Normal force

Classical Mechanics Level 2

The above picture is a 1000 1000 kg car passing the top of a hill with radius of curvature 250 250 m at a speed of 50 50 m/s. Find the magnitude of normal force on the car (in N).

Gravitational acceleration is g = 10 g= 10 m/s 2 ^{2} .


The answer is 0.

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Srinivasa Satti by Srinivasa Satti

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2 solutions

Varun Goenka
Apr 25, 2014

The resultant of weight and normal reaction force provides centripetal acceleration to the car towards the center of the curvature.

So, mg-N= m ac, where ac= v^2/r implies mg-N= mv^2/r, so N= m g - m v^2/r = 100 (10- 50^2/250)=0

Normal Reaction = 0N...Hence the passengers inside the car will feel weightless at the top of the trajectory, a condition similar to free fall.

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The acceleration of the car can be computed as follows: a = v2/R = 50x50/250 = 10m/sec square F =mxa =1000x10 =10000 N

Fnet = Fnorm -Fgrav Fgrav = mxg = 1000x10 =10000 Fnet = 10000-10000 =0 answer

Shafia Jathol - 7 years ago
Karan Joisher
Jul 6, 2014

The two opposite vertical forces acting on the car are normal reaction and the weight... The required centripetal force is given by the weight.. Thus we get:- mg - N = mv²/r N = m(g - v²/r) N = m(10-10) = 0 N

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