Trailer swing

Classical Mechanics Level 3

Jackknifing is a dreaded accident that may happen with tractor-trailers or similar vehicles. Broadly speaking, jackknifing describes when the front (the tractor) and the back side (the trailer) folds in a manner similar to a partially folded pocket knife. It usually occurs when the driver applies the breaks, the vehicle is slowing down, and a set of tires loses traction.

One possible cause of jackknifing is "trailer swing". In a trailer swing accident, the front part of the vehicle keeps the correct track (at least initially), but the rear wheels are sliding, and the back side swings violently forward. To investigate the conditions leading to the accident, let us assume that initially the rig is moving forward and decelerates (slowing down) with $a= \nu g$ , where $g$ is the acceleration of gravity and $\nu$ is a number between zero and one (it depends on how strongly the breaks are applied, and how strong the friction forces at the wheels are). The trailer has a length of $l$ , it has a uniform mass distribution, and half of its weight is supported by the rear wheels. The rear wheels are sliding on the road and the friction force acting on them is $F=\mu N$ with $N=\frac {1}{2} Mg$ , where $M$ is the mass of the trailer. The direction of this force is opposite to the velocity. Initially, the trailer makes a small angle (for example, $\phi_0=$ 1 degree) with the tractor, and as the tractor moves straight forward, the trailer also makes the same angle with the direction of the motion.

Initially the angle of the trailer will evolve with time as $\displaystyle \phi= \frac{\phi_0 e^t}{\tau}$ . The time constant $\tau$ tells us how much time the driver has to make a corrective move (for example, stop breaking or steer the front in the direction of the swing). Mark the correct expression for $\tau$ . (All expressions include $f$ , a numerical factor of order of unity. )

Notes: The moment of inertia of the trailer around its center of mass is approximated as $I = \frac{1}{12} M l^2$ . For simplicity, we do not include the width of the trailer as a separate parameter.

[Photo by Drew Simon](https://www.gobytrucknews.com/new-braking-system-reduces-jack-knifing/123) Photo by Drew Simon

From Wikipedia : A semi-trailer truck, more commonly called a semi truck (often shortened to just "semi"), is the combination of a tractor unit and one or more semi-trailers to carry freight. It is variously known as a transport (truck) in Canada; semi or single in Australia and New Zealand; semi, tractor-trailer, big rig, or eighteen-wheeler in the United States; and articulated lorry, abbreviated artic, in Great Britain and Ireland.

f\sqrt \frac{l}{\mu g}

f\sqrt \frac{\nu l}{\mu g}

f\sqrt \frac{\mu l}{\nu g}

f\sqrt \frac{(\nu-\mu)l}{g}

f\sqrt \frac{l}{\nu g}

f\sqrt \frac{l}{(\nu-\mu)g}

f\sqrt \frac{l}{g}

1 solution

Laszlo Mihaly
Oct 24, 2017

The trailer is pushed back ( and a bit sideways) at the axis where it is joined with the tractor. We represent that force with its components, $P_x$ and $P_y$ . We replace the friction forces acting on the rear wheels (green on the Figure) with a single force at the center of the two wheels (blue on the Figure).

We take a system of reference where the $y$ axis is opposite to the velocity and the $x$ axis is perpendicular to it and the origin is at the connecting axle between the tractor and the trailer. To calculate the positions, we will use the small angle approximation when $\sin \phi \approx \phi$ and $\cos \phi \approx 1$ . Initially the center of mas of the trailer is at $y=l/2$ and $x=(l/2)\phi$ . This leads to the relationship between the angular acceleration of the trailer $\beta=\ddot \phi$ and the acceleration of the center of mass $a_x$ :

$a_x=\frac{l}{2} \beta$

In the $y$ direction, in the small angle approximation the acceleration of the trailer is the same as the acceleration of the front part, $a_y=\nu g$ .

The friction force is in the positive $y$ direction, with magnitude $\frac{1}{2} \mu Mg$ . Its line of action is at a distance of $l\phi$ from the origin and a distance of $(l/2)\phi$ from the center of mass. In the small angle approximation the rotational equation of motion around the center of mass is:

$P_y \frac{l}{2}\phi -\frac{1}{2} \mu Mg \frac{l}{2} \phi - P_x \frac{l}{2}= \beta \frac {1}{12} M l^2$

The equations of motion for the center of mass are:

$P_y+\frac{1}{2}\mu Mg = Ma_y=M \nu g$

$P_x=Ma_x$

We can combine these into a single equation that we need to solve for $\phi$ :

$(M \nu g - \frac{1}{2}\mu Mg) \frac{l}{2} \phi - \frac{1}{2} \mu Mg \frac{l}{2} \phi -M \beta \left(\frac{l}{2}\right)^2 = \beta \frac {1}{12} M l^2$

$(\nu - \mu)g \phi = \frac{4}{3} \ddot \phi$

Assuming $\nu > \mu$ , the solution is

$\phi= \phi_0 \exp t/\tau$

with

$\tau= \sqrt {\frac{4 g}{3(\nu-\mu)l}}$

The less is the friction force at the rear ( $\mu$ ), the shorter is the time to react. The easiest way to stabilize the rig is to reduce the breaking (making $\nu$ smaller). If $\nu < \mu$ , the rig is stable, and the rear end returns to the straight direction, with oscillations.

Trailer swing

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