Given The system shown below supports a 225 lb weight Rope O

Given: The system shown below supports a 225- lb weight. Rope OD lies in the x- y plane. Find: Determine the tension force developed in ropes OA, OC, and OD if the spring in OB is stretched by 2 inches due to the weight supported at O. Solution:

Solution

Force in OB due to deflection of 2 in = k * 2 = 40 lb

Basically the forces can be represented by the vectors as follows:

OA is (3,-4,2); OB is (-2,-3,3); OC is (-4,4,4)

The unit vector for OA can be obtained by dividing 3,-4,2 each by [3^2 + (-4)^2 + 2^2]^(0.5) = 29

So the unit vector OA can be represented by (3/29i - 4/29j + 2/29k)

Likewise OB is (-2/22i,- 3/22j, 3/22k); OC is (-4/48i, 4/48j, 448k); OD is (sin 30, cos 30)

Now force balance in z direction gives

225 = 40*3/22 + OC*4/48 + OA*2/29

Force balance in y direction gives

40*3/22 + OA*4/29 = OC*4/48 + ODcos30

Force balance in x direction gives

40*2/22 + OC*4/48 = OA*3/29 + ODsin30

So we have 3 equations and 3 unknowns. Solving them we get the tension in OA 201.6 lb; OC 215.7 lb; OD 58.6 lb