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a 4500 kg truck is parked on a 7.0∘ slope. how big is the friction force on the truck?

A 4500 kg truck is parked on a 7.0∘ slope. How big is the friction force on the truck?

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asked by Elias
Nov 12, 2017
M*g = 4500 * 9.8 = 44,100N. = Wt. of the truck.

Fp = 44,100*sin7 = 5374 N. = Force parallel with the slope = Friction force required to prevent the truck from moving.

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posted by Henry
Nov 12, 2017
thanks guys

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posted by Finna Clash
Nov 21, 2017

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