A curve of radius 161 m is banked at an angle of 12°. An 840-kg car negotiates the curve at 94 km/h without skidding. Neglect the effects of air drag and rolling friction. Find the following.
(a) the normal force exerted by the pavement on the tires
(b) the frictional force exerted by the pavement on the tires
(c) the minimum coefficient of static friction between the pavement and the tires
centripetal acceleration along the slope:
m(VcosTheta)/r
friction force:
mg*mu*cosTheta
gravity down the slope:
mg (sinTheta)
set gravity down+friction=centripetal all along the slope, of course.
set them equal
Ah I see thank you, I had tried a different setup before which got me the wrong answer. But this helps a bunch thanks!
To find the required values, we can use the concepts of circular motion and the forces acting on the car.
Let's start by finding the normal force exerted by the pavement on the tires (a).
(a) The normal force, denoted as N, is the force exerted perpendicular to the surface of contact between the car and the pavement. In this case, it acts in the upward direction to balance the weight of the car and provide the necessary centripetal force for circular motion.
To determine the normal force, we first need to analyze the forces acting on the car. The forces involved are:
1. Weight (mg) acting vertically downwards, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Normal force (N) acting vertically upwards.
3. Frictional force (f) acting horizontally inwards towards the center of the curve.
In the absence of skidding, the frictional force provides the centripetal force required for circular motion:
f = m * (v^2 / r),
where v is the velocity of the car and r is the radius of the curve.
Substituting the given values:
m = 840 kg,
v = 94 km/h = 94,000 m/3600 s ≈ 26.11 m/s,
r = 161 m,
we can calculate the frictional force f.
Since the car is not skidding, the frictional force f is equal to the total necessary centripetal force for circular motion:
f = m * (v^2 / r) = N.
Now we can substitute the values and calculate the normal force:
N = m * (v^2 / r) = 840 kg * (26.11 m/s)^2 / 161 m = 34,244.99 N ≈ 34,245 N.
Therefore, the normal force exerted by the pavement on the tires is approximately 34,245 Newtons.
(b) To find the frictional force exerted by the pavement on the tires, we can use the same equation used to determine the centripetal force in part (a).
The frictional force (f) is given by:
f = m * (v^2 / r) = 840 kg * (26.11 m/s)^2 / 161 m = 34,244.99 N ≈ 34,245 N.
Therefore, the frictional force exerted by the pavement on the tires is approximately 34,245 Newtons.
(c) Now, let's find the minimum coefficient of static friction between the pavement and the tires.
The coefficient of static friction, denoted as μs, represents the maximum value of friction that can act between two surfaces to prevent slipping or skidding.
The equation for the coefficient of static friction is:
μs = f / N,
where f is the frictional force and N is the normal force.
Substituting the given values:
μs = (34,245 N) / (34,245 N) = 1.
Therefore, the minimum coefficient of static friction between the pavement and the tires is 1.