F}_y=0$

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

$-mg+2T\cos \theta =0$

T = Tension in the cable.

$\theta$= 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 $\theta$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 =$147.15N$.