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**Physics, Astronomy & Cosmology / Re: How much wind to blow over a truck?**

« **on:**13/02/2019 23:59:53 »

The wind force F on a flat surface is ½ρv

ρ = air density

v = wind speed

A = surface area.

To tip a truck over, you need to move its center of gravity outside the wheelbase.

Consider the wind force F to act through the midpoint of the side of the truck. This gives a turning moment of hF/2 where h is the height of the truck. The restoring moment is Ww/2 where W is the weight of the truck and w is the wheelbase width. If hF> Ww, the windward wheel will lift.

Counterintuitively, it may be easier to tip a loaded truck than an empty one, because adding load raises the CG. If W is constant, you need the same force F to lift the windward wheel off the ground, but with a higher CG you don't have to lift it so far in order to tip the vehicle over.

In practice, things get more complicated. A gust can cause an articulated truck to snake, and the resultant centrifugal force can tip it even though the initial gust couldn't.

^{2}A whereρ = air density

v = wind speed

A = surface area.

To tip a truck over, you need to move its center of gravity outside the wheelbase.

Consider the wind force F to act through the midpoint of the side of the truck. This gives a turning moment of hF/2 where h is the height of the truck. The restoring moment is Ww/2 where W is the weight of the truck and w is the wheelbase width. If hF> Ww, the windward wheel will lift.

Counterintuitively, it may be easier to tip a loaded truck than an empty one, because adding load raises the CG. If W is constant, you need the same force F to lift the windward wheel off the ground, but with a higher CG you don't have to lift it so far in order to tip the vehicle over.

In practice, things get more complicated. A gust can cause an articulated truck to snake, and the resultant centrifugal force can tip it even though the initial gust couldn't.

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