Who would hit the ground harder?

[GEEKfromTX]

Person A: Falling from the top of a 20' ladder.

Or

Person B: Falling with a 20' ladder that is falling over. (from the top)


Give me your answer and tell me why? This one is driving me crazy!

 

[ranger]

I'm [only] guessing you could use Newton’s Second Law, F = ma. Of course I am not taking into account air resistance. So Persion B would hit the ground with more force since the combined mass of the person and the ladder would be more. The acceleration remains constant at g and therefore velocity is increasing 9.80m for every second the object is in the air.

Do correct me if I am wrong.

 

[russ_watters]

The problem may be trying (poorly) to illustrate how the acceleration at the end of the ladder can go above g if you include the weight of the ladder. Regardless, acceleration would start at zero and increase as the ladder supports less and less of the weight of the person. So if you ignore the weight of the ladder, the person's ending speed would be lower. I'm not sure how to work that out mathematically or if you are supposed to ignore the weight of the ladder.

edit: Hmm - now I'm not sure. Since there is circular motion, there is more than one acceleration - perhaps if you look at it according to potential energy, the end result has to be no difference whether the ladder is there or not, since the ladder does no work on the person.

 

[GEEKfromTX]

I'm [only] guessing you could use Newton’s Second Law, F = ma. Of course I am not taking into account air resistance. So Persion B would hit the ground with more force since the combined mass of the person and the ladder would be more. The acceleration remains constant at g and therefore velocity is increasing 9.80m for every second the object is in the air.

Do correct me if I am wrong.

Acceleration will not be at a constant for person B. You need to take in account the ladder. Person B will be supported by the ladder during the fall.

 

[berkeman]

Person A: Falling from the top of a 20' ladder.
Or
Person B: Falling with a 20' ladder that is falling over. (from the top)
Give me your answer and tell me why? This one is driving me crazy!

Pretty interesting problem. At first glance, you'd think that the person hits with equal force, since it seems that there is no other place for their initial stored potential energy to go except into their terminal velocity. Yeah, it takes longer to hit the ground for the guy swinging down with the ladder, but he could still have the same terminal velocity.

But as I was drawing the problem up on a Post-It note to see if I could work out the math quickly, I noticed that there is someplace else where the swinging ladder guy's energy can go. Some energy is transferred sideways into the ground, so less of the ladder guy's initial potential energy will be converted into his terminal velocity, so he will hit the ground a little slower than the free-falling guy. If you sketch out the vertical and horizontal velocity vectors versus time for the ladder guy, you'll see that he has a net area under his horizontal velocity vector curve, and hence net work gets done in the horizontal direction. the horizontal force starts at zero, then increases as the ladder swings down, and then goes back to zero right at impact because the guy's motion is all downward at that instant. Since the horizontal force has a positive net, the ladder and guy did work on the Earth sideways during the fall. If you connect the base of the ladder with a hinge on the top of a platform on a frictionless plate, you'd see the sideways motion of the base of the ladder in the direction opposite of the guy's fall.

I still need to work out how much less speed the ladder guy has at impact, but I'm out of room on this Post-It, and I need to get back to work...

 

[Danger]

I'm out of room on this Post-It, and I need to get back to work...

You'll have to requisition larger Post-Its if you expect to be of any further use to us.

 

[berkeman]

You'll have to requisition larger Post-Its if you expect to be of any further use to us.

 

[Tide]

Apply energy conservation including the motion of the person and the ladder. I think the question implies the person is fixed at one end of the ladder and that the base of the ladder is fixed to the ground. The person's speed of impact with the ground will be greater than that of a person free falling from the same height - as long as the mass of the ladder is not zero.

 

[berkeman]

Apply energy conservation including the motion of the person and the ladder. I think the question implies the person is fixed at one end of the ladder and that the base of the ladder is fixed to the ground. The person's speed of impact with the ground will be greater than that of a person free falling from the same height - as long as the mass of the ladder is not zero.

You sure? If two objects of different masses fall at the same rate in a vacuum, and if my comments above about some of the ladder guy's potential energy going into horizontal work, then how could the ladder guy hit with a higher velocity than free-fall guy?

 

[russ_watters]

You sure? If two objects of different masses fall at the same rate in a vacuum, and if my comments above about some of the ladder guy's potential energy going into horizontal work, then how could the ladder guy hit with a higher velocity than free-fall guy?

Same rate, different mass = higher energy. The question askes about which hits harder, not which is going faster.

