1.

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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