The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m>s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at the instant it reaches point A. Neglect the size of the car.

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Question

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 . 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

Recalling the fact that the statement can be related to that of geometry as represented in the diagram and illustration below

Therefore, referencing the question and calling the knowledge of geometry, we have:

dy / dx = - 0.00625x and

d²y / dx² =  - 0.00625x .

Also, from the diagrammatic illustration, we can see that the slope angle tan θ at point A is given by

tan θ = dy / dx ║ ₓ = ₈₀ₙ  = - 0.00625(80)  

Therefore, tan θ =  - 26.57°

also, if we consider the radius. The radius of curvature at point A is

ρ = [ 1 + (dx / dy )² ] ³⁺² / d² y / dx²  = [ 1 + ( -0.00625x)² ] ³⁺²║ ₓ = ₈₀ₙ

Therefore, ρ = 223.61 m

Also, recalling the equation of Motion

We then apply the equation to  Applying Eq. 13–8 with  θ = 26.57° and ρ = 223.61 m,

we then have,  

∑Ft = Mat;  800 (9.81) sin 26.57° - Ff = 800 (3)

Ff = 1109.73 N

= 1.11 kN

∑Fn = Man; 800(9.81) cos 26.57° - N = 800 (9² / 223.61)

N = 6729.67 N

= 6.73 kN

Therefore, the resultant normal force = 1.11 kN and the resultant minimal force = 6.73 kN

compass can be used day and night, cloudy or sunny

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000