Question

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coefficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?

Answer 1

Answer:

$f_{fr}=1590.85 N$

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

$f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r}$ (1)

• f(fr) is the friction force
• N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

$Ncos(25)-mg-f_{fr}sin(25)=0$ (2)          

Let's combine (1) and (2) to find f(fr)

$f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}$  

$f_{fr}=1590.85 N$

I hope it helps you!

Answer 2

The magnitude of the force will be "7543 N".

Given:

Mass,

• m = 600 kg

Angle,

• $\Theta$ = 25°

Coefficient of static friction,

• $\mu_s$ = 0.30

By using the net force equation, we get

→ $N cos \Theta -mg - \mu_s N sin \Theta =0$

or,

→                                     $N = \frac{mg}{cos \Theta- \mu_s sin \Theta}$

By substituting the values, we get

→                                          $= \frac{600\times 9.8}{cos 25^{\circ}- 0.30\times sin 25^{\circ}}$

→                                          $= \frac{5880}{cos 25^{\circ}- 0.30\times sin 25^{\circ}}$

→                                          $= 7543 \ N$

Thus the response above is right.

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