Question

a loudspeaker of mass 21.0 kg is suspended a distance of h = 2.00 m below the ceiling by two cables that make equal angles with the ceiling. each cable has a length of l = 4.00 m. (figure 1) part a what is the tension t in each of the cables? use 9.80 m/s² for the magnitude of the free - fall acceleration. view available hint(s)

Explanation:

Step1: Calculate the vertical - component of tension

The mass of the loudspeaker is $m = 21.0\ kg$, and the acceleration due to gravity $g=9.80\ m/s^{2}$. The weight of the loudspeaker is $W = mg$. Since the loudspeaker is in equilibrium, the sum of the vertical components of the tension in the two cables equals the weight of the loudspeaker. Let the tension in each cable be $T$. The vertical - component of the tension in each cable is $T_y=T\sin\theta$. The sum of the vertical components of the two cables is $2T_y = 2T\sin\theta$, and this must equal the weight $W$ of the loudspeaker, so $2T\sin\theta=mg$.

First, we need to find the angle $\theta$ that the cable makes with the vertical. We know the length of the cable $l = 4.00\ m$ and the vertical distance $h = 2.00\ m$. Using the right - triangle relationship $\sin\theta=\frac{\sqrt{l^{2}-h^{2}}}{l}$.
\[

$$\begin{align*} \sin\theta&=\frac{\sqrt{4.00^{2}-2.00^{2}}}{4.00}\\ &=\frac{\sqrt{16 - 4}}{4.00}\\ &=\frac{\sqrt{12}}{4.00}\\ &=\frac{2\sqrt{3}}{4.00}=\frac{\sqrt{3}}{2} \end{align*}$$

\]

Step2: Solve for the tension $T$

From $2T\sin\theta=mg$, we can solve for $T$. We know that $m = 21.0\ kg$ and $g = 9.80\ m/s^{2}$ and $\sin\theta=\frac{\sqrt{3}}{2}$.
\[

$$\begin{align*} T&=\frac{mg}{2\sin\theta}\\ &=\frac{21.0\times9.80}{2\times\frac{\sqrt{3}}{2}}\\ &=\frac{21.0\times9.80}{\sqrt{3}}\\ &\approx\frac{205.8}{\sqrt{3}}\\ &\approx119\ N \end{align*}$$

\]

Answer:

$T\approx119\ N$