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# Car on a sloping road

A car is parked on a sloping road with the handbrake on. The handbrake acts on the rear wheels. What is the minimum coefficient of static friction required to prevent the car from slipping?

## Solution

The following video is in German language, but English subtitles are available.

The coordinate system will be turned by the angle Alpha for the solution.

Balance of forces in x-direction

$\tag{1} 0={F_{\mathit{Bx}}}-G \sin{\left( \alpha \right) }$

Balance of forces in y-direction

$\tag{2} 0=-G \cos{\left( \alpha \right) }+{F_{\mathit{By}}}+{F_A}$

Balance of forces around A (left-turning moments are positive)

$\tag{3} 0=G a \sin{\left( \alpha \right) }-G b \cos{\left( \alpha \right) }+{F_{\mathit{By}}} \left( c+b\right)$

Relationship between tangential and normal force on the rear wheel

$\tag{4} {F_{\mathit{Bx}}}={F_{\mathit{By}}} {µ_0}$

Solved for FBy and used in the balance of moments

$\tag{5} {F_{\mathit{By}}}=\frac{{F_{\mathit{Bx}}}}{{µ_0}}$

$\tag{6} 0=G a \sin{\left( \alpha \right) }-G b \cos{\left( \alpha \right) }+\frac{{F_{\mathit{Bx}}} \left( c+b\right) }{{µ_0}}$

Equation 1 solved for FBx and used in the previous result

$\tag{7} {F_{\mathit{Bx}}}=G \sin{\left( \alpha \right) }$

$\tag{8} 0=\frac{G\, \left( c+b\right) \sin{\left( \alpha \right) }}{{µ_0}}+G a \sin{\left( \alpha \right) }-G b \cos{\left( \alpha \right) }$

The minimum value for the coefficient of static friction is

$\tag{9} {µ_0}=\frac{\left( c+b\right) \sin{\left( \alpha \right) }}{b \cos{\left( \alpha \right) -}a \sin{\left( \alpha \right) }}$

It is obvious that the weight does not play a role in the determination of the required coefficient of static friction, i.e. a discussion of the distribution of the weight force over several wheels per axle is superfluous.