Question

2-P A tugboat pulls on cable AB with a force of 50 kN. Determine the forces developed in cables BC and BD necessary for equilibrium. (Note: All three cables lie in the same plane.) 址 -30° 20° 50 kN
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Answer #1

BCsin30 FBC FBCcos30 FAB 30 deg 20 deg FBDcos20 FBD VFBDsin20

Tension in the cables are acting at angles, Split them into their components

Consider all vertical forces

0 = 1

FBcsin30 – Fopsin 20 = 0

FBcsin30 = Fopsin 20

FBC =) Fbdsin 20 sin 30

FBC = 0.684FBD------------(1)

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Consider all horizontal forces

FAB – FBCCOs30 – FBpcos20 = 0

FAB = FBCCOs30 + FBpcos20

Put (1) in the above equation

FAB = 0.684FBD * cos30 + FBDcos20

FAB= (0.684 * cos30+ cos20) FBD

50kN = 1.532 * FBD

50kN = FBD 1.532

ANSWER: FBD = 32.637kN

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Put value of F_{BD} in (1)

FBC = 0.684 * 32.637kN

ANSWER: FBC = 22.324kN

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