F(fr) = k•N =k•m•g =0.93•1300•9.8 =11848 N,
friction force = centripetal force to keep the car on road
mv^2/R =1300•(16.5)^2/57.7 =6134 N
Net force = 11848 – 6134 =5714 N
When you take your 1300-kg car out for a spin, you go around a corner of radius 57.7 m with a speed of 16.5 m/s. The coefficient of static friction between the car and the road is 0.93. Assuming your car doesn't skid, what is the force exerted on it by static friction?
3 answers
thanks
The answer above is wrong ^^^
The question stats that you do not skid so therefor you use Newtons second law formula ( Fcp = m ( v^2 / r ) ) only and the coefficient is not needed.
Also, Centripetal force is any