Homework Answers
3.46
FOC= FOC(6i+4j+12k/sqrt(6^2+4^2+12^2))=(3/7)FOCi+(2/7)FOCj+(6/7)FOCk
FOA=-FOAj
FOB=-FOBi
F=-196.2K
SumF=0
Equating I,j,k components
(3/7)FOC-FOB=0
(2/7)FOC-FOA=0
(6/7)FOC-196.2=0
gives,
FOC=228.9N
FOB= 98.1N
FOA=65.4N
By Hooke’s law F=kx
x= F/k
xOB=98.1/300=0.327m
xOA=65.4/300=0.218m
Add Answer to:
3-46. Determine the stretch in each of the two springs required to hold the 20-kg crate...
Each spring has an unstretched length of 2 m and a stiffness of k = 250...
Each spring has an unstretched length of 2 m and a stiffness of
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A.) Determine the stretch in OA spring required to hold
the 20-kg crate in the equilibrium position shown.
B.) Determine the stretch in OB spring required to hold
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Each spring has an unstretched length of 2 m and a stiffness of
k = 250 N/m
A.) Determine the stretch in OA spring required to hold
the 20-kg crate in the equilibrium position shown.
B.) Determine the stretch in OB spring required to hold
the 20-kg crate in the equilibrium position shown.
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Determine the stretch in OA and OB spring required to
hold the 19-kg
Each spring has an unstretched length of 2 m and a stiffness of k = 320 N/m.
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very much urgent !
(REQUIRED) Determine the unstretched length of D8 to hold the 40-kg crate in the position shown. Take k 1. 180 N/m. 2 im
Determine the stretch in each spring for for equilibrium of the
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A)Determine the stretch in spring AD
B)Determine the stretch in spring AC.
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in the figure. Calculate in this condition the unstretched length
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