3.A car with a mass of 1500 kg is parked on a hill with an incline of 20°.
a.What is the weight of the car?
b.What is the normal force the car experiences? (hint: balance the forces parallel and perpendicular to the hill using Newton’s second law)
c.What is the magnitude of the force of static friction that keeps the car from sliding down the hill? (hint: balance the forces parallel and perpendicular to the hill)
d.What is the coefficient of static friction? (hint: use the answers you found in b and c to solve for mu. If you did not find answers for b and c, assume N=10000N and f=5000N – note that these are not the correct numbers. They are only here so you can work part d, even if you can’t get b and c)
a. The weight of the car can be calculated using the formula w = m * g, where m is the mass of the car and g is the acceleration due to gravity (approximately equal to 9.8 m/s^2). So, w = 1500 kg * 9.8 m/s^2 = 14700 N.
b. The normal force the car experiences is equal to the component of the weight perpendicular to the hill. Therefore, the normal force can be calculated as N = w * cos(θ), where θ is the angle of incline (20°). N = 14700 N * cos(20°) = 13693 N.
c. The magnitude of the force of static friction that keeps the car from sliding down the hill can be calculated as the component of the weight parallel to the hill. Therefore, the force of static friction can be calculated as f = w * sin(θ), where θ is the angle of incline (20°). f = 14700 N * sin(20°) = 5016 N.
d. The coefficient of static friction can be determined by dividing the force of static friction by the normal force. Therefore, the coefficient of static friction (μ) is given by μ = f / N. μ = 5016 N / 13693 N = 0.3665 (approximately).
Note: These calculations are based on the assumption that the car is parked and stationary on the hill.
a. To find the weight of the car, we can use the formula:
Weight = mass × gravitational acceleration
Given that the mass of the car is 1500 kg and the value of gravitational acceleration is typically 9.8 m/s², we can calculate the weight:
Weight = 1500 kg × 9.8 m/s² = 14,700 N (Newtons)
Therefore, the weight of the car is 14,700 Newtons.
b. The normal force experienced by the car is the force exerted by the hill perpendicular to its surface. To calculate it, we can use the formula:
Normal force = weight × cos(angle of incline)
In this case, the angle of incline is 20°. Using the weight calculated in part a:
Normal force = 14,700 N × cos(20°) ≈ 13,859 N (rounded to three significant figures)
Therefore, the normal force experienced by the car is approximately 13,859 Newtons.
c. The force of static friction that keeps the car from sliding down the hill can be calculated using the formula:
Force of static friction = coefficient of static friction × normal force
We need to find the force of static friction, given that the normal force is the same as the value calculated in part b. Let's assume it is 13,859 N for simplicity. Therefore, we have:
Force of static friction = coefficient of static friction × 13,859 N
We do not have any information to calculate the coefficient of static friction, so we cannot determine its exact value.
d. In order to determine the coefficient of static friction, we need the exact value of the force of static friction, which we do not have. Without this information, we cannot calculate the coefficient of static friction accurately.
To answer these questions, we need to consider the forces acting on the car while it is parked on the hill. Here's how you can find the answers:
a. Weight of the car:
The weight of an object is given by the equation: Weight = mass * gravitational acceleration. In this case, the mass of the car is 1500 kg, and the gravitational acceleration is approximately 9.8 m/s². Therefore, the weight of the car is: Weight = 1500 kg * 9.8 m/s² = 14,700 N.
b. Normal force on the car:
The normal force is the force exerted by a surface to support the weight of an object resting on it. Since the car is parked on an inclined hill, we need to consider the forces acting parallel and perpendicular to the hill. The normal force cancels out the component of the weight perpendicular to the hill. Using Newton's second law (ΣF = m * a), we can find the normal force. In this case, the car is parked, hence not accelerating. So, the acceleration (a) is 0. Therefore, using ΣF = m * a, the normal force is equal to the weight, which is 14,700 N.
c. Force of static friction:
The force of static friction acts parallel to the hill and prevents the car from sliding down. When the car is parked, the frictional force balances the component of the weight parallel to the hill. So, the force of static friction (f) is equal to the weight component parallel to the hill. To find this, we need to calculate the sine of the angle of the incline and multiply it by the weight of the car.
Weight component parallel to the hill = Weight * sin(θ)
θ is the angle of the incline, which is 20° in this case.
Weight component parallel to the hill = 14,700 N * sin(20°) = 4,990.3 N
Therefore, the magnitude of the force of static friction that keeps the car from sliding down the hill is approximately 4,990.3 N.
d. Coefficient of static friction (μ):
The coefficient of static friction (μ) is the ratio of the force of static friction to the normal force (μ = f/N). Using the values we found for the force of static friction (f = 4,990.3 N) and the normal force (N = 14,700 N), we can calculate the coefficient of static friction.
μ = f/N = 4,990.3 N / 14,700 N ≈ 0.339
Therefore, the coefficient of static friction is approximately 0.339.
a. Fc = mg = 1500kg * 9.8N/kg = 14700N.
Wc = 1500kg / 0.454kg/lb = 3304lbs. =
weight of car in lbs.
b. Fv = mgcos20 = 14700cos20 = 13,814N.
= Force perpendicular to the hill.
c. Ff = Fp = Force parallel to the hill.
Ff = Fp = mg*sin20 = 14,700sin20 = 5028N.
d. u = Ff / Fv = 5028 / 13814 = 0.364.