Canton avenue in pittsburgh, pa is considered to be one of the steepest roads in the world with a grade of 37%.
a. assuming no friction on a particularly icy day, what would be the acceleration of a 1,000 kg car with only
gravity acting on it?
b. the force due to friction is equal to the product of the force perpendicular to the road and the coefficient of
friction for icy roads and a non-moving vehicle, assume the coefficient of friction is = 0.3. find the force
due to friction for the car above. if the car is in park, will it begin sliding down canton avenue if the road is this
icy?
c. assume the coefficient of friction for moving cars on icy roads is = 0.2. what is the maximum angle of road
that the car will be able to stop on?

a)  a = 3.53 m/sec²

b ) As force in the sense of movement is bigger than Fr  (that is  m*g*sin21 > Fr ) the car will slide down.

c) β = 11⁰

Step-by-step explanation: We must start calculating the angle of the road.

37% means that inclination is such that for 100 m of horizontal distance we will be up the path a 37 m, then tan α = 0.37  and α = 21⁰

a) According to Newtons second law

∑ Fx  =  ma    ⇒   if only gravity is acting

m*g* sinα = ma     ⇒  1000 Kg*9.8 m/sec² *sin 21°  = 1000 Kg*a

9.8 * 0,36 = a      ⇒   a = 3.53 m/sec²

b) Fr  =  μ*Fn      

∑ Fy  = 0  

m*g*cos21° - Fn  =  0        Fn  =  1000*9.8*cos21°     Fn  = 9143.4 [N]

Then  Fr = 9143.4 * 0,3     ⇒  Fr = 2743.02 [N]

And  

m*g*sinα   = m*g*sin21⁰   =  1000* 9.8* 0.36  =  3528 [N]

As force in the sense of movement is bigger than Fr  (that is  m*g*sin21 > Fr ) the car will road down.

c) If μk  =  0.2

m*g*sinβ   = 9800* sinβ - 9143.4*0.2 = 0

sinβ  = 1828.68/9800     ⇒   sinβ  =  0.1866   ⇒ β = 11⁰

no idea

step-by-step explanation: