Question

Given that the statistical data for falls occurring in constructoás that approximately 50% ofall workers that fall from height to a lower surface at or ahove eleven (1l) feet are killed, let's see what kind of forces (F) n pounds is ganerated at impact. We will also need to calaulate the velocity in feet second and miles hour of the falling body For purposes of this exercise, we will use a body weight of 225 pounds Two criteria for deceleration will be: aj allowing the lanyard deceleration device to deploy after six (6) feet of fall and decelerate the fall in 14 inches and b) striking a concrete surface affer ten (10) feet of fall where a deceleration distance of three (3) inches resulted from the compression of the bedy. Answer the following questions using our statics or physics expertise related to principles for bodies in motion. Use back of sheet if more space is needed. What is the maximum velocity of the falling body from six (6) feet ft sec mph What is the maximum velocity of the falling body at impact from tem(10)feet?2 c) tsec mph ) Given the deceleration distance of 14 inches from a six (6) foor fall whar is the force generated to the nearest pound in the body harness eg straps thus imto the body? (3) Givan the deceleration distance of three (3) inches from a ten (10) foot fall, what is the force generated to the nearest pound into the body itself when striking the coocrete surface without a lanyard and body hamess? (3)

Answer 1

Given that the purpose of this exercise we will use a body weight 225 weight pounds g = 32.15 ft/s^2 free fall v^2 = u^2 + 2as v^2 = 0 + 2 times 32.15 times 6 v^2 = 385.8 therefore v = 19.64 ft/s v = 19.64 times 0.682 therefore v = 13.39 mgh \r\n maximum value of the fallowing body from 6 v = 19.64 fts rightarrow (a) v = 13.39 mgh rightarrow (b) what is the maximum velocity of the falling body of impact from the feet v = [u^2 + 2as]^1/2 = [0 + 2 times 32.15 times 10]^1/2 implies 25.36 ft/s = 25.36 times 0.682 mgh given the deceleration distance of 14 inches from six foot ball for generated = rate of change of maximum = m[v - u]/t = m[v - u]/t = m[v/t] = ma = m[u^2/2h] therefore v^2 = u^2 - 2as theta^2 = u^2 - 2as f = 225 times [19.64^2/2 times [14/12]] therefore f = 37.1919lb