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 weight of the object

The weight of the object is given by $W = mg$, where $m = 210\ kg$ and $g=9.80\ m/s^{2}$. So $W=210\times9.80 = 2058\ N$.

Step2: Analyze the vertical - component of the tension

Since the object is in equilibrium and is suspended by two cables making equal angles with the ceiling, the sum of the vertical - components of the tension in the two cables equals the weight of the object. Let the tension in each cable be $T$. The vertical - component of the tension in each cable is $T_y=T\sin\theta$. In the vertical direction, $2T_y = W$, or $2T\sin\theta=W$. First, we need to find the angle $\theta$. If the horizontal distance between the suspension points is $l = 4.00\ m$ and the vertical distance from the ceiling to the object is $h = 2.00\ m$, then $\tan\theta=\frac{l/2}{h}=\frac{4.00/2}{2.00}=1$, so $\theta = 45^{\circ}$ and $\sin\theta=\cos\theta=\frac{\sqrt{2}}{2}$.
From $2T\sin\theta=W$, we can solve for $T$. Substituting $\sin\theta=\frac{\sqrt{2}}{2}$ and $W = 2058\ N$ into the equation $2T\sin\theta=W$, we get $2T\times\frac{\sqrt{2}}{2}=2058\ N$, which simplifies to $T\sqrt{2}=2058\ N$.

Step3: Solve for the tension $T$

$T=\frac{2058}{\sqrt{2}}\ N\approx1455\ N$.

Answer:

$T\approx1455\ N$