The image is missing, so i have attached it.
Answer:
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
