What is the force on the safety strap when stopping a fall at 1 meter?

[Postal Pete]

Homework Statement

A 100 kg man is walking on a tight cable. His safety harness has a strap that is attached to an overhead cable that will allow him to fall 1 meter before stopping him from falling any further. What is the force on the strap when it stops his fall at 1 meter.

Homework Equations

v = at
v^2 = 2ad
Fd = KE = .5mv^2
a = 9.81 m/s^2
Weight = ma = m*9.81

The Attempt at a Solution

v = sqrt (2*9.81*1) = 4.43 m/s
KE = .5*100*(4.43)^2 = 981 J
Fd = F*1 = KE = 981 J
Force on the strap at the end of the 1 m fall is 981 N
The man's weight is ma = 100*9.81 = 981 N
Shouldn't the force on the strap after the one meter fall be greater than the man's weight?

 

[andrewkirk]

There's a missing piece of the question, which is how long the strap takes to arrest his fall. That will be a function of how much elasticity there is in the strap and in the cable from which it hangs.

Note that it is impossible for them to be infinitely stiff, and hence the fall cannot be arrested instantaneously. If it were, the deceleration, and hence the force, would be infinite, and the man would be pulverised.

In practice the force on the strap will vary over time, going from near zero when it first begins to take his weight, increasing to a maximum, and then reducing until it just equals his weight.

What they probably want you to give is the average force, given a period of time t for the deceleration. THat force will be the man's weight, plus an extra force that decelerates him.

If you work out the momentum he has when he has fallen one metre then you can work out the average force using the equation p (momentum) = F t.

 

[haruspex]

I concur with andrewkirk (except that there may be some bounce, so the force might oscillate before settling out at equal to the man's weight).
But you may be puzzled as to what was wrong with your own solution

v = sqrt (2*9.81*1) = 4.43 m/s
KE = .5*100*(4.43)^2 = 981 J
Fd = F*1 = KE = 981 J

It will be clearer if we stick to symbols. You computed v from $\frac 12mv^2=mgh$, then set $\frac 12mv^2=Fh$. Clearly that will produce F=mg. You effectively treated the retarding force as constant over the whole metre of descent.

 

[Postal Pete]

There's a missing piece of the question, which is how long the strap takes to arrest his fall. That will be a function of how much elasticity there is in the strap and in the cable from which it hangs.

Note that it is impossible for them to be infinitely stiff, and hence the fall cannot be arrested instantaneously. If it were, the deceleration, and hence the force, would be infinite, and the man would be pulverised.

In practice the force on the strap will vary over time, going from near zero when it first begins to take his weight, increasing to a maximum, and then reducing until it just equals his weight.

What they probably want you to give is the average force, given a period of time t for the deceleration. THat force will be the man's weight, plus an extra force that decelerates him.

If you work out the momentum he has when he has fallen one metre then you can work out the average force using the equation p (momentum) = F t.

Thanks for your response. I had not thought about the elasticity factor. The reason that I was interested in this problem comes from being on what was called an "Adventure Ropes Course" (although you were actually walking on steel cables) and being told by the guy who got you into your harness that the breaking point of the strap was 5000 pounds and you don't weigh 5000 pounds so don't worry about the strap breaking. But I knew that if I did fall the length of the strap the force on the strap would be greater than my weight. The time of fall would be from d = .5at^2. Solving for t, t = sqrt (2d/a) = sqrt (2*1/9.81) = .45 s. From p = mv = Ft, solving for F, F = mv/t = 100*4.43/.45 = 984 N, which is only 3 N more than the man's static weight but which I think is correct.

 

[Postal Pete]

I concur with andrewkirk (except that there may be some bounce, so the force might oscillate before settling out at equal to the man's weight).
But you may be puzzled as to what was wrong with your own solution

It will be clearer if we stick to symbols. You computed v from $\frac 12mv^2=mgh$, then set $\frac 12mv^2=Fh$. Clearly that will produce F=mg. You effectively treated the retarding force as constant over the whole metre of descent.

Thanks for your response. Expressing the problem using just symbols clarifies my error.

 

[andrewkirk]

The time of fall would be from d = .5at^2. Solving for t, t = sqrt (2d/a) = sqrt (2*1/9.81) = .45 s. From p = mv = Ft, solving for F, F = mv/t = 100*4.43/.45 = 984 N, which is only 3 N more than the man's static weight but which I think is correct.

No it's much more than that. There are two errors in the calc.

Firstly, the time you must divide the momentum by to get the deceleration force is the time of deceleration, not the time of the fall. The deceleration time will be much less, unless there's a lot of give in the overhead cable. I would guess at a typical deceleration time of about one-third of the fall time - ie about 0.15s rather than 0.45s - so you need to triple your 984N.

Secondly, you have only included the force of deceleration. You need to add to that the man's weight to get the total force.

So the final force may be about four times the man's weight. Fortunately, that's still a very long way short of the breaking strain of the strap!

 

[Postal Pete]

Thanks for your response. I had not thought about the elasticity factor. The reason that I was interested in this problem comes from being on what was called an "Adventure Ropes Course" (although you were actually walking on steel cables) and being told by the guy who got you into your harness that the breaking point of the strap was 5000 pounds and you don't weigh 5000 pounds so don't worry about the strap breaking. But I knew that if I did fall the length of the strap the force on the strap would be greater than my weight. The time of fall would be from d = .5at^2. Solving for t, t = sqrt (2d/a) = sqrt (2*1/9.81) = .45 s. From p = mv = Ft, solving for F, F = mv/t = 100*4.43/.45 = 984 N, which is only 3 N more than the man's static weight but which I think is correct.

 

[Postal Pete]

No it's much more than that. There are two errors in the calc.

Firstly, the time you must divide the momentum by to get the deceleration force is the time of deceleration, not the time of the fall. The deceleration time will be much less, unless there's a lot of give in the overhead cable. I would guess at a typical deceleration time of about one-third of the fall time - ie about 0.15s rather than 0.45s - so you need to triple your 984N.

Secondly, you have only included the force of deceleration. You need to add to that the man's weight to get the total force.

So the final force may be about four times the man's weight. Fortunately, that's still a very long way short of the breaking strain of the strap!

Thanks for your interest and help with this problem. I can see why I could not find the solution in my first year physics text.