Forces and Safety Factors
Name: Barry C.
I am having trouble with a concept related to gravity,
speed of acceleration due to gravity, and the imposed forces measured
when stopping a falling object. A second part of my problem has to do
with the force doubling. We need to be able to explain and measure the
doubled forces, in that we often look towards a 2x safety factor when
working with catching falling objects.
I will attempt to help you with your questions in order. My thoughts are
below your questions below:
1. Simply put, if a 200 lb object (say a weight on a steel cable) is
dropped freely 6', and then stopped suddenly, with a load cell in
line, we might see a load indicating what? __________ (*This is the
F(force)=M (Mass (weight sitting still)) x A(acceleration due to
gravity). We measure force on the load cell in lbs.
The crucial factor is "How suddenly is the weight stopped?" Let's solve the
general problem: If a mass m falls a distance h under the influence of
gravity (force mg), gravity does an amount of work given by mgh which is
converted into the kinetic energy of the weight. If we then wish to stop it
in a distance d, we must do an equal amount of negative work on the weight.
So if the net force stopping the weight is F,
Fd = mgh or F = mgh/d
Notice especially that if d = 0 (the weight is stopped instantaneously), d=0
and the force F is infinity. Obviously this does not happen! For
completeness, the net force on the weight F is composed of the tension in
the cable upwards, T, and the weight of the weight downward, mg, so F = T -
T = mg(h/d + 1)
With your numbers, mg=200 lb and h= 6 ft. If I estimate d to be 0.06 ft,
T = 20200 lb. If d = 1 ft, T = 1400 lb.
2. Then, how far up in the air do we have to raise the same weight and
same cable attachment, in order to get 2x the force. In my mind the 'A'
component has a "squaring effect",and it is not required that you would
raise the weight to allow a 12' free fall. Would you? If the weight is
going more than 2x as fast in 12' free fall, you'd only have to raise it
somewhat higher, but not a full 2x higher in order to get a 2x force
right? What is the factors that would let me know how to double speed,
therefore doubling the force.
To double the force, you simply double h (ignoring the 1 in the equation).
Using the equation for constant acceleration, the speed of the weight after
falling h ft is given by
v^2 = 2 g h. So if you double the height, the weight gains in speed by a
factor of the square root of two. Notice that this doubles the kinetic
energy since the kinetic energy is proportional to v^2. Notice that this
(almost) doubles the T needed to stop the weight in the same distance d.
Notice also that doubling the speed quadruples the kinetic energy and also
quadruples the T needed to stop the weight in the same distance d. Doubling
the speed does NOT double the force!
3. Is it fair to say that using the exact same free fall, but doubling
weight would get us an actual 2x force measurement? Or.. does the gravity
work on the two weights differently?
No, that is exactly correct (except for the 1).
In all cases I understand that we have some effect from air friction, and
some effect from shock absorbancy of the steel cable and any deflection
from the weight, but theoretically, we are working in a vacuum with the
same negative acceleration distance.
By the way, this is why mountain climbers use nylon ropes to protect
themselves from falls. If they fall, since the nylon ropes are very elastic
and stretch easily, the rope slows the climber gradually after a fall,
minimizing the force exerted on his body. A steel cable could cause serious
injuries as it stops the climber almost instantaneously after a fall.
Best, Dick Plano, Professor of Physics emeritus, Rutgers University.
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Update: June 2012