Question

A crate is hung by ropes attached to a steel ring. Assume that the center of gravity (CG) of the crate shown lies at its geometric center, and that ring A lies directly above the crate's CG. The top surface of the crate is horizontal, and ring A lies 39 inches above it. Other dimensions are given as:

a = 40 in
b = 33 in
c = 7 in

A load cell measures the tension in rope AC to be TACTAC = 157 lb. Find the weight of the crate and the tension in the other two ropes.

2013 Michael Swanbom C030 BY NC SA S1 ZA g

WcrateWcrate =

TABTAB =

TADTAD =

Answer 1

Solution:- (0,0,0) = TPA + F . (0,0,-39 W Z Joroba , h) =(-23, 20, -39) Werate Location og point B is -, - , -1) = (-3, -20, -39) location of point c R (+ ,-(-6), - h) = (+32, -3, -39) Now force in rope BA is give by, in vector form. FBO = TBA = Tension in rope BA B À = unit vector in the direction of PBA IBA force FEA. FRA = F / 20-1-3)} +{0-6-2013 + {0-(-39)} Å ) IBAT FA - FOA | 3 1 +27 +39 4 20² + 392 1

For = Fon (0.35232 î +0.427057 À + 0,8327613 ) Similarly Fen - For ( a ) FCA = FCA ( (0-3)i + (0-(-3)) } + (0-(-39)) i ) (1 = 44.29729 Fon = Fcm (-0.372483 î + 0.2934716 +0.8804158) DA Foa ( (--)2 + ( -2); + -(139)) DA I DÀ for ( 13 46. 85214 15 -20 +398) 46.83214 From = FPM (0.352322 (0.352322 î - 0.427057 À + 018327613 ) Ã å because direction of Warate is from A to G or downword. RO + + Co-or Waste = W ( e permanece en el momento Wait = w / 0-3@ +60+358 +(3-0)8) = w (09+of-sť) 399 (

Now for equilibrium of whole chate. 55x=o: Fyzo ; F = o Matoutony = point & for Ex=0 For (0-35232) + FOA (0.352322) - FAX0:372483 +Wx0 =0 0.352322 FRA + 0.352322 FPA -0.372483 FeA = → 0.352722 FRA + 0.352322 FOA = 0.372483 *157 [lbgiven EFCA = 157|| FOA + FOA = 165.984 lb - → for & Fy=0 0.427057 FBA - 0.427057 FPA + D.2934716 FCA = 0 → FBA - FOA = – 107. 88967 Solving equation & and & we get. FRA = 29.047 lb FA = 136.9368 lb for EF =0 0.8327613 Fe + 0.8327613 FPA +0.880415 FCA -W=o [W=276.45 lb

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