Suppose the coefficient of static friction between the road and the tires on a car is 0.749 and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of 31.7 m radius?
Ff = mv^2/r but Ff = (mu)mg so:
v = sqrt((mu)gr)
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From Newton's 2nd Law, we have ΣF=ma.
In the horizontal direction, the only force acting on the car is the force of friction. Since the car is moving in a circle, we have:
ΣF_x=f=ma_c
where f=force of friction and a_c=centripetal acceleration.
The formula for centripetal acceleration is a_c=v^2/r and the formula for maximum static friction is f_s,max=μN, where μ is the coefficient of static friction and N is normal force. Since ΣF_y=N-mg=0, the magnitude of N must be equal to the magnitude of mg.
Therefore, we have the equation:
μmg=mv^2/r,
v^2=μgr,
⋙ v=sqrt(μgr)
Using μ=0.739, g=9.81 m/s^2, and r=31.7m, we get:
v=sqrt(0.739*9.81*31.7)=15.1595515435 ≈ 15.2 m/s (ans)
To determine the speed at which the car is on the verge of sliding as it rounds a level curve, we can use the concept of centripetal force and friction.
The centripetal force required to keep an object moving in a circular path is given by the equation:
Fc = (m * v^2) / r
where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the curve.
In this case, the centripetal force is provided by the friction between the road and the car's tires:
Fc = μ * N
where μ is the coefficient of static friction and N is the normal force between the car and the road.
Since the car has no negative lift, the normal force (N) is equal to its weight (mg), where g is the acceleration due to gravity.
Setting the centripetal force equation equal to the friction equation, we have:
(μ * N) = (m * v^2) / r
Substituting N = mg:
μ * mg = (m * v^2) / r
Simplifying the equation and solving for v, we get:
v^2 = μ * g * r
v = √(μ * g * r)
Now we can substitute the known values into the equation to find the speed:
μ = 0.749 (coefficient of static friction)
g = 9.8 m/s^2 (acceleration due to gravity)
r = 31.7 m (radius of the curve)
v = √(0.749 * 9.8 * 31.7)
v ≈ √231.837
v ≈ 15.228 m/s
Therefore, the speed that will put the car on the verge of sliding as it rounds the level curve of 31.7 m radius is approximately 15.228 m/s.