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Newton's Second Law and Normal Force Calculation, Exercícios de Engenharia Elétrica

Calculations of the normal force acting on a car moving in a circular path using newton's second law. Two examples with different velocities and explains the significance of the positive and negative signs in the normal force calculation.

Tipologia: Exercícios

Antes de 2010

Compartilhado em 08/10/2007

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38. We will start by assuming that the normal force (on the car from the rail) points up. Note that gravity
points down, and the yaxis is chosen positive upwards. Also, the direction to the center of the circle
(the direction of centripetal acceleration) is down. Thus, Newton’s second law leads to
Nmg =mv2
r.
(a) When v= 11 m/s, we obtain N=3.7×103N. The fact that this answer is positive means that
N
does indeed point upward as we had assumed.
(b) When v= 14 m/s, we obtain N=1.3×103N. The fact that this answer is negative means that
Npoints opposite to what we had assumed. Thus, the magnitude of
Nis 1.3 kN and its direction
is down.

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  1. We will start by assuming that the normal force (on the car from the rail) points up. Note that gravity

points down, and the y axis is chosen positive upwards. Also, the direction to the center of the circle

(the direction of centripetal acceleration) is down. Thus, Newton’s second law leads to

N − mg = m

v

2

r

(a) When v = 11 m/s, we obtain N = 3. 7 × 10

3 N. The fact that this answer is positive means that

N

does indeed point upward as we had assumed.

(b) When v = 14 m/s, we obtain N = − 1. 3 × 10

3

N. The fact that this answer is negative means that

N points opposite to what we had assumed. Thus, the magnitude of

N is 1.3 kN and its direction

is down.