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11 December, 05:26

A 4800 kg truck is parked on a 12 ∘ slope. how big is the friction force on the truck? the coefficient of static friction between the tires and the road is 0.90.

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  1. 11 December, 06:42
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    The amount of friction on truck required to stop is = mass*g*sin (12)

    = 4800*9.8*sin (12)

    = 9780.16 newton

    The maximum friction = coefficient of friction * force between tyres and road

    Since maximum friction is more than required force to park the truck. Only required amount of friction will act.
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