A 27-kg chandelier hangs from a ceiling on a vertical 4.0-m-long wire. (a) What horizontal force would be necessary to displace its position 0.15 m to one side? (b) What will be the tension in the wire?
3 Answers
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arctan (0.15/4) = 2.15°
Weight = 27*9.8 = 264.6 N
Horizontal force is 246.4 sin(2.15°) = 9.24 N
Tension is sqrt(264.6² + 9.24²) = 264.8 N
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Chandelier M
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Fy = Tcos(theta) = mg Fx = Tsin(theta) 3.58sin(theta) = 0.1m => theta = a million.60 ranges a. What horizontal stress could be mandatory to displace its place 0.10 m to one ingredient? Fx/Fy = (Tsin(theta))/(Tcos(theta)) = tan(theta) Fy = mg => Fx = mgtan(theta) = (30.7)(9.80 one)tan(a million.60) Fx = 8.4N b. what could be the stress in the cord? T = Fx/sin(theta) = 300N (2 s.f)
