apparent mass?

Question

These questions are for fall arrest systems.
1. A 180LB man has a 10' rope tying him to a bolt in a wall. The man free falls for 10'. How much force (lbs) is applied to the bolt when he comes to a sudden stop?
2. Now, the last 3' of the 10' rope is set-up to decelerate the 180LB man. (Man free falls for 7' then decelerates for 3'). How much force (lbs) is applied to the bolt at 7' (point of deceleration) and at 10' (end of 3' deceleration)?

please post answers in Lbs.

Answer 1

How sudden?

looking at conservation of energy
.5*m*v^2=f*d
and
.5*m*v^2=m*g*h

Let's assume he stops in 0.1 ft (about 1 inch)

180*10=f*.1
18000 lbs of force get applied

2) Although you don't state it, the rope should act as a spring where F=k*x

It looks like you are assuming constant deceleration, which is impractical.

Let's look at both cases

first, constant deceleration
after 7 ft, the man has gained KE equal to 180*7

which means his speed is
v=sqrt(180*7*2*32/180)
v=8*sqrt(7)
21.17 ft/sec
the constant deceleration must satisfy
0=21.17-a*t
and
3=21.17*t-.5*a*t^2
solve for a

3=.5*21.17*t
t=6/21.17
a=21.17^2/6
=74.7 ft/sec^2
and since F=m*a
F=180*74.7/32
F=420 lbs

Check: using conservation of energy
180*7=f*3
f=420 lbs
This is a constant force, the same at 7 ft as at 10 ft
Notice how conservation of energy was much easier to use than the equations of motion!

for the spring
The way to set this up is that the spring would have a latch such that it would arrest at 10 ft even though there is energy transferred to the spring.

180*10=.5*k*9
k=2*180*10/9
k=400 lbs/ft

so at 7 feet the fore is zero+, and at 10 ft it is 4000 lbs
note that if the spring is allowed to reach equilibrium supporting 180 lbs, the extension is only
400/180 ft or 2.22 ft

j