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When a car negotiates a curve, the norma...

When a car negotiates a curve, the normal force exerted on the inner and outer wheels are `N_(2)` and `N_(1)` respectively. Then `N_(1)//N_(2)` is

A

1

B

lt 1

C

gt 1

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
Referring to the free body diagram we obtain,
`sum F_(y) = N_(1) + N_(2) - mg = 0`
`rArr" "N_(1) + N_(2) = mg" "...(1)`
`sum F_(x) = ma_(x)`
`rArr" "f = (mv^(2))/(r)" "...(2)`
`sum tau_(G) = 0`
`rArr" "N_(1) (d)/(2) + fh - N_(2)(d)/(2) = 0" "...(3)`
`rArr" "N_(2) - N_(1) = (2h)/(d) ((mv^(2))/(r))`
`rArr" "N_(2) = mg//2 + (hmv^(2))/(dr)`
and `N_(1) = (mg)/(2) - (hmv^(2))/(dr) rArr (N_(1))/(N_(2)) lt 1`
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