Determine the tension developed in cables OD and OB and the strut OC, required to support the 500-lb crate. The spring OA has an unstretched length of 0.2 ft and a stiffness of kOA

Fob= 86.2 lb, Foc= 175 lb, Fod= 506 lb

Explanation:firstly we need to get the spring stretch by subtracting the unstretched length from 1. The result will be (1-0.2=0.8ft). We then get the spring force by multiplying kOA by 0.8ft. therefore the spring force will be {350x0.8}= 280 lb. After that we can resolve each force as shown in the attachment.

i think if it were me i whould divide 300 dide 15

step-by-step explanation: