A 1400 kg car rounds a curve of 57 m banked at an angle of 14 degrees
If the car is traveling at 98 km/h, will a friction force be required? If so, in what direction?
May someone please walk me through this problem step by step? Thank you
To determine if a friction force will be required and its direction, we need to analyze the forces acting on the car.
Step 1: Calculate the centripetal force:
The centripetal force is the force acting towards the center of the circular path and is given by the equation: Fc = (m * v^2) / r
Where:
- m is the mass of the car (1400 kg in this case)
- v is the velocity of the car (98 km/h)
- r is the radius of the curve (57 m)
First, we need to convert the velocity from km/h to m/s:
98 km/h * (1000 m/1 km) * (1 h/3600 s) = 27.2 m/s
Now we can calculate the centripetal force:
Fc = (1400 kg * (27.2 m/s)^2) / 57 m = 17,548.07 N
Step 2: Analyze the horizontal and vertical forces:
The forces acting on the car are the gravitational force (mg) acting vertically downward and the normal force (N) from the banked curve acting vertically upward.
Since the curve is banked, the horizontal component of the normal force (N horizontal) will provide the necessary centripetal force. The friction force (f) will also act horizontally and may assist or oppose the motion of the car depending on its direction.
Step 3: Determine the direction and magnitude of the friction force:
To find out if a friction force is required, we can compare the horizontal component of the normal force (N horizontal) with the centripetal force.
N horizontal = N * sin(θ)
Where θ is the angle of banking (14 degrees)
Therefore, N horizontal = N * sin(14°)
If N horizontal is greater than the centripetal force (17,548.07N), no friction force will be needed. However, if N horizontal is less than the centripetal force, a friction force will be required.
Step 4: Calculate the magnitude of N horizontal:
Since the car is not slipping vertically on the curve, the vertical forces must be balanced. The vertical component of the normal force must equal the weight of the car (mg).
N * cos(θ) = mg
N = (mg) / cos(θ)
Substituting the known values:
N = (1400 kg * 9.8 m/s^2) / cos(14°) = 10,453.50 N
Now, we can calculate N horizontal:
N horizontal = N * sin(14°) = 10,453.50 N * sin(14°) = 2,945.42 N
Step 5: Compare N horizontal with the centripetal force:
N horizontal (2,945.42 N) is less than the centripetal force (17,548.07 N). Therefore, a friction force will be required to provide the remaining centripetal force.
Since the car is going to the left (neglecting any other external factors), the friction force should act to the right, opposing the motion of the car.
In conclusion, a friction force will be required to provide the remaining centripetal force, and it should act in the direction opposite to the motion of the car.
Let the x-axis point toward the center of curvature and the y-axis point upward. Use Newton’s second law
ΣFy = N• cos θ = mg, N
ΣFx = N• sin θ = m•v²/R,
Dividing the 2nd equation by the 1st equation, we obtain
tan θ = v² /R•g
v=sqrt(R•g•tan θ)=sqrt(57•9.8•tan14)=11.8 m/s,
V = 98 km/h =98000/3600 =27.2 m/s
V>v => the friction force is required (directed to the center of the curvature)