newton’s Law Question HELP 2!?

Question

A highway curve with a radius of curvature of 100 meters is banked at an angle of 15°. 
a. Calculate the vehicle speed for which this curve is suitable if there is no friction between the road and the tires of the vehicle.
On a dry day, when friction is present, an automobile successfully navigates the curve at a speed of 25 m/s. 
b. Determine the minimum coefficient of friction required to prevent this automobile from sliding while going around the curve.

Anonymous · Accepted Answer

first case, no frictionwe use a  standard x, y coordinate system even though the track is bankedthe forces on the car are the horizontal and vertical components of the normal force (N) and the weightin the vertical direction, the vertical component of N balances the weight since the car does not accelerate vertically, soN cos(theta) = mg (eq. 1)  draw a diagram showing the normal force (which is perpendicular to the track) and be sure you see what the vertical component is N cos(theta) where theta is the bank angle)the horizontal component of the normal force must match the centripetal force, so we haveN sin(theta)=mv^/r  (eq.2)now, divide eq. 2 by e1. 1:Nsin(theta)/Ncos(theta)=v^2/rgtan(theta)=v^2/rgfor r=100m and theta =15 deg, we have that v=sqrt[ r g tan(theta)]v=16.2 m/ssecond case, with frictionto keep the car from sliding up the track at 25m/s, there must be friction opposing the motionthere is both a horizontal component of friction and a vertical component of friction; the horizontal component of friction acts toward the center of the curve and has magnitude f cos(theta) where f is the total frictional force, the vertical component acts down with magnitude f sin(theta)remembering that f=  u N where u is the coefficient of friction, our force laws become:in the x direction: N sin(theta) + f cos(theta) = mv^2/r                      Nsin(theta)-+u N cos(theta)=mv^2/r                      N[sin(theta)+u cos(theta)]=mv^2/r (eq.3)in the y direction: N cos(theta)-mg &ndash; fsin(theta)=0                      N[cos(theta)- u sin(theta)]=mg  (eq.4)divide eq. 3 by e1 4N[sin (theta) + u cos(theta)]/N[cos(theta)- u sin(theta)]=mv^2/r/(mg)[sin(theta)+u cos(theta)]/[cos(theta)-u sin(theta)=v^2/rgknowing theta =15 deg, v=25m/s, r=100 m and g=9.8m/s/s, you can solve for u

Dr. Virgie Abshire Sr.

newton’s Law Question HELP 2!?

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