*Originally posted by Nemesio*

**Generally, when I see those grade % signs, it's accompanied by a maximum speed limit for
**

trucks weighing over a certain amount. I suspect it's because the increased angle can lead to

losing traction with the ground and slipping.

How much more traction does a body need to have to keep from slipping when it's on a hill

10% steep? 20%?

Nemesio

If I screw this one up, my high school physics teacher is going to kill me from

*beyond the grave*.

For a truck moving at constant velocity applying its brakes on a flat surface, we have the following:

Fn = Fg

mu = Fb / Fn

where:

Fn = normal force

Fg = gravitational force

Fb = force due to braking

mu = coefficient of friction (either rolling or static, whichever is appropriate)

Therefore, the maximum braking force that can be applied without slipping is Fb = mu*Fn.

For a truck initially moving at constant velocity applying its brakes on a downward slope with angle of elevation theta, where theta = Tan^(-1)(grade in % ), we have:

Fn = Fg*cos(theta)

Fa = Fg*sin(theta)

where Fa = force of acceleration due to the angle of elevation. Also:

Fb = mu*Fn = mu*Fg*cos(theta)

This is the maximum amount of braking force that can be applied, which is strictly less than the amount that can be applied on a flat surface. Also, the braking becomes more ineffective and the push from the ramp becomes more effective as the angle of elevation increases, according to the formula for the resultant force (in the "down-the-ramp" direction):

Fr = Fa - Fb = Fg*sin(theta) - mu*Fg*cos(theta) = Fg*(sin(theta) - mu*cos(theta))

Letting Fr = 0 (the break-even point where the truck starts going down the hill regardless of braking power), we can see that:

sin(theta) = mu*cos(theta)

mu = tan(theta)

So for a given mu, an angle of elevation exceeding Tan(-1)(mu) will result in horrible, firey death. ðŸ™‚