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# Determination of the reaction forces

A railway wagon stands on a sloping track on a (frictionless) buffer stop. Which reaction forces occur on the wheels and on the buffer stop?

## Solution

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#### Solution approach

$\require{cancel}$ $\newcommand{\myvec}[1]{{\begin{pmatrix}#1\end{pmatrix}}}$

The coordinate system is rotated by the angle α to solve the problem. Left turning moments are positive.

The equilibrium of forces in the x-direction is

$\tag{1} \sum F_x = 0 = F_3 - G \cdot \sin \alpha$

$\tag{2} F_3 = G \cdot \sin \alpha$

The equilibrium of forces in the y-direction is

$\tag{3} \sum F_y = 0 = F_1 - G \cdot \cos \alpha + F_2$

Due to the geometric specifications, the equilibrium of moments can only be considered around G and leads to

$\tag{4} \sum M(G) = 0 = F_3 \cdot h - F_1 \cdot l + F_2 \cdot l$

$\tag{5} 0 = G \cdot \sin \alpha \cdot h - F_1 \cdot l + F_2 \cdot l$

$\tag{6} F_2 = \frac{F_1 \cdot l - G \cdot \sin \alpha \cdot h}{l}$

From equation 3 follows

$\tag{7} 0 = F_1 - G \cdot \cos \alpha + \frac{F_1 \cdot l - G \cdot \sin \alpha \cdot h}{l}$

$\tag{8} G \cdot \cos \alpha = F_1 + \frac{F_1 \ \cdot \bcancel{l}}{\bcancel{l}} - \frac{G \cdot \sin \alpha \cdot h}{l}$

$\tag{9} F_1 = \frac{G}{2} \cdot \left( \cos \alpha + \frac{\sin \alpha \cdot h}{l} \right)$

F2 follows from equation 6

$\tag{10} F_2 = \frac{\frac{G}{2} \cdot \left( \cos \alpha + \frac{\sin \alpha \cdot h}{l} \right) \cdot l - G \cdot \sin \alpha \cdot h}{l}$

$\tag{11} F_2 = \frac{ \frac{G}{2} \cdot \cos \alpha \cdot l + \frac{G}{2} \cdot \frac{\sin \alpha \cdot h}{\bcancel{l}} \cdot \bcancel{l} - G \cdot \sin \alpha \cdot h}{l}$

$\tag{12} F_2 = \frac{ G \cdot \cos \alpha \cdot l - G \cdot \sin \alpha \cdot h}{2 \cdot l}$