Minimum Coefficient of static friction

Question

A curve that has a radius of 95m is banked at an angle of 10.2 degrees If a 1000kg car navigates the curve at 70km/h without skidding, what is the minimum coefficient of static friction  between the pavement and the tires?

Answer 1

Given:

r=95m

θ=10.2

m=1000kg

v=70km/hr (convert to m/s) 70km/hr(1000m/1km)*(1hr/3600s)=19.4m/s

Question:

µ_s=?

First draw a free body diagram what are your forces?

F_g pulls straight down

F_n is perpendicular to the surface

F_fs is going to point along the surface "down" the angle (or towards the center of the circle

If we define x as along the surface and y as perpendicular to it:

F_n has no x-component and F_fs has no y-component so we only have to break gravity up.

Draw a triangle and you should get:

F_gx=F_gsinθ

F_gy=F_gcosθ

Now that we make Fnet equations:

Fnet_y=F_n-F_gy Since there is no acceleration in the y plug in zero F_n=F_gy

The x is a little trickier, are F_gx and F_fs pointing in the same or opposite directions? They both point "down" the incline, or towards the center of the circle.

Fnet_x=F_gx+F_fs

Now the net force in the x provides our acceleration, since our acceleration is centripetal not tangential the net force is the centripetal force.

F_c=F_gx+F_fs

Then substitute in

mv²/r=mgsinθ+µ_sF_n

mv²/r=mgsinθ+µ_sF_gy

mv²/r=mgsinθ+µ_smgcosθ

Solve for μ_s

μ_s=(mv²/r-mgsinθ)/(mgcosθ)

plug in values

μ_s=((1000)(19.4)²/(95)-(1000)(9.81)cos(10.2))/((1000)(9.81)sin(10.2))

μ_s=0.23