|Class 11 Physics Rotational Motion||Centre of Gravity|
Centre of Gravity
Example: A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solution: Equating forces in vertical direction (translational equilibrium) - Rf + Rb = mg - (1)
Equating the torques at CG (Rotational equilibrium) - Rf (1.05) = Rb (1.8-1.05) - (2)
Note: We have chosen CG as the torque of gravitational forces is zero at this point.
Solving Eqn (1)&(2) we can find Rf & Rb .