Let me start off with why I do this. First, I find it fun. Second, why does Ford say things like "An Actual Demonstration" and "30,000 lbs. of Force"? What are they trying to say. I think they are trying to say "hey, its me! trust me!". So, here is what I aim to explore: Is it reasonable that the box (or whatever it is) exerts 30,000 lbs of force on the truck? (in case you have not seen it, here is my analysis of another F150 commercial)

So, what do I have to work with? I have the truck, and essentially two video clips of importance from the commercial. The first clip shows the box falling from the helicopter in a zoomed out view. The second clip is closer up and shows the box landing on the truck (this second clip is in some type of slow motion).

What do I NOT have? I don't know the mass of the box, and I don't know the frame rate of the second clip. So, here is the plan: From the first clip, I can essentially find the speed of the box as it hits the truck. I can use this speed to find the frame rate of the second clip and then I can estimate the forces exerted on the truck and see if it is near 30,000 lbs of force (at least force correctly labeled lbs as a unit of force. The English unit for mass is the slug - really).

Analysis of Clip 1:
Here the box is shown falling from the helicopter. I scaled the video using the length of the Ford F-150 Supercrew. Really all I need from this clip is the distance that the box drops. With that I can calculate the speed it hits the truck (assuming there is negligible air resistance). Using Tracker Video Analysis software (free), I obtained the following:
Note: video analysis only gives position and time data from each frame. This is a plot of vertical velocity and it is obtained by looking at how the position changes with time. There are several methods to calculate this (or you could even fit a quadratic function to the position data), but I used Logger Pro (by Vernier) derivative function. Since the acceleration of an object is
This would mean that the slope of a velocity time graph should be the acceleration. A free falling object (meaning only the force of gravity) should have an acceleration of 9.8 m/s² in the downward direction. The slope of the above line is around 9.8 m/s². The result is that the scale I have chosen (by using the length of the truck) is perhaps valid.
Now, if the scale is correct (and it appears ignoring air resistance is ok), I can use this scale to find out how far the box was dropped.
So, 13 meters (approximately). So, how fast would an object be going if it fell 13 meters? There are a couple of methods for determining this, I will use my favorite - the work-energy method. The work energy method is very useful (and easy) when you know the distance over which some force is applied (like this case). The work-energy method says the following:
- basically the work done on an object is equal to its change in energy. For this case, there are two types of energy that can change: kinetic energy (energy of motion) and gravitational potential energy (energy of position).

In this case, there is no work done on the falling box (I will not count the work done by gravity because I am using gravitational potential energy). You can't have your cake and eat it too. Putting this all together:
Note that the mass canceled. Also, there is a negative under the square root sign, but that is ok since Δy is also negative (it is moving down). This gives a final velocity of:
- ok, I am done with this part of the video.

Analysis of clip 2:
Clip 2 shows a close up of the box impacting with the truck. I can still get the distance scale by using the dimensions of the truck, but now the time scale is unknown.
I have two methods for determining the time scale. First, this clip shows the box still in free fall before the collision. I already know that this speed should be 16 m/s so I can just adjust the time scale to give the same velocity. Here is the data from analysis of that video:
This data shows several things. First it shows the speed at which the box hits the truck (from the graph this would be around 10 m/cs - where cs is a "clip second" not to be confused with a real second). Next it shows the box interacting with the truck. During this time the speed slows down (becomes less negative) and even has a positive speed. After that the box leaves the truck bed and is once again in free fall. This second free fall will also be used. So, what is the conversion between seconds and cs? If the impact velocity is the same, then:

Now what about the other free fall data? This should have an acceleration of 9.8 m/s² but it is listed as 5.082 m/cs². If these two are equivalent, then:
- ok, so two possible time scales (they are too far apart, that is good).

