# How fast would a truck have to go to pull a pedestrian onto the road? [closed]

Let's suppose a pedestrian P is walking or standing next to a highway.
Suppose a truck T will drive by the said pedestrian at speed V leaving distance L between the two.

Assuming L is a reasonably small finite size how would one calculate the force acting upon the pedestrian from both the general differences in pressure and turbulence?

Shortly put, how fast would the truck have to go in practice so that the air flow left in its wake would pull the pedestrian onto the road?

• Depends on their size, weight and balance. In general 70mph winds are hard to stand upright in. Often 100mph is considered enough to knock most people down. The bow/stern wave is a smaller portion of the vehicle's velocity so the truck would probably have to be moving very fast. – user6972 Mar 28 '14 at 20:37
• @user6972 And how slippery the sidewalk is. – Bernhard Mar 30 '14 at 14:29
• While admittedly this question is humorous, it is still very clearly a homework-like question. Please refer to our FAQ page for guidelines on posting homework-like questions. As it stands, this question is off-topic for the site – Jim Apr 22 '14 at 13:31
• I updated the question so that it hopefully fits the theme of this site better, if you have any additional pointers in that regard that would be much appreciated. – Nit Apr 22 '14 at 16:41
• What would really help this question is if you were to show some effort to solve it yourself. In a nutshell, we tend to react badly to questions that say only "calculate this thing for me" or "tell me how to do this" (which is kind of what you're doing). A better question would be if you say something like "I searched Google to see if this had been done before, I tried calculating it using method X but that didn't work because Y, and then I tried method Z but that didn't work because Z'. Is there some other method that will work?" – David Z Apr 22 '14 at 16:56

The truck will have in its wake some unknown mass of air almost moving with a speed $v$ comparable to the truck's speed $\bf V$. The pressure behind the truck will be lower than the pressure at the sidewalk because air pressure follows the Bernoulli equation, $$P_\mathbf{P} = P_\text{road} + \frac{1}{2}\rho v^2,$$ where $\rho \approx 1~$kg/m$^3$ is the air density. For a truck at 60 mph $\approx$ 30 m/s, this is a pressure difference of 450 Pa, or 0.004 atm.
If the pedestrian is very near the wake of the truck (note that this wake will extend partway into the buffer zone $\bf L$), they might feel this pressure difference $\rho v^2/2$ across their torso. The force depends on the orientation and area of the pedestrian. My torso is about a meter long and about 0.4 meter wide, so I could feel a force as large as 180 N. That's equivalent to about 20% of my body weight, or the lateral force that I'd feel leaning at an angle of about 10º. I can recover from a 10º lean, but I have to be ready for it.
All of these numbers should be taken as coarse approximations, because I've ignored turbulence. Turbulence will play a huge role in the dynamics here, especially for finite $\bf L$. However the order of magnitude is probably correct.