Asked by Sandhya

A 600-kg car is going around a curve with a radius of 120 m that is banked at an angle of 25.0° with a speed of 30.0 m/s.The coefficient of static friction between the car and the road is 0.300. What is the force exerted by friction on the car?

Answered by Damon
Inward force needed =m a = m v^2/R
= 600 (900)/120 = 4500 Newtons inward
down slope component = 4500 cos 25

Force up slope = Ff friction
weight down slope = m g sin 25
= 2488

Net force down slope = 2488 - Ff

so
2488-Ff = 4500 cos 25
Ff = 2488 - 4078
Ff = -1590 N so 1590 DOWN the slope

Can that .3 coef of friction provide the 1590 N ?
normal component of weight = 600 * 9.81 * cos 25 = 5334, yes, good, we do not slip up or down
Answered by Sierra
What does force up slope? Do you have a free-body diagram to better explain your forces?
Answered by Onkar
6. A 600 kg car traveling at 30m / s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coefficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car? (tan 25 deg = 0.47)

(b) 3430 N

(d) 7820 N

7

10.

(c) 7240 N

(a) 1590 N
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