Asked by jjh
A crazed armadillo with a mass of 60kg stands at the center of a rope which was initially strung horizontally between two poles 50m apart. Its weight causes the rope to sag 1.20m. What is the tension in the rope
Solution:
2Tsin(angle) - mg = 0
Tan-1() = 1.2/25 = 2.75°
2Tsin(2.75°) - mg = 0
T = mg/2Sin(2.75°)
T = 6200 N
My questions is regarding the first part of the solution and how would I know I have to set 2Tsin(angle) - mg equal to 0.
Please help me understand this problem.
Solution:
2Tsin(angle) - mg = 0
Tan-1() = 1.2/25 = 2.75°
2Tsin(2.75°) - mg = 0
T = mg/2Sin(2.75°)
T = 6200 N
My questions is regarding the first part of the solution and how would I know I have to set 2Tsin(angle) - mg equal to 0.
Please help me understand this problem.
Answers
Answered by
Damon
the rope is evidently mass less.
total force up = total force down because nothing is accelerating.
Total force down = m g
Tension in rope is T on each side of mass
horizontal component of T = T cos angle above horizontal right and left
vertical component of T = T sin angle, twice because from both sides of mass
so
total up force = 2 T sin angle
so 2 T sin angle = m g
to get angle simple geometry
tan angle = 1.2/(.5*50)
total force up = total force down because nothing is accelerating.
Total force down = m g
Tension in rope is T on each side of mass
horizontal component of T = T cos angle above horizontal right and left
vertical component of T = T sin angle, twice because from both sides of mass
so
total up force = 2 T sin angle
so 2 T sin angle = m g
to get angle simple geometry
tan angle = 1.2/(.5*50)
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