A 19.0Kg loudspeaker is suspended 1.00m below the ceiling by two 2.50m long cables that angle outward at equal angles. What is the tension on each rope?
2006-09-30 10:06:47 · 2 answers · asked by joshuaewell 1 in Science & Mathematics ➔ Physics
I assume the anchor point for the cables is 1.00 m below the ceiling, or at the top of the speaker. For angle from the horizontal,
theta = arcsin(1/2.5) = 23.6 deg
Each cable carries half the load, or 9.5 kg. The vertical component of the cable tension offsets this downward load. Since gravitational acceleration is 9.8 m/s2:
tension = 9.5 * 9.8 / sin(theta) = 233 N
2006-09-30 10:45:21 · answer #1 · answered by SAN 5 · 0⤊ 0⤋
Total downward pull in newtons equals 9.8*19=186.2 newtons; but since both ropes share this the downward tension in each rope is only 93.1newtons. The angle of the rope is arcsine(1/2.5)=23.57 degrees. The actual tension in the rope is 93.1/sin(23.570=232.82 newtons.
2006-09-30 18:23:44 · answer #2 · answered by bruinfan 7 · 0⤊ 0⤋