Calculating the force.
Here is a force diagram for the block while it is interacting with the truck:
During this time, there are two forces acting on the box - the truck pushing up and the gravitational force of the Earth pulling down. Since the box is slowing down while moving down, the truck must be exerting a greater force. One thing to note: you might be saying "hey! why are we looking at the forces on the box. Who cares about the box! I am not considering buying the box on that F150!" Well, those are good points. But if you examine forces, you will find that forces are an interaction between two objects (this is always the case). So, for every force there are "two sides". The amount the truck pushes on the box is exactly equal (but in the opposite direction) as the force the box pushes on the truck. You may recognize this as Newton's 3rd law.
So, back to forces. One form of Newton's second law says (in one-dimension):

In this case, with the two forces:

So the force the truck exerts on the box is:
All I need now is the mass of the box and the acceleration. I can get the acceleration from the above plot of velocity. During that phase, the acceleration is 30.8 m/cs². Depending on which time scale I use, this means either:
Giving Ford the benefit of the doubt, I will use the greater acceleration.
For the mass, I will start with the maximum payload capacity of the truck (they even state this in the commercial):
In case this is too small to read, it says "Properly equipped Full-Size Pickups under 8500 lbs GVWR. F-150 FX4 Supercrew shown 1,630 lbs payload". I am not sure what "properly equipped" means. Also, it i probably not standard to have holes going through the bed of the truck for cables. So, I have the acceleration, I have the mass (1,630 lbs = 739 kg - I am not going to talk about the difference between mass and weight now). Plugging this into the equation above:

Converting: 65,478 Newtons = 14,720 lbs.
Where did Ford come up with 30,000 lbs? At first I thought this project would not be possible. I knew I had a good chance of calculating the average force the box exerts on the truck, but not the max. This is because what really matters when stopping the box is the integral of force over time. This being the case, the following two forces could have the same effect:

The blue force line has a much larger max force. However, looking at the data from the video, it seems like a force similar to the red curve is very reasonable as the acceleration of the box as it is stopping is nearly constant.

So, why even say 30,000 lbs? Does it just look cool? Could it really be 30,000 lbs? Of course, perhaps there is still some spike in there, but its not obvious. Perhaps the dropped box weighs more than the recommended load (I don't think so - the truck drives away and does not look overly loaded). Is the Ford F150 a tough truck, I think so. Look at this picture:

I don't think I would buy this truck after this little stunt (but I don't know alot about trucks and engineering). This sucker took a hit. The gas cap lid even comes loose.

So, this is really the end of the story. By the best estimates, the force the box exerts is still less than 30,000 lbs? Would it matter anyway? What if they said 15,000 lbs - does that really mean anything to the average Joe that is think of getting a Toyota Tacoma instead (I have problems with their insurance fraud commercials also). Why not just have 30 seconds of slow motion video showing this truck taking a big hit. 30,000 lbs - does anyone really have a good idea of that kind of force anyway? Ford, who do you think we are? Patsies? (what does that mean?)
See - its right there in the commercial (30,000 lbs of force - as opposed to 30,000 lbs of feathers which would be much better).

Maybe Ford is thinking of something else:
Perhaps Ford is talking about the total force on the truck? That still would not be 30,000 lbs. Here is a force diagram for both the truck and the box during the interaction:

As I said early, the force F-box and F-truck are really two ends of the same force. This means they have the same magnitude (but opposite direction). The rest of the forces on the truck have to add up to zero since it is essentially not accelerating (yes, the body moves, but not the tires). These are really vector quantities, but without going into all that stuff, I can say:
Where the F-weight and F-box are negative because they are pointing down. This would make the force the ground pushes on the truck to be about 20,000 lbs (since the truck is about 5,000 lbs). If you look at the TOTAL force on the truck, it would be zero (these are vector quantities, you can't just add them up as numbers you have to take into account the direction).
I am sorry Ford, I tried to find a way that would make you look good.


Bonus stuff:
Here are some other great snap shots:
aption: Closed airfield. Do not attempt. - DAMN I was planning this for my weekend project. Oh well.
Caption: Actual demonstration. Meaning its not fake? I am almost sure this is fake. I saw it on that show Reno 911.

Tags: video, modeling, momentum, forces, kinematics