 

[turbo]

The person at the top of the ladder has lots of potential energy and little kinetic energy. I think it is fair to say that a person clinging to a falling ladder will convert his potential energy to kinetic energy at a slower rate (especially near the top of the arc) than a person falling off the ladder. If it were me, I would hang onto that ladder and ride it down. That way, my rate of acceleration would not approach that of free-fall until the ladder was close to horizontal, and the lower rate of acceleration earlier would result in a slower speed at impact and less injury.

 

[Tide]

You sure? If two objects of different masses fall at the same rate in a vacuum, and if my comments above about some of the ladder guy's potential energy going into horizontal work, then how could the ladder guy hit with a higher velocity than free-fall guy?

You have to consider that some parts of the ladder strike the ground with very little speed while parts of the ladder near the oher end strike at much greater speed! I suggest doing the calculation. :)

 

[berkeman]

You have to consider that some parts of the ladder strike the ground with very little speed while parts of the ladder near the oher end strike at much greater speed! I suggest doing the calculation. :)

I agree it's time to play with some numbers -- I'm about to head home for the day, so maybe I'll make it my evening project in front of the TV

BTW, I don't think the ladder hitting the ground is relavent to the OP's question. The ladder could be weightless, as long as it is rigid enough to hold up the ladder guy as he falls. It converts the vertical downward force of gravity into vertical cos(theta) and a horizontal sine(theta) (maybe) opposing components, where theta is measured between the vertical and the ladder. When the ladder guy is balanced on top, the cos(0) component exactly balances his weight W and holds him up. When the ladder has swung down to 45 degrees, the ladder guy's weight vector points down, but the ladder is pushing radially outward some, which applies force vectors in the vertical and horizontal directions.

Fun problem. I'll bet somebody answers it quantitatively by the morning. Cheers, all. -Mike-

 

[pervect]

I could use a double check of this, but I'm getting a (oops) faster impact speed, with the assumption that the mass of the ladder is uniformly distributed along its length.

Let the mass of the person be m
Let the mass of the ladder be u*m
Let the length of the ladder be L

Then the potential energy of the system is
$$V = m g L (1+\frac{u}{2}) sin \theta$$

The kinetic energy of the system is

$$T = \frac{1}{2} m (L \dot{\theta})^2 +\frac{1}{2} \int_{x=0}^L \frac{m u}{L} (x \dot{\theta})^2 dx = \frac{1}{2} m v^2 + \frac{1}{6} u m v^2 = \frac{1}{2} m (1+\frac{1}{3} u) v^2$$

If the ladder starts out with $\theta=90$ and winds up with $\theta=0$, we have

$$v^2 = 2 g L \frac{1 + \frac{u}{2}} {1 + \frac{u}{3}}$$

 

[Tide]

Pervect,

That is correct so that as long as the ladder has any mass he'll hit the ground harder.

 

[pervect]

I agree, I also get a faster impact (I had to correct some arithmetical errors as well as some Latex issues, hence the 'oops').

My original post is probably a bit terse, let me fill in some of the coordinates and discussionP
.
.
.
.
.$\theta$
---------

the P is the person of mass M on the ladder. He is at a height L above the ground with the ladder vertical as shown.

The bottom of the ladder is assumed to be a fixed pivot point.

$\theta$ is the angle of the ladder with respect to the ground, 90 degrees when the ladder is vertical

$\dot{\theta}$ is [tex]\frac{d\theta}{dt}[/itex], the rate at which the angle $\theta$ is changing.<br /> <br /> The velocity of the person at the end of the ladder is v, and has a magnitude of $r \dot{\theta}$<br /> <br /> Note that with the ladder vertical, the center of mass of the ladder is at height L/2 above the ground. This is the origin of the factor of 1/2 in the first section of the post for the potential energy of the system.<br /> <br /> Different parts of the ladder have different velocities. The total kinetic energy is the intergal of .5*v(x)^2*dm, where v(x) is the velocity of the ladder element which is a function of position (x), and dm is the mass of the element of the ladder at position x.<br /> <br /> v(x) is just $x \dot{\theta}$, where x is the distance from the bottom of the ladder to the mass element.<br /> <br /> The mass of the ladder is u*M = <br /> $$\int dm = \int_{0}^{L} \frac{u M}{L} dx$$<br /> <br /> thus dm = (u M / L) dx<br /> <br /> Putting this all together gives the factor of u/3 in the bottom of the equation.[/tex]

 

[berkeman]

Pervect,
That is correct so that as long as the ladder has any mass he'll hit the ground harder.

