Question

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

Answer 1

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.

Answer 2

thanks guys

Answer 3

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