HW9-phy2048-Spring-2014-solution

Therefore t3 t 2 t 2 57 n x y d the angle

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Unformatted text preview: leads to T2 = T1 sin 35° = (49N)sin 35° = 28 N. (c) We denote the components of T3 as Tx (rightward) and Ty (upward). Analyzing horizontal forces where string 2 and string 3 meet, we find Tx = T2 = 28 N. From the vertical forces there, we conclude Ty = wB = 50 N. Therefore, T3 = T 2 + T 2 = 57 N. x y (d) The angle of string 3 (measured from vertical) is ⎛T ⎞ ⎛ 28 ⎞ θ = tan −1 ⎜ x ⎟ = tan −1 ⎜ ⎟ = 29°. ⎜T ⎟ ⎝ 50 ⎠ ⎝ y⎠ 1 PHY2048 Spring 2014 HW9 12P21. (a) We note that the angle between the cable and the strut is α =θ – φ = 45º – 30º = 15º. The angle between the strut and any vertical force (like the weights in the problem) is β = 90º – 45º = 45º. Denoting M = 225 kg and m = 45.0 kg,...
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This document was uploaded on 03/24/2014.

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