Equilibrium-solutions

# Equilibrium-solutions - le (txl65) Equilibrium mohanty...

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le (txl65) – Equilibrium – mohanty – (81440) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A letter A is Formed From two uniForm pieces oF metal each oF weight 22 . 7 N and length 1 . 6 m, hinged at the top and held together by a horizontal wire oF length 1 . 18 m. The structure rests on a Frictionless surFace. d l 1 l 2 IF the wire is connected at points a dis- tance oF 0 . 743 m From the top oF the letter, determine the tension in the wire. Correct answer: 31 . 9314 N. Explanation: Let : W = 22 . 7 N , 1 = 1 . 6 m , 2 = 1 . 18 m , and d = 0 . 743 m . N A N B A α O B 2 w ±or each bar cos α = 2 2 d = 2 2 d α = arccos p 2 2 d P = cos 1 b 1 . 18 m 2 (0 . 743 m) B = 37 . 4317 . Choose the whole Frame as the object. Tak- ing torques about point A, its leFt contact with the ground, s v τ = (2 W ) 1 cos α - N B (2 1 cos α ) = 0 N B = W = 22 . 7 N . l /2 O d sin α T l 2 _ cos α α w 1 ±or the right bar, apply torques about its top end: T d sin α + W p 1 2 P cos α - W 1 cos α = 0 T = W 1 cos α 2 d sin α = W 1 2 d tan α = (22 . 7 N) (1 . 6 m) 2 (0 . 743 m) tan37 . 4317 = 31 . 9314 N . 002 10.0 points Three hollow metal rods, oF equal length and negligible mass, are connected by pins to Form an triangular Frame as shown. The Frame stands upright, and a weight is hung From the vertex oF the triangle Frame. 3 . 6 m 260 N What is the tension in the horizontal rod which Forms the base oF the triangular Frame? Correct answer: 75 . 0555 N. Explanation: Let : θ = 60 , = 3 . 6 m , and W = 260 N .

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le (txl65) – Equilibrium – mohanty – (81440) 2 Let F be the force exerted by right-hand strut, W 2 the normal force at the right-hand corner (since the object is symmetric the force on each corner will be equal), and T the ten- sion in the base. Consider the forces on the bottom right-hand corner of the triangle. F T W 2 θ The forces in the horizontal and vertical direction, respectively, are F sin θ - 1 2 W = 0 and F cos θ - T = 0 . Dividing, F sin θ F cos θ = 1 2 W T = W 2 T T = W 2 tan θ = 260 N 2 tan 60 = 75 . 0555 N . 003 10.0 points Two ropes support a load of 667 kg. The two ropes are perpendicular to each other, and the tension in the Frst rope is 2 times that of the second rope. ±ind the tension in the second rope. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 2923 . 26 N. Explanation: Let : m = 667 kg , f = 2 , and g = 9 . 8 m / s 2 .
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## This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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Equilibrium-solutions - le (txl65) Equilibrium mohanty...

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