Answer :
The force P on the cable if the spring is compressed 0.5 in with a stiffness of k = 800 lb/ft is 46.91 lb
k = Fe / δ
k = Stiffness of spring
Fe = Force on spring
k = 800 lb / ft = 6.67 lb / in
δ = 0.5 in
Fe = k δ
Fe = 6.67 * 0.5
Fe = 33.33 lb
At joint A,
∑ MA = 0
( - Bx * 6 ) + ( - By * 6 ) + ( 33.33 * 30 ) = 0
Bx + By = 166.67 → ( 1 )
At joint D,
∑ MD = 0
( - By * 6 ) + ( P * 4 ) = 0
By = 0.067 P → ( 2 )
∑ Fx = 0
- Bx + FCD cos 30° = 0 → ( 3 )
At joint B,
∑ MB = 0
( - FCD sin 30° * 6 ) + ( P * 10 ) = 0
FCD = 3.33 P
Sub FCD in ( 3 ),
- Bx + 3.33 P cos 30° = 0
Bx = 2.8867 P
Sub Bx and By in ( 1 ),
2.8867 P + 0.667 P = 166.67
P = 46.91 lb
Therefore, the force P on the cable if the spring is 46.91 lb
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