Determine the Reactions at O and the Cable Tensions

1. Oct 6, 2012

Northbysouth

1. The problem statement, all variables and given/known data

The light right-angle boom which supports the 520-kg cylinder is supported by three cables and a ball-and-socket joint at O attached to the vertical x-y surface. Determine the reactions at O and the cable tensions.

2. Relevant equations

3. The attempt at a solution
I'm a little unsure if what I should do with this question. When I spoke with my TA, he said to find the unit vectors of the cables, multiply those unit vectors by their appropriate tensions (which are unknown) and then to take the cross product of these with the vector from O to a point of action on the line.

0 = rOAXTAC$\hat{}n$AC + rOBXTBD$\hat{}n$BD + rOBXTBE$\hat{}n$BE - 520*9.8

My calculations are as follows:

$\hat{}n$AC = <-1.25, 1, -2.60>/$\sqrt{}9.3225$

$\hat{}n$BD = <0, 2, -2.60>/$\sqrt{}10.76$

$\hat{}n$BE = <0, 0, -2.60>/$\sqrt{}2.60$

TAC$\hat{}n$AC = TAC<-1.25/$\sqrt{}9.3225$, 1/$\sqrt{}9.3225$, -2.60/$\sqrt{}9.3225$>

TBD$\hat{}n$BD = TBD<0, 2$\sqrt{}10.76$, -2.60/$\sqrt{}10.76$>

TBE$\hat{}n$BE = TBE<0, 0, -2.60/2.60>

rOAXTAC$\hat{}n$AC =

<0, 0, 2.60> X <-1.25TAC/$\sqrt{}9.3225$, TAC$\sqrt{}9.3225$, -2.60TAC/$\sqrt{}9.3225$

rOAXTAC$\hat{}n$AC = <-2.60TAC/$\sqrt{}9.3225$, -3.25TAC/$\sqrt{}9.3225$, 0>

rOBXTBD$\hat{}n$BD = <1.80, 0, 2.60> X<0, 2TBD/$\sqrt{}10.76$, -2.60TBD/$\sqrt{}10.76$>

= <-5.2TBD/$\sqrt{}10.76$, 4.68TBD/$\sqrt{}10.76$, 3.60TBD/$\sqrt{}10.76$>

rOBXTBE$\hat{}n$BE = <1.80, 0, 2.60> X <0, 0, -TBE>

= <0, 1.80TBE, 0>

ROF = <0.90, 0, 2.60>

ROF X <0, -5096, 0> = < 4586.4, 0, 0>

MOx = -2.60TA/$\sqrt{}9.3225$ - 5.2TBD/$\sqrt{}10.76$ + 4586.4

MOy = -3.25TAC/$\sqrt{}9.3225$ + 4.68TBD/$\sqrt{}10.76$ + 1.80TBE

MOz = 3.60TBD/$\sqrt{}10.76$

Thus I get:

TAC = 5385.98
TBD = 0
TBE = 3185

But this doesn't make any sense. TBD can't be zero.

Help would be greatly appreciated. Thanks

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