A car of mass 1490 kg makes a 23.0 m radius turn at 7.85 m/s on flat ground. What is the (minimum) coefficient of static friction?

Answer 1

Given

#m->"mass of car"=1490kg#
#r->"radius of circular turn"=23m#
#v->"velocity of car on the turn"=7.85" m/s"#

Let

#mu->"coefficient of static friction"=?#
The centripetal force #F_c# required for turning of the car of mass m in the circular path of radius r is given by
#F_c=(mv^2)/r#
Here the force of static friction will provide the required centripetal force and resist the tendency of skidding and this force of static friction #F_s# is given by
#F_s=muxx"normal reaction"#
#=>F_s=muxxmg#

By the condition of the problem

#F_s>=F_c#
#=>mumg>=(mv^2)/r#
#=>mu>=v^2/(r*g)=7.85^2/(23*9.8)#
#=>mu>=0.27#
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Answer 2

The minimum coefficient of static friction required for the car to make the turn is approximately 0.360.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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