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?
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
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
Well, we're dealing with some physics here, so let's see if I can help you out! To find the force exerted by friction, we need to consider a couple of things.
First, let's figure out the gravitational force acting on the car. That's just the mass of the car (600 kg) multiplied by the acceleration due to gravity (which is approximately 9.8 m/s^2). So the gravitational force is about 5880 N.
Next, we need to calculate the normal force acting on the car. This force is perpendicular to the surface of the road and is equal to the gravitational force, but pointing in the opposite direction. So the normal force is also 5880 N.
Now, for the fun part - friction! Since the car is going around a curved surface, there needs to be a centripetal force acting on it to keep it in motion. And this centripetal force is provided by the force of friction between the car's tires and the road.
The formula for the centripetal force is Fc = mv^2/r, where m is the mass of the car, v is the velocity, and r is the radius of the curve. Plugging in the values we have, we get Fc = (600 kg)(30.0 m/s)^2 / 120 m, which is equal to 45000 N.
But we're not done yet! We need to determine the maximum force of static friction that the surface can provide. The formula for static friction is Fs = μsN, where μs is the coefficient of static friction and N is the normal force. Plugging in our values, we get Fs = (0.300)(5880 N), which is equal to 1764 N.
Since the centripetal force (Fc) is less than the maximum force of static friction (Fs), the force of friction will be equal to the centripetal force, which is 45000 N.
So, the force exerted by friction on the car is approximately 45000 N. But remember, this is just a calculated value and might differ in reality. Alright, that was quite a ride! I hope I was able to help.
To find the force of friction, we need to consider two forces acting on the car: the gravitational force (weight) and the normal force.
First, let's find the gravitational force acting on the car:
Weight = mass * gravity
Weight = 600 kg * 9.8 m/s^2
Weight = 5880 N
Next, we need to find the normal force. The normal force is the component of the weight perpendicular to the inclined surface.
Normal force = Weight * cos(angle of inclination)
Normal force = 5880 N * cos(25°)
Normal force = 5339 N
Now, we can find the force of friction. The force of friction acts in the opposite direction to the direction of motion, so it points inward towards the center of the curve.
Force of friction = Normal force * coefficient of static friction
Force of friction = 5339 N * 0.300
Force of friction = 1601.7 N
Therefore, the force exerted by friction on the car is 1601.7 N.