HW_13-solutions

# M 1 m 2 l 1 l 2 θ m joseph jrj2569 hw 13 hoffmann

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M 1 M 2 L 1 L 2 θ m

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joseph (jrj2569) – HW 13 – hoffmann – (58485) 3 For equilibrium, summationdisplay vector F = 0 and summationdisplay vector τ = 0 . The sum of the torques about the pivot is T L 2 sin θ - m g L 2 2 - M 1 g L 1 - M 2 g L 2 = 0 T = m g 2 sin θ + M 1 g L 1 L 2 sin θ + M 2 g sin θ = (24 kg)(9 . 8 m / s 2 ) 2 sin 21 + (55 kg) (9 . 8 m / s 2 ) (3 . 1 m) (4 . 6 m) sin 21 + (57 kg)(9 . 8 m / s 2 ) sin 21 = 2900 . 48 N = 2 . 90048 kN . 006 10.0points The bars are in equilibrium, each 7 m long, and each weighing 98 N. The string pulling down on the two bars is attached 0 . 9 m from the fulcrum on the leftmost bar and 0 . 8 m from the left end of the rightmost bar. The spring (of constant 20 N / cm) is attached at an angle 36 at the left end of the upper bar. 7 m 0 . 9 m 7 m 0 . 8 m 20 N / cm 36 7 kg If the suspended mass is 7 kg, by how much will the spring stretch? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 95 . 5614 cm. Explanation: Let : W = 98 N , m = 7 kg , = 7 m , a = 0 . 9 m , b = 0 . 8 m , k = 20 N / cm , and g = 9 . 8 m / s 2 . a b k θ m Find the tension T 1 in the string connecting the levers by applying rotational equilibrium on the leftmost bar. The suspended mass weighs m g and causes a tension of 2 m g on the string attached to the end of the leftmost bar. m g m g 2 m g 2 m g W T 1 a From rotational equilibrium on the leftmost lever, the tension 2 m g acts down at a dis- tance - a from the fulcrum, the tension T 1 acts down at a distance a from the fulcrum, and the weight W of the bar acts down at a distance 2 - a from the fulcrum, so T 1 a - 2 m g ( - a ) - W parenleftbigg 2 - a parenrightbigg = 0 2 T 1 a - 4 m g ( - a ) - W ( - 2 a ) = 0
joseph (jrj2569) – HW 13 – hoffmann – (58485) 4 T 1 = 4 m g ( - a ) + W ( - 2 a ) 2 a = 2 (7 kg) (9 . 8 m / s 2 ) (7 m - 0 . 9 m) 0 . 9 m + (98 N) [7 m - 2 (0 . 9 m)] 2 (0 . 9 m) = 1213 . 02 N . W T 1 b k Δ x θ From rotational equilibrium on the right- most lever, the weight W acts down at a distance 2 from the fulcrum, the tension T 1 acts down at a perpendicular distance - b from the fulcrum, and the tension k Δ x acts up at a perpendicular distance sin θ from the fulcrum, so k Δ x ( sin θ ) - T 1 ( - b ) - W parenleftbigg 2 parenrightbigg = 0 2 k Δ x ℓ sin θ - 2 T 1 ( - b ) - W = 0 Δ x = 2 T 1 ( - b ) + W 2 k ℓ sin θ = (1213 . 02 N) (7 m - 0 . 8 m) (20 N / cm) (7 m) sin 36 + (98 N) (7 m) 2 (20 N / cm) (7 m) sin 36 = 95 . 5614 cm . 007 10.0points Consider two ladders that are identical in all aspects except for the difference described be- low. The lengths of the sides AO = BO. A mass of weight W is attached to the mid-point P(AP = PO). The masses of both ladders are negligible compared to W , and both ladders rest on a frictionless horizontal floor. Their differences are: For ladder (1): the junction at O is rigid. For ladder (2): the junction at O is loose,and there is a connecting rope DC to secure the spread. W P rope D C P' F 2 A O B W P P' F 1 A O B B 4 Which expressions are correct?

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