Question

Given: A 90-load is suspended from the hook and supported by two cables and a spring. Find: The force in the cables and the stretch of the spring at equilibrium. k = 500 lb/ft 00000001 Plan: 90 lb 1. Draw a FBD for point A. 2. Apply the E-of-E to solve for the forces in ropes AB and AC.

Answer 1

Equation of equilibrium

Σ Fx = 0

FD sin 30° - (4/5)FC = 0 (a)

ΣFy = 0

-FD cos 30° + FB = 0 (b)

ΣFz = 0

(3/5) FC – 90 lb = 0 (c)

Solving Eq. (c) for FC, then Eq (a) for FD, and finally Eq. (b) for FB, yields,

FC = 150 lb

FD = 240 lb

FB = 208 lb

The strech of spring is therefore

FB = k X sAB

208 lb = 500 lb/ft X (sAB)

sAB = 0.416 ft
TOTAL Fz=0 3/5 FC- 90*32 = 0 because F=m*g and 90 is just the mass , not the weight

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Now Solve fources clong Touris fr (co3 36 39 ) - Fap sim 3o (66741) (0%) I E 0.5 = FAD Now fapa 1067.86N Solve forces elong - creis Fap COS90 FAB FARE (1067.86 ) C0.866) =924.8N FAC= 667.4) N 5150.0397 lbr FAD=1067.86N = 240.064 db6 FAB9248 N = 207.903 107 - Foore on cleflection of spsrima 6 = fosce om spring for SHFF moss of Sporing 6: 207.903 To 0.4158 Ft 500 deflection of sporing 5= 0.4158 Ft
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