Equilibrium - Excercise 1

A car is placed on a ramp.
Given: G=10kN, α=30°
Searched: Reaction Forces A, B

Let's cut the car out of the system to see the bearing forces. Apart from the two supports A and B, there is also the rope force S.

To avoid the calculation of rope force S, we choose:
$\mathrm{\Sigma }\mathrm{F}\uparrow =0$
$\mathrm{\Sigma }{\mathrm{M}}_{\mathrm{S}}=0$
Using the forces from the free-body diagram leads to:
$\mathrm{\Sigma }{\mathrm{M}}_{\mathrm{S}}=0={\mathrm{G}}_{\mathrm{N}}·2\mathrm{a}-\mathrm{B}·\mathrm{a}-\mathrm{A}·3\mathrm{a}$
$\mathrm{\Sigma }\mathrm{F}\uparrow =0=-\mathrm{G}+{\mathrm{A}}_{\mathrm{V}}+{\mathrm{B}}_{\mathrm{V}}$
This is a system of equations with 2 equations. If we translate the components into A, B and G, then it will be a system with 2 equations and 2 unknown variables.

${\mathrm{G}}_{\mathrm{N}}=\mathrm{G}·\sin \left(60°\right)$
${\mathrm{A}}_{\mathrm{V}}=\mathrm{A}·\sin \left(60°\right)$

Result:
A = 0,291 G = 2,91 kN
B = 0,859 G = 8,59 kN