Think about how this relates to (c).Ihe center of the circle?2. A 1500 kg car is driving on a concrete highway, on a sunny day. The road makes a sharp turn, with a radius ofcurvature of 50. meters. What is the maximum speed at which the car can make the turn without sliding sideways??The centripetal force is provided by the friction force that the concrete road exerts on the rubber tires. Calculate thefriction force, then plug that into F.-mv/r, as FM 9EESE

Answered: 2. A 1500 kg car is driving on a…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Think about how this relates to (c).
Ihe center of the circle?
2. A 1500 kg car is driving on a concrete highway, on a sunny day. The road makes a sharp turn, with a radius of
curvature of 50. meters. What is the maximum speed at which the car can make the turn without sliding sideways??
The centripetal force is provided by the friction force that the concrete road exerts on the rubber tires. Calculate the
friction force, then plug that into F.-mv/r, as F
M 9
EESE

Expert Solution

Step 1 If there is no provision of super elevation on road

If there is no super elevation then centripetal force will be the function of friction only

Friction which is generally consider to be generated on b/w tyre and road in lateral direction is 0.15

$F r i c t i o n f o r c e F c = f r i c t i o n c o e f f i c i e n t \times N o r m a l f o r c e W h e r e N o r m a l f o r c e = m g = 1500 \times 9.81 F r i c t i o n f o r c e F_{c} = 0.15 \times 1500 \times 9.81 = 2207.25 N N o w F_{c} = \frac{m v^{2}}{r}; W h e r e R a d i u s r = 50 m a n d m a s s m = 1500 k g 2207.25 = \frac{1500 \times v^{2}}{50} v = 30.87 K m p h (v e l o c i t y o f w i t h o u t s l i d i n g)$

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