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:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is , and the steam properties as, A-4E:
Using the energy balance for the system:
Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:
Entropy generation in the heat exchanger could be defined using the entropy balance on the system:
Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R