Go Back   Physics Help Forum > Physics Forums > Physics

Physics Physics Forum - General Physics Discussion and Physics News

Like Tree1Likes
  • 1 Post By ChipB
LinkBack Thread Tools Display Modes
Old Sep 17th 2018, 01:02 AM   #1
Junior Member
Join Date: Sep 2018
Posts: 2
Question Falling Lever - Impact Force

So as I lay here contemplating back surgery and trying to avoid sneezing the remains of a kidney through assorted orifices, I contemplate a problem - well, the primary "problem" is how long ago College was, but the problem I'm here to explore today is the force generated by a falling lever.

Now, I could simply have dropped this on my head, and the math is pretty trivial ("perform it in your head as a parlor trick"-simple)

So let's imagine that we have a ladder, and this ladder has one end in contact with the surface that someone is standing upon. The other end is 14.6m in the air. The ladder weighs 63.5kg overall.

The ladder becomes unsettled and falls through its arc, and the woman reflexively reaches overhead and catches the end around 14m from the pivot. Let's say she stands 1.8m tall, and the easily compressible forearms put the initial point of impact at around 2.05m

So assume our heroine (moron?) foolishly forgot her cape and boots that morning. What kind of impact force are we looking at?

I'm going to try attaching an image here - but please bear in mind that I'm much better at math than art... Although illustrations make these things easier for me, I'm not terribly good at 'em...

I'm happy to simplify this - let's agree to ignore cross section, wind resistance, energy lost in material flexibility, ... If you want to bounce some massless balls on frictionless planes - knock yourself out. This is a magnitude problem for me.

TIA for any insights!

[P.S. If it's unclear - it's a practical problem, not a homework problem ]
Attached Thumbnails
Falling Lever - Impact Force-falling-ladder-landscape.jpg  
jsquared is offline   Reply With Quote
Old Sep 18th 2018, 02:20 AM   #2
Junior Member
Join Date: Sep 2018
Posts: 2
Laying here still contemplating...

The only force here is gravity * mass. The ladder's center of gravity is going to roughly be its midpoint.

So the force acting is 63.5 * [9.8] = 622.3
The Torque is then Force * Distance to center of gravity from the pivot, right?

T ~ 4542.75 Nm

At this point, I lose focus and forgot why I cared about Torque. I'm sure it'll come back.

But I need a moment of Inertia in there, 'I', and I = (1/3) * m * l^2
I = 1/3 * 63.5kg * (14.6^2) ~ 309 kg*m^2 which seems like a standard unit...

So I'll leave this note here in case it tickles someone else, and revisit it after a nap otherwise...


And T=Iα right
So then α ~ 14.717 rad/s^2

That seemed an important memory to me a moment ago...

Last edited by jsquared; Sep 18th 2018 at 02:43 AM.
jsquared is offline   Reply With Quote
Old Sep 18th 2018, 06:30 AM   #3
Physics Team
ChipB's Avatar
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,347
You've figured out how to determine the ladder's acceleration at the moment that it is parallel to the ground (i.e., the moment arm is half of 14.6m). But since it tips from vertical, as it rotates the moment arm that gravity is acting on changes from 0m to 7.3m. Hence the rotational acceleration increases from 0 rad/s^2 to some value a little less than 14.717 rad/s^2 as it falls (a little less because the woman catches it before it hits the ground). Once you have a formula for acceleration as a function of time you would have to perform an integration to determine the rotational velocity at the instant the woman catches it. That's a pretty difficult calculation.

There is a better way to solve this, using energy principles. As the ladder falls its initial gravitational potential energy (PE) is converted to kinetic energy (KE). Then as the woman slows the ladder's fall she does work on the ladder, and KE goes to zero. From conservation of energy we know that the work she does is equal to the change in PE of the ladder.

$ W = \Delta PE = mg \Delta h $.

Here delta h is the change in the height of the center of mass from beginning (h1=7.3m) to when the ladder stops moving. You haven't told us what the distance is that the woman requires to stop the ladder's movement. It's not instantaneous, so let's assume that she arrests its movement at height 1.5m. Then h2= 1/2(1.5 m x 14.6/14) = 0.72m. So delta h for the center of mass of the ladder is h2-h1 = 0.72m - 7.3m = -6.58m.

The work she performs in stopping the ladder is calculated from W = Fd, where F is the force she applies to the end of the ladder and d is the distance over which she applies that force. Using the values I suggested above, d= 1.5 - 2.05 = -0.55m.

So now we have: $W = \Delta PE$
$ F \times \ -0.55m = 65 kg \ \times \ 9.8\ m/s^2 \ \times \ -6.58m$

Solve for F. Does the result seem reasonable? Should we consider using a different values for 'h' and 'd'?
jsquared likes this.

Last edited by ChipB; Sep 18th 2018 at 09:01 AM.
ChipB is offline   Reply With Quote

  Physics Help Forum > Physics Forums > Physics

falling, force, impact, lever

Thread Tools
Display Modes

Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Falling object force for starting lever builder89 General Physics 13 Feb 18th 2018 07:52 PM
Pendulum impact force cpoulsen2 Kinematics and Dynamics 5 May 25th 2017 08:26 AM
Force of impact of a stud on a studded tyre LeBirk Kinematics and Dynamics 6 Sep 2nd 2013 06:29 AM
Impact force pmlapl Kinematics and Dynamics 6 Mar 9th 2013 10:06 AM
Impact force measurement over my head General Physics 2 Aug 13th 2009 08:47 PM

Facebook Twitter Google+ RSS Feed