Two identical car $A$ and $B$ are moving at $36 k m p h$ . A goes on a bridge, convex upward and $B$ on concave upward. If the radius of curvature of bridge is $20 m$ , the ratio of normal forces exerted on the cars when they are at the middle of bridges $(g = 10 m s^{- 2})$

1 : 3

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1 : 2

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2 : 3

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1 : 5

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Solution

The correct option is A $1 : 3$
Velocity, $v = 36 \times \frac{5}{18} = 10 m s^{- 1}$

Centripetal force, $F_{c} = \frac{m v^{2}}{r} = \frac{m \times 10^{2}}{20} = 5 m$

For A bridge is convex upward, normal reaction, $N_{A} = m g - F_{c} = 10 m - 5 m = 5 m$

For B bridge is concave upward, normal reaction, $N_{B} = m g + F_{c} = 10 m + 5 m = 15 m$

$N_{A} : N_{B} = 5 m : 15 m = 1 : 3$

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Two identical car A and B are moving at 36 kmph. A goes on a bridge, convex upward and B on concave upward. If the radius of curvature of bridge is 20 m, the ratio of normal forces exerted on the cars when they are at the middle of bridges (g=10 ms−2)

The correct option is A 1:3Velocity, v=36×518=10ms−1 Centripetal force, Fc=mv2r=m×10220=5m For A bridge is convex upward, normal reaction,NA=mg−Fc=10m−5m=5m For B bridge is concave upward, normal reaction,NB=mg+Fc=10m+5m=15m NA:NB=5m:15m=1:3

Two identical car $A$ and $B$ are moving at $36 k m p h$ . A goes on a bridge, convex upward and $B$ on concave upward. If the radius of curvature of bridge is $20 m$ , the ratio of normal forces exerted on the cars when they are at the middle of bridges $(g = 10 m s^{- 2})$