A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping? (g = 9.8 m/s2)
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µ = v^2/ Rg
To solve this problem, we can use the centripetal force equation:
F = (m * v^2) / r
where
F = centripetal force
m = mass of the car
v = velocity of the car
r = radius of the curve
We also need to consider the maximum static friction force (Fs) that can act on the car to prevent it from slipping.
The maximum static friction force is given by:
Fs = μs * N
Where
Fs = maximum static friction force
μs = coefficient of static friction
N = normal force
In this case, the normal force (N) is equal to the weight of the car (mg).
Therefore, the equation becomes:
Fs = μs * mg
Since we want to find the minimum coefficient of friction, we need to consider the case where the car is at the verge of slipping. Therefore, the static friction force will be equal to the maximum static friction force:
Fs = F
Substituting the values into the equations:
Fs = F
μs * mg = (m * v^2) / r
Simplifying the equation:
μs = (m * v^2) / (r * g)
Now, we can substitute the given values into the equation:
μs = (1,500 * 12^2) / (52 * 9.8)
Calculating the value:
μs = 0.372
Therefore, the minimum coefficient of friction that must exist between the road and tires is approximately 0.372.
To find the minimum coefficient of friction needed to prevent the car from slipping, we can use the centripetal force formula and consider the different forces acting on the car.
First, let's consider the forces acting on the car while rounding the curve:
1. The gravitational force (weight) acting on the car is given by:
F_gravity = m * g
where m is the mass of the car (1,500 kg) and g is the acceleration due to gravity (9.8 m/s^2).
2. The centripetal force acting on the car is given by:
F_centripetal = (m * v^2) / r
where v is the velocity of the car (12 m/s) and r is the radius of the curve (52 m).
3. The frictional force between the tires and the road opposes the motion of the car and provides the centripetal force:
F_friction = μ * F_normal
where μ is the coefficient of friction and F_normal is the normal force acting on the car.
In this case, the normal force is equal to the gravitational force:
F_normal = F_gravity = m * g
Now, to prevent the car from slipping, the frictional force should be equal to or greater than the centripetal force:
F_friction ≥ F_centripetal
Substituting the formulas mentioned above, we get:
μ * F_normal ≥ (m * v^2) / r
Plugging in the values, we have:
μ * (m * g) ≥ (m * v^2) / r
Simplifying the equation:
μ ≥ (v^2) / (r * g)
Now we can calculate the minimum coefficient of friction needed to prevent the car from slipping by plugging in the given values:
μ ≥ (12^2) / (52 * 9.8)
μ ≥ 0.267
Therefore, the minimum coefficient of friction required for the car to prevent slipping is 0.267.