A car with mass (452) kg attempts to make a turn with a radius of (31.8) m at a speed of (21.5) m/s on a horizontal surface. What is the minimum value for the static coefficient of friction between the tires and the roadway necessary for the car to make the turn without sliding? Give your answer with 3 significant figures.
set friction force = centripetal force, solve for mu.
I got (mu)(4434)=(425)(21.5^2)/31.8
Would the answer be 1.39 ?? Thank you
To determine the minimum value for the static coefficient of friction required, we can calculate the maximum centripetal force acting on the car.
The centripetal force (Fc) required to keep an object of mass (m) moving in a circle with radius (r) and speed (v) can be calculated using the following formula:
Fc = (m * v^2) / r
Substituting the given values:
Fc = (452 kg * (21.5 m/s)^2) / 31.8 m
Fc ≈ 650.696 kg m/s^2
The centripetal force required is approximately 650.696 kg m/s^2.
The frictional force (Ff) between the tires and the roadway must provide this centripetal force to prevent the car from sliding. The maximum frictional force can be determined using the formula:
Ff = μ * N
where μ is the static coefficient of friction, and N is the normal force acting on the car.
Since the car is on a horizontal surface, the normal force (N) is equal to the weight of the car (mg), where g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = m * g
N = 452 kg * 9.8 m/s^2
N ≈ 4433.6 kg m/s^2
Substituting the values into the equation for the maximum frictional force:
Ff = μ * N
650.696 kg m/s^2 = μ * 4433.6 kg m/s^2
Rearranging the equation to solve for μ:
μ = (650.696 kg m/s^2) / (4433.6 kg m/s^2)
μ ≈ 0.147
The minimum value for the static coefficient of friction required for the car to make the turn without sliding is approximately 0.147.
To solve this problem, we need to use the concept of centripetal force. In order for the car to make the turn without sliding, the static friction between the tires and the roadway must provide the necessary centripetal force.
The centripetal force (Fc) is given by the formula:
Fc = (m * v^2) / r
where:
- m is the mass of the car
- v is the velocity of the car
- r is the radius of the turn
Substituting the given values into the formula:
Fc = (452 kg * (21.5 m/s)^2) / 31.8 m
Calculating Fc:
Fc = (452 kg * 462.25 m^2/s^2) / 31.8 m
= 6623.9 N (rounded to four significant figures)
Since the maximum static friction force (Fs max) is equal to the coefficient of static friction (µs) multiplied by the normal force (N), we need to find the normal force acting on the car. On a horizontal surface, the normal force (N) is equal to the weight of the car (mg).
The weight of the car (mg) is given by:
mg = (452 kg) * (9.8 m/s^2)
= 4429.6 N (rounded to four significant figures)
Now, let's calculate the minimum value for the static coefficient of friction (µs) needed:
Fs max = µs * N
Since Fs max is equal to the centripetal force (Fc), we can substitute the values:
6623.9 N = µs * 4429.6 N
Solving for µs:
µs = 6623.9 N / 4429.6 N
= 1.494 (rounded to three significant figures)
Therefore, the minimum value for the static coefficient of friction between the tires and the roadway necessary for the car to make the turn without sliding is 1.494.