That's really interesting. So the ladder mass helps to add some velocity to the final impact for the ladder guy, and if the ladder is massless, the impact velocity is the same as for free-fall. Cool.

I also think I see where my argument about some horizontal work being done wasn't quite right for the situation given. Since the Earth is so heavy compared to the ladder and guy, the Earth moves very little sideways from the sideways force component, so the work done (and energy lost) is negligible. But if you put the ladder base on a plane with moderate friction, so that the base of the ladder displaces a few meters sideways during the fall, then the work done and energy lost would count, I would think. So with a massless ladder in this scenario with the friction plate and displacement of the bottom of the ladder, I think the guy would have a lower impact velocity compared to the free-fall guy.

Cool problem.

 

[simple123]

Hu..no need to think complicity here. Think the basics. Your question seems to be dealing with different masses falling and which one would fall harder if they feel from the same distance? If I'm correct, then just think of a simple analogy..this should help. Imagine dropping a car and a ant from 20ft above..which would hit harder on the ground? It should be clear to see now.

 

[Tide]

Presumably, Person A and Person B have the same mass and we're asked which of them hits the ground "harder." I think it's reasonable to infer the intent of the question is to ascertain which person hits the ground with greater speed.

 

[Art]

A person free falling travels the 20 metres to the ground sooner than the person traveling the longer route (of around 31 metres) by clinging to the ladder as it scribes an arc through the air and so discounting the weight of the ladder the free falling person hits the ground harder as his vertical velocity is greater.

 

[BobG]

Have you ever watched a smoke stack fall over? The tangential acceleration of the top of smoke stack is greater than acceleration due to the force of gravity -- so much so that the smoke stack will break apart before it can hit the ground.

I'm thinking you need to find the center of mass and the moment of inertia. The base of the ladder will be the pivot point. The center of mass and the ladder's orientation relative to the ground will determine the angular acceleration of the ladder and the person. The end of the ladder will accelerate faster than the center of mass.

In the case of a smoke stack, especially if the smoke stack is slightly tapered, there will be a big difference between the center of mass and the top of the smoke stack. In the case of a ladder, the center of mass should be much closer to the guy at the top of the ladder, but at least slightly closer to the base of the ladder.

If you're losing no energy due to air resistance, etc, the kinetic energy at the bottom of the fall has to balance the loss of potential energy. The man at the ladder started out accelerating slower due to accelerating nearly parallel with the ground, but the acceleration increases and lasts for a longer amount of time. The linear tangential velocity (and the kinetic energy) of the portions of the ladder between the center of mass and the base is slower than that for the center of mass. The linear tangential velocity of everything beyond the center of mass is greater than that for the center of mass.

Therefore, the man at the top of the ladder, who is at least slightly further away from the base than the center of mass, hits the ground at a higher velocity than the free falling man.

Edit: If the ladder were massless, the man would be the center of mass and would hit at the same speed as the free falling man. Assuming the ladder has mass, the only way the man help himself to very rapidly climb down the ladder as it falls, hoping to put himself between the center of mass and the ground. (It's taken a lot of practice, but I've gotten pretty good at this).

 

[Art]

Acceleration due to gravity is constant whatever the mass. The fact the top of the ladder (and thus the man) has to travel a greater distance to reach the ground than the free falling man means he must be traveling slower and therefore hits the ground with less force. (ignoring the mass of the ladder itself)

Take a pen and stand it on end and a small object and let both go at the same time and you will see the free falling object hits the table before the top of the pen As they have both traveled the same vertical distance but the top of the pen takes longer it is obviously falling slower (distance/time)

Although both start off with the same potential energy in the case of the falling ladder some of this energy is expended in moving the man 20 metres horizontally from the starting point.

 

[berkeman]

Acceleration due to gravity is constant whatever the mass. The fact the top of the ladder (and thus the man) has to travel a greater distance to reach the ground than the free falling man means he must be traveling slower and therefore hits the ground with less force. (ignoring the mass of the ladder itself)
Take a pen and stand it on end and a small object and let both go at the same time and you will see the free falling object hits the table before the top of the pen As they have both traveled the same vertical distance but the top of the pen takes longer it is obviously falling slower (distance/time)
Although both start off with the same potential energy in the case of the falling ladder some of this energy is expended in moving the man 20 metres horizontally from the starting point.

And the error in the posted math is... what?

 

[pervect]

Acceleration due to gravity is constant whatever the mass. The fact the top of the ladder (and thus the man) has to travel a greater distance to reach the ground than the free falling man means he must be traveling slower and therefore hits the ground with less force. (ignoring the mass of the ladder itself)

Nope - see the previous posts for a more detailed analysis.

Or consider a very simple case for a "sanity check" - a man on the end of a massless ladder with mass M, and a bead with mass m that can be placed in various positions along the massless ladder.

Put the bead halfway up the ladder, for instance. The total energy of the system will be M*g*L + m*g*L/2 when the system is vertical.

The velocity of the bead will always be 1/2 the velocity at the end of the ladder. This means that the energy of the bead will always be (m/4M) times the energy of the falling man. (Why? because energy is proportional to velocity^2).

Solve for the resulting velocity v of the man using the conservation of energy, or note that the bead is multiplying the potential energy of the man, g*M*L, by 1+.5*(m/M), while dividing the kinetic energy by a factor of (1+.25*(m/M)).

Note that a pivot conserves energy, because force*distance = 0.

Take a pen and stand it on end and a small object and let both go at the same time and you will see the free falling object hits the table before the top of the pen

True, but irrelevant, because the pen standing on its end is not moving at a constant velocity. It may start out more slowly, but that says nothing about how fast it is moving at the end.

Although both start off with the same potential energy in the case of the falling ladder some of this energy is expended in moving the man 20 metres horizontally from the starting point.

It doesn't take any energy to move someone 20 meters horizontally. Work = force*distance, there is no force opposing the horizontal motion.

 

[BobG]

Acceleration due to gravity is constant whatever the mass. The fact the top of the ladder (and thus the man) has to travel a greater distance to reach the ground than the free falling man means he must be traveling slower and therefore hits the ground with less force. (ignoring the mass of the ladder itself)
Take a pen and stand it on end and a small object and let both go at the same time and you will see the free falling object hits the table before the top of the pen As they have both traveled the same vertical distance but the top of the pen takes longer it is obviously falling slower (distance/time)
Although both start off with the same potential energy in the case of the falling ladder some of this energy is expended in moving the man 20 metres horizontally from the starting point.

Acceleration for the free falling man is constant. Acceleration for the man at the end of the ladder is constantly changing ($$a=9.8 sin \theta$$. You could almost conclude that the man on the ladder must be traveling slower since his acceleration only reached the same rate as the free falling man at impact - except you'd be overlooking one other relationship:
$$v_f=v_i + at$$
The man on the ladder is accelerating for a longer period of time - in fact, if the man on the ladder had constant acceleration of 9.8 m/s^2, he would be going much faster at impact, since he'd have been accelerating longer.

Because there's two things happening (non-constant rate of acceleration ranging from 0 to 9.8 m/sec^2 and a longer amount of time) you can't make the simple analogy you did. The slower rate of acceleration will cancel out the increased time and the center of mass of both systems will hit the ground at the exact same speed.

Any difference in speed between the free falling man and the ladder man is due to the ladder man's location on the ladder relative to the center of mass. All parts of the ladder have to have the same angular velocity or the ladder will break, so anything further away from the pivot point than the center of mass must have a linear velocity greater than the center of mass. (real world, the ladder would probably have some flex to it that would increase as the ladder's angular acceleration increased, but without being given the stiffness of the ladder, I think you'd have to assume it's perfectly rigid - plus, that last little acceleration as the ladder unflexed isn't going to help the ladder man any.)

Edit: That's not quite right. I overlooked the fact that the center of mass starts out closer to the ground, so the center of mass of the ladder-man hits the ground slower. The man on the ladder still hits the ground faster than the free falling man, though.

The free falling man hits the ground $$\sqrt{\frac{r_{end}}{r_{cm}}}$$ times as fast as the center of mass of the ladder-man, but the man at the end of the ladder hits the ground $$\frac{r_{end}}{r_{cm}}$$ times as fast as the center of mass.

 

[Art]

Thanks BobG. So if I understand it correctly the lesson is to wait until the ladder has fallen about 45 degrees and then jump off preferably onto a soft grassy patch

 

[BobG]

Thanks BobG. So if I understand it correctly the lesson is to wait until the ladder has fallen about 45 degrees and then jump off preferably onto a soft grassy patch

A better idea would be to climb down the ladder very fast. Once you climb past the halfway point, you will hit the ground even slower than the center of mass. (Calculating how fast the climber has to climb down the ladder in order to hit at the same speed as the free-falling man would be a pretty challenging problem.)

The exception, of course, would be if the man were climbing the "Ladder of Divine Knowledge". One would have to decide if the increased chances of survival were worth the descent back down (so many variables that could be tossed into this problem ).