## Universal Physics Journal   Question 2: Why does gravity behave so forcefully at times?

Hi Ethan:  I am enjoying your articles.  Keep them coming.  My question is about gravity.  Most times it seems easy to handle.  But at other times, the force of gravity can be brutal.  Like when I was younger and fell from a ladder and hit my head on a sidewalk.  I'll never forget how violent that blow felt to my head.  I didn't black out, but it was some time before I felt okay.  Why does gravity behave so forcefully at times like this and not at others?      R.S.  Austin, TX, USA

Hello R.S.:
Hope you had no lasting effects from your impact with the sidewalk.  Protecting one's head from impact (sudden negative acceleration) is so important.  I can't stress that enough.  Pardon me if I digress for a moment to warn our readers about the latest scooter craze.  The ones with the small front wheels can be a real hazard if the small front wheel drops into a depression in the street or sidewalk.  The scooter may quickly stop, while providing little force to stop (negatively accelerate) any portion of the rider's body above the knees.  Thus the rider's feet are taken out from under the rider as the rider falls over the scooter's steering bars.  I hear that many serious head and body injuries are being caused in this manner.  My friend's son broke his hip while falling from a small-wheeled scooter.  He ended up needing an expensive and uncomfortable hip joint replacement, at 15 years of age!

On our city bus there is a safety ad that shows three images; one, a human brain; two, a young helmetless person riding a small-wheeled scooter; and three, a well-padded helmet.  The caption under the brain reads "If you have one of these...".  The caption under the scooter reads: "And you ride one of these...".  The caption under the helmet reads: "...Then wear one of these."

US units         Force causing negative acceleration for an unprotected head impacting concrete.
Velocity1 (V1) = 20 ft/sec down.
Velocity2 (V2) = 0 ft/sec down.
Average Velocity during impact = (V1+V2) / 2 = 20 ft/sec + 0 ft/sec / 2 = 10 ft/sec down.
Distance of acceleration = 0.02 ft
Elapsed Time = Distance / Average Velocity = 0.02 ft / 10 ft/sec =  0.00,2 sec
Acceleration = (V2 - V1) / Elapsed Time = (0 ft/sec - 20 ft/sec) / 0.00,2 sec =  -10,000 ft/secē
Absolute Force (Poundal) = Mass of object x Acceleration of object = 10 lb.m  x  -10,000 ft/secē =  -100,000 Poundal
Acceleration/Action Force (lb.f) = Absolute Force (P) / g = -100,000 P / 32.17 ft/secē = -3,108 lb.force!

Now to the shockingly high upward-directed Type 3 bipole external acceleration/Action stacking force of -3,108 lb., that is causing upward-directed acceleration for your head, you need to add the relatively insignificant upward Type 3 external 10 lb.force of Earth's gravitational weight against your head.  The total -3,118 lb.f is as much force as the weight of a mid-sized automobile yet as you can plainly see, gravitation's role is not even 1 % of the forces present during this event. Of course the upward-directed acceleration/Action (abbrev. a/A) force from Earth is mutually supported by the downward-directed acceleration/Reaction (a/R) force of 3,118 lb.force from your head, including all of its thoughtful contents, with this large reaction force being slightly increased by the downward Type 3 external 10 lb. action force from your head's gravitational weight against Earth.

Before having a look at the reduced forces of acceleration present when wearing a safety helmet, I will repeat the above helmetless calculations in SI units.

SI units         Force causing negative acceleration for an unprotected head impacting concrete.
Velocity1 (V1) = 6.1 meters/sec down.
Velocity2 (V2) = 0 m/sec down.
Average Velocity during impact = (V1+V2) / 2 = 6.1 m/sec + 0 m/sec / 2 = 3.5 m/sec down.
Distance of acceleration = 0.00635 meters
Elapsed Time = Distance / Average Velocity = 0.00635 m  / 3.5 m/sec =  0.00,2 sec
Acceleration = (V2 - V1) / Elapsed Time = (0 m/sec - 6.1 m/sec) / 0.00,2 sec =  -3,050 m/secē.
Acceleration/Action Force (Newton) = Mass of object  x  Acceleration of object = 4.5 kg  x -3,050 m/secē  =   -13,725 Newton

Now let us calculate the acceleration/Action and mutual supporting acceleration/Reaction forces that are present when your 10 lb.mass head, sporting a well-padded safety helmet, impacts with and is thereby accelerated by the concrete sidewalk.  Because of the helmet's padding, this time your head will be accelerated over a 1 in or 0.08 ft, or 0.02,54 meter distance which is 4 times the previous distance of 1/4".

US units         Force causing negative acceleration for a helmeted head impacting concrete.
Velocity1 (V1) = 20 ft/sec down.
Velocity2 (V2) = 0 ft/sec down.
Average Velocity during impact = (V1+V2) / 2 = 20 ft/sec + 0 ft/sec / 2 = 10 ft/sec down.
Distance of acceleration = 0.08 ft
Elapsed Time = Distance / Average Velocity = 0.08 ft / 10 ft/sec =  0.00,8 sec
Acceleration = (V2 - V1) / Elapsed Time = (0 ft/sec - 20 ft/sec) / 0.00,8 sec =  -2,500 ft/secē
Absolute Force (Poundal) = Mass of object  x  Acceleration of object = 10 lb.m  x -2,500 ft/secē =  25,000 Poundal
Acceleration/Action Force (lb.f) = Absolute Force (P) / g = 25,000 P / 32.17 ft/secē = -777 lb.force

After adding the -10 lb.force of Earth's gravitational toward your head, the total a/A force reaches -787 lb.force.  While this equals the force of the weight of 4 large adults, which would you rather have impressed against your head, an impact force momentarily equal to the weight of  4 large adults or one that is momentarily equal to the weight of a mid-sized automobile?

While you are thinking about these facts, R.S., consider that whatever the magnitude of the external acceleration/Action contact force against your head, it will be equaled at the point of contact by the supporting internal acceleration/Reaction forces from all the components of matter contained within your head.  This means that during impact, your head, including the fragile brain inside, is serving the same role as the head of a hammer!

Again I will make the same calculation in the SI standard:

SI units         Force causing negative acceleration for a helmeted head impacting concrete.
Velocity1 (V1) = 6.1 meters/sec down.
Velocity2 (V2) = 0 m/sec down.
Average Velocity during impact = (V1+V2) / 2 = 6.1 m/sec + 0 m/sec/2 = 3.5 m/sec down.
Distance of acceleration = 0.02,54 m
Elapsed Time = Distance / Average Velocity = 0.02,54 m  / 3.5 m/sec =  0.00,726 sec
Acceleration = (V2 - V1) / Elapsed Time = (0 m/sec - 6.1 m/sec) / 0.00,726 sec =  -840.22 m/secē.
Acceleration/Action Force (Newton) = Mass of object  x  Acceleration of object = 4.5 kg  x -840.22 m/secē  =   -3,780.99 Newton

As you can see, wearing a helmet increases the distance over which the acceleration occurs which dramatically reduces the rate of acceleration for your head and the magnitude of the mutual a/A forces acting between your head and the concrete sidewalk to 1/4 their former helmetless value.  Meanwhile the forces of gravitation are limited to only about 1% of the forces present during this head-impacting event.  I hope by reading this answer to your question, R.S., many parents around the world will benefit by insisting that their children always wear safety helmets complete with an effective chin strap anytime there is a risk of impact to the head.

On a personal note, I have extensive experience in this area for I have fallen many times from riding my off-road motorcycle without serious injury by tucking my head, arms, and legs into a "ball" before impacting the ground which greatly increases the distance over which negative acceleration occurs to my rolling body.  And yes, I always wear my Bell, full-coverage, Snell Approved, fiberglass helmet with built-in chin guard when riding.  I hope you do the same, no matter what type of sport vehicle you ride!

Ethan Skyler

P.S.  If you are going to buy a scooter, for certain make sure it comes with large-diameter wheels, and always wear a well-padded helmet, complete with a rigid chin guard, every time you ride!

Copyright Đ 2001 - 2013 by Ethan Skyler.  All rights reserved.  The author grants each visitor to The Universal Physics Journal the right to make one copy of Question 2 for his or her own personal archive as long as the author's copyright notice is permanently affixed to the archive copy.