Ff = mv^2/r but Ff = (mu)mg so:
v = sqrt((mu)gr)
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?
3 answers
jjh
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)
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)