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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
Answer:
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
