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Unformatted text preview: karna (pk4534) – HW 13 – coker – (58245) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This is the last chapter (14) to be covered on Quiz 3, to be given on the evening of April 15. Quiz 3 covers Chs. 10, 11, 12, 13 and 14. 001 (part 1 of 3) 10.0 points A floodlight with a mass of 48.0 kg is used to illuminate the parking lot in front of a library. The floodlight is supported at the end of a horizontal beam that is hinged to a vertical pole, as shown. A cable that makes an angle of 19 . ◦ with the beam is attached to the pole to help support the floodlight. Assume the mass of the beam is negligible when compared with the mass of the floodlight. The acceleration of gravity is 9 . 81 m / s 2 . 48 kg 19 ◦ Note: Figure is not drawn to scale a) Find the force F T provided by the cable. Correct answer: 1446 . 33 N. Explanation: Basic Concept: The second (rotational) condition of equi librium (axis of rotation at the pole and length of beam ℓ ): F T (sin θ ) ℓ mgℓ = 0 Given: m = 48 . 0 kg θ = 19 . ◦ g = 9 . 81 m / s 2 Solution: F T (sin θ ) = mg F T = mg sin θ = (48 kg)(9 . 81 m / s 2 ) sin 19 ◦ = 1446 . 33 N 002 (part 2 of 3) 10.0 points b) Find the horizontal force exerted on the beam by the pole. Correct answer: 1367 . 53 N. Explanation: Basic Concept: The first (translational) condition of equi librium: Horizontally, F x = R x F T (cos θ ) = 0 Solution: R x = F T (cos θ ) = (1446 . 33 N)(cos19 ◦ ) = 1367 . 53 N 003 (part 3 of 3) 10.0 points c) Find the vertical force exerted on the beam by the pole. Correct answer: 0 N. Explanation: Basic Concept: The first (translational) condition of equi librium: Vertically, F y = R y + F T (sin θ ) mg = 0 Solution: R y = mg F T (sin θ ) = mg mg = 0 N 004 (part 1 of 2) 10.0 points The uniform diving board shown in figure has a mass of 28 kg . The acceleration of gravity is 9 . 81 m / s 2 . karna (pk4534) – HW 13 – coker – (58245) 2 A B 1 . 2 m 3 . 6 m Find the force on the support A when a 74 kg diver stands at the end of the diving board. Correct answer: 2 . 4525 kN. Explanation: Let : m = 28 kg , M = 74 kg , ℓ = 4 . 8 m , x 1 = 1 . 2 m , and x d = 3 . 6 m . Consider rotational equilibrium about B : A F A F B B mg M g x 1 x 2 x 3 x d x 2 = ℓ 2 x 1 = 1 . 2 m Applying rotational equilibrium at point B , summationdisplay vector τ = F A x 1 mg x 2 M g x d = 0 F A = parenleftbigg x 2 m + x d M x 1 parenrightbigg g = bracketleftbigg (1 . 2 m) (28 kg) + (3 . 6 m) (74 kg) 1 . 2 m bracketrightbigg × (9 . 81 m / s 2 ) × 1 kN 1000 N = 2 . 4525 kN . The direction of F A is downward, so it is tension....
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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