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
1 answer:
Answer: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.
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Answer:
stone A is diamond.
Explanation:
given,
Volume of the two stone = 0.15 cm³
Mass of stone A = 0.52 g
Mass of stone B = 0.42 g
Density of the diamond = 3.5 g/cm³
So, to find which stone is gold we have to calculate the density of both the stone.
We know,
density of stone A
density of stone B.
Hence, the density of the stone A is the equal to Diamond then stone A is diamond.