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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asked Oct 11, 2019 77.7k views

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Pavel Polivka · Answer 1 · 2019-10-18T02:09:14+0000

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.

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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