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A car making this turn is pulled downward by its own weight, and pushed up by the road at an angle of 45°, so by Newton's second law,
• the net horizontal force on the car is
∑ F = N cos(45°) = m a = m v ² / R
• the net vertical force on the car is
∑ F = N sin(45°) - m g = 0
where
• N = magnitude of the normal force
• m = mass of the car
• a = v ² / R = centripetal acceleration of the car
• v = tangential speed of the car
• R = 100 m = radius of curvature
• g = 9.8 m/s² = acceleration due to gravity
From the net vertical force equation, we get
N = m g / sin(45°)
and substituting this into the net horizontal force equation and solving for v gives
(m g / sin(45°)) cos(45°) = m v ² / R
v = √(R g cos(45°) / sin(45°)) ≈ 31 m/s
We have that A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of
V=32m/s
From the question we are told
a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:
Generally the equation for the Velocity is mathematically given as
[tex]V=\sqrt{rgtan\theta}[/tex]
Therefore
[tex]V=\sqrt{rgtan\theta}\\\\V=\sqrt{100*9.8*tan45}\\\\V=32m/s[/tex]
Therefore
A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of
V=32m/s
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