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Kim has her coffee cup on her car's dash when she takes a corner with radius 4 m and 20 km/h.
What is the minimum coefficient of static friction which would allow the coffee cup to stay there
without slipping?

Sagot :

The minimum coefficient of static friction is  0.807

What is static friction?

Static friction is friction between two or more solid objects that are not moving relative to each other.

According to the question,

When an object turns on a curve path at a specific speed, two forces will be acting on the object which balance each other:

Centripetal force and friction force.

The friction coefficient between the object and the path is independent of the object's mass.

Given,

The radius of a corner is, r=4.16m

The turning speed of the car is, v

[tex]20km/h * (\frac{\frac{5}{18}m/s }{1km/h} )[/tex]

= 5.72 m/s

To avoid slipping a coffee cup on the car's dashboard, the centripetal force would balance the friction force.

So,  Fc=Ff

[tex]\frac{mv^2}{r}[/tex]=μmg

μ=[tex]\frac{v^2}{rg}[/tex]

Here,

Fc represents the centripetal force.

Ff represents the friction force.

μ represents the coefficient of static friction.

g represents the gravitational acceleration whose value is 9.8m/s²

By Substituting the values in the above formula, we get:

μ =[tex]\frac{(5.72m/s)^2}{(4m)(9.8m/s^2)} \\[/tex]  ≈ 0.807

Therefore,

The minimum coefficient of static friction is  0.807

Learn more about static friction here:

https://brainly.com/question/13000653

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