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Unformatted text preview: Niu (qn269) – HW08 – Tsoi – (58020) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A weight (with a mass of 28 kg) is suspended from a point on a uniform boom whose mass is unknown. To support this boom, a cable (with a tension of 338 N) runs from this same point on the boom to a point on the wall (with vertical coordinate at a height of 10 m) and a pivot is on its lower end (at the origin of the coordinate axes). 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 VerticalHeight(m) Horizontal Distance (m) M b 3 3 8 N 28 kg Figure: Drawn to scale. Fine the weight of the boom. The accelera tion of gravity is 9 . 8 m / s 2 . Your answer must be within ± 3.0% Correct answer: 176 . 225 N. Explanation: Let : m = 28 kg , T = 338 N , and g = 9 . 8 m / s 2 . 1 2 3 4 5 6 7 8 9 1021 1 2 3 4 5 6 7 8 9 10 VerticalHeight(m) Horizontal Distance (m) ( x, y ) W b M b α (0 , h ) (0 , y ) W w T x T T y θ From the figure, for a boom of length ℓ , ( x, y ) = (8 . 5 m , 6 . 5 m) , h = 10 m , ℓ x = 9 m , x b = ℓ x 2 = 4 . 5 m , and tan θ = h y x . The cable makes an angle θ with the hori zontal, so θ = arctan parenleftbigg h y x parenrightbigg = arctan parenleftbigg 10 m 6 . 5 m 8 . 5 m parenrightbigg = 22 . 3801 ◦ , T y = T sin θ = (338 N) sin22 . 3801 ◦ = 128 . 693 N , T x = T cos θ = (338 N) cos 22 . 3801 ◦ = 312 . 541 N . Using the pivot at the origin as the fulcrum, and applying rotational equilibrium, summationdisplay τ cw = summationdisplay τ ccw W b x b + W w x = T y x + T x y Niu (qn269) – HW08 – Tsoi – (58020) 2 W b = T y x + T x y mg x x b = (128 . 693 N)(8...
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 Spring '09
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 Physics, Mass

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