A loudspeaker of mass 19.0 kg is suspended a distance of \( h=2.00 \mathrm{~m} \) below the ceiling by two cables that make equal angles with the celing. Each cable has a length of \( l \) \( =3.30 \mathrm{~m} \). (Figure 1) \( \begin{array}{l}\text { Use } 9.80 \mathrm{~m} / \mathrm{s}^{2} \text { for the magnitude of the free-fall acceleration. }\end{array} \)

We start by drawing a free‐body diagram for the loudspeaker. The speaker of mass m = 19.0 kg hangs at rest, so its weight mg (with g = 9.80 m/s²) is balanced by the vertical components of the tensions in the two cables. Since the two cables are identical and symmetrically arranged, each cable has the same tension T. Step 1. Determine the angle between a cable and the ceiling. Each cable has length l = 3.30 m and is attached to the speaker which is a vertical distance h = 2.00 m below the ceiling. This vertical drop is the side opposite the angle between the cable and the ceiling. Therefore, if we call this angle θ (the angle the cable makes with the horizontal, i.e. with the ceiling), then   sin θ = (vertical drop)/(length of cable) = h/l = 2.00/3.30 ≈ 0.6061 Thus,   θ = arcsin(0.6061) ≈ 37.4° Step 2. Write the vertical force balance. Each cable’s tension T has a vertical component T_vertical = T sin θ (since the cable makes an angle θ with the horizontal). With two cables, the upward force is   2T sin θ This must balance the weight mg:   2T sin θ = mg Step 3. Solve for the tension T. Rearrange the equation:   T = mg/(2 sin θ) First, calculate the weight:   mg = 19.0 kg × 9.80 m/s² = 186.2 N Now substitute:   T = 186.2 N / (2 × 0.6061) ≈ 186.2 N / 1.2122 ≈ 153.6 N Thus, the tension in each cable is approximately 154 N. Summary: • Angle with the ceiling: θ ≈ 37.4° • Tension in each cable: T ≈ 1.54×10² N This completes the solution.