If a car is taveling with a speed 6 and comes to a curve in a flat road with radius 13.5 m, what is the minumum value the coefficient

3 answers
Question:

If a car is taveling with a speed 6 and comes to a curve in a flat road with radius 13.5 m, what is the minumum value the coefficient of friction must be so the car doesn’t slide of the road?

Answers

The minimum friction coefficient required is 0.3

(friction coefficients have no units)

Explanation:

To find a force we need to know something about a mass, and we haven't been told the mass of the car. Let's just call it 'm' and leave it at that for the moment, because it will cancel out in the end.

The centripetal force is given by F = ma = mv2/r

We have values for the velocity and the radius, so:

Fcent=m×6 × 6/13.5 = 2. 667m N

The frictional force must be equal to or greater than this force in order for the car to successfully make it around the curve without sliding out.

The frictional force will be given by:

Ffrict = μFnorm

Where Fnorm is the normal force, equal to mg.

We can equate these two forces, the frictional force and the centripetal force:

Fcent = Ffrict

2.667m=μmg

We can cancel out a factor of m in both sides and rearrange to make μ the subject:

μ = 2.667g

Substituting in the value g=9.8 ms−2,

μ = 2.667/9.8 = 0.27

Approximately = 0.3

Explanation:

Below is an attachment containing the solution.


[tex]If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, wha[/tex]

[tex]\mu_s \geq 0.27[/tex]

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

[tex]f_s=\frac{mv^{2}}{R}[/tex]

But we know that:

[tex]f_s\leq \mu_s N[/tex]

And the normal force is given by the sum of the forces in the vertical direction:

[tex]N-mg=0 \implies N=mg[/tex]

Finally, we have:

[tex]f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27[/tex]

So, the minimum value for the coefficient of friction is 0.27.

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