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_{g}sinθ

F_{gy}=F_{g}cosθ

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

mv^{2}/r=mgsinθ+µ_{s}F_{n}

mv^{2}/r=mgsinθ+µ_{s}F_{gy}

mv^{2}/r=mgsinθ+µ_{s}mgcosθ

Solve for μ_{s}

μ_{s}=(mv^{2}/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