A 4500 kg truck is parked on a 19° slope. How big is the friction force on the truck?. Created by Dan. science-mathematics-en - physics-en

mgsin(19)-f=malet the truck is at rest so a=0mgsin(19)=ff=4500*9.8*0.325 =14357.55

A 4500 kg truck is parked on a 19° slope. How big is the friction force on the truck?

Dan @Dan

May 2021 2 35 Report

A 4500 kg truck is parked on a 19° slope. How big is the friction force on the truck?

A 4500 kg truck is parked on a 19° slope. How big is the friction force on the truck?

Science & Mathematics / Physics

oldprof

As the truck is not accelerating, we have the net force f = ma = 0 = w - F = W sin(theta) - F; where W = mg = 4500*9.8 = ? is the weight; so that the friction force F = w = mg sin(theta) = 4500*9.8*sin(19) = 14357.55561 = 14357.6 Newtons. ANS.

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