For purposes of this exercise, we will use a body weight of 225 pounds. Two crit

Question

For purposes of this exercise, we will use a body weight of 225 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 our statics or physics expertise related to principles for bodies in motion (equations are below). Use back of sheet if more space is needed. gh= 1/2 mv^2 = 2mgh/m=v^2 = 2gh=v^2 = v=2gh

1) What is the maximum velocity of the falling body from six (6) feet? (2)

a) _______________ ft./sec

b) _______________ mph

2) What is the maximum velocity of the falling body at impact from ten (10) feet? (2)

c) _______________ ft./sec

d) _______________ mph

3) Given the deceleration distance of 14 inches from a six (6) foot fall, what is the force generated to the nearst 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

1 )

V = ( 2 g h )1/2

= ( 2 X 32.17 X 6 )1/2

= 19.65 ft/sec

V = 19.65 ft/sec = 21561.5 mph

2 )

V' = ( 2 g h' )1/2

= ( 2 X 32.17 X 10 )1/2

= 25.36 ft/sec

V ' = 25.36 ft/sec = 27827.02 mph

3 )

a = V2 / 2 d

a = 19.652 / 2 X 14 X 0.083

a = 166.15 ft/sec2

F = W a / g

F = 225 X 166.15 / 32.17

F = 1161.9 lbs

4 )

a' = V'2 / 2 d'

a' = 25.362 / 2 X 3

a ' = 107.18 ft/sec2

F ' = W a ' / g

F ' = 225 X 107.18 / 32.17

F ' = 749.53 lbs

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.