The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A, it is traveling at 9 m/s and increasing its speed at 3 m/s2 . Determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at this instant. Neglect the size of the car.

The image is missing, so i have attached it.

resultant normal force; N= 6727.9 N

resultant frictional force;F_f = -1144.33 N

Explanation:

From the image,

y = 20(1 - x²/6400)

Expanding, we have;

y = 20 - 20x²/6400)

dy/dx = -40x/6400

From the diagram, x = 80

At x = 80,

dy/dx = -40(80)/6400

dy/dx = -0.5

Also, d²y/dx² = -40/6400

d²y/dx² = -1/160

Now,

The radius of curvature is;

R = [(1 + (dy/dx)²)^(3/2)]/(d²y/dx²)

Plugging in the relevant values;

R = [(1 + (-0.5)²)^(3/2)]/(-1/160)

R = -223.61m

But we'll take the absolute value as radius cannot be negative.

Thus;

R = 223.61m

We know that acceleration (a_n) = v²/R

Thus, a_n = 9²/223.61

a_n = 81/223.61

a_n = 0.3622 m/s²

Now, to get the resultant normal force. From the diagram, resolving forces, gives;

N = W•cosθ - m•a_n

We are given; m = 0.8 x 10³ kg

Now, tan θ = dy/dx

And dy/dx at the distance of 80 = -1.5

Thus,tanθ = - 1.5

θ = tan^(-1)(-1.5)

θ = 26.6°

(the negative sign was ignored)

Thus;

ΣF_n;

N = mg•cos26.6 - m•a_n

N = (0.8 x 10³ x 9.81 x 0.8942) - (0.8 x 10³ x 0.3622)

N = 6727.9 N

Now for the resultant frictional force;

ΣF_t;

F_f + Wsin26.6 = m(v/t)

Where;

F_f is resultant frictional force

v/t is rate of change of velocity which is given as 3 m/s²

Thus;

F_f = m(v/t) - Wsin26.6

F_f = (0.8 x 10³ x 3) - (0.8 x 10³ x 9.81 x 0.4478)

F_f = 2400 - 3514.33

F_f = -1144.33 N

found in the nucleus

has mass of one amu