Given that the statistical data for falls occurring in construction show that ap

Question

Given that the statistical data for falls occurring in construction show that approximately 50% of all workers that fall from height to a lower surface at or above eleven (11) feet are killed, let's see what kind of force (F) in pounds is generated at impact. We will also need to calculate the velocity in feet second and miles/hour of the falling body For purposes of this exercise, we will use a body weight of 175 pounds. Two criteria for deceleration will be: a) 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 after ten (10) feet of fall where a deceleration distance of three (3) inches resulted from the compression of the body. Answer the following questions using your statics or physics expertise related to principles for bodies in motion. Show all of your work for full credit!! 1) What is the maximum velocity of the falling body from six (6) feet? (2) a) ftusec b) 2) What is the maximum velocity of the falling body at impact from ten (10) feet? (2) fsec 3) Given the deceleration distance of 14 inches from a six (6) foot fall, what is the force generated to the nearest pound in the body harness leg straps thus into the body? (3) 4) Given 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 concrete surface without a lanyard and body harness? (3)

Explanation / Answer

by third equation of motion

v^2 = u^2 + 2as

initial velocity = 0 so,

v^2 = 0 + 2 * 32.174 * 6

v = 19.649 ft/sec

1) maximum velocity = 19.649 ft/sec or 13.397 mph

similarily for 10 feet fall

v^2 = 0 + 2 * 32.174 * 10

v = 25.367 ft/sec

2) maximum velocity = 25.367 ft/sec or 17.295 mph

now body stops at distance 14 inches or 1.167 ft

so,

0 = 19.649^2 - 2 * a * 1.167

a = 165.417 ft/sec^2

force = mass * acceleration

force = 175 * 165.417

3) force = 28947.975 pound . ft/s^2

now body stops at distance 3 inches or 0.25 ft

0 = 25.367^2 - 2 * a * 0.25

a = 1286.969 ft/sec^2

force = 175 * 1286.969

4) force = 225219.575 pound . ft/s^2

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