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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Solution :
Given weight of Kathy = 82 kg
Her speed before striking the water, = 5.50 m/s
Her speed after entering the water, = 1.1 m/s
Time = 1.65 s
Using equation of impulse,
Here, F = the force ,
dT = time interval over which the force is applied for
= 1.65 s
dP = change in momentum
dP = m x dV
= 82 x (1.1 - 5.5)
= -360 kg
∴ the net force acting will be
= 218 N