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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 and back wheel |
Answer» Solution : For TRANSLATION equilibrium of CAR `N_(F) + N_(B) = W = 1800 XX 9.8 = 17640 N` For rotational equilibrium of car `1.05 N_(F) = 0.75 N_(B)` `1.05 N_(F) = 0.75 (17640 - N_(F))` `1.8 N_(F) = 1320` `N_(F) = 1320//1.8 = 7350 N` `N_(B) = 17640 - 7350 = 10290` N Force on each front wheel `=(7350)/2 = 3675` N Force on each back wheel `=(10290)/2 = 5145` N |
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