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The hook and spring assembly shown in the figure are equilibrium, calculate: F. lb k lb/ft А (1) Falb В 02 2 ft F3 lb COLLAPS

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Drawing ewing FBD of given body. Fi sind, F2 Ficos y E Fn = F + F, cosa, + Fa cosa F₂ cos &₂ +98 + 110 cos 21°+ 52 cos 35° ItChange in le length of spring - Fn Kg » K 8 Fn = 243-29 - 243-29 243-29 K 1506 &- 0.161 ftPlease like the answer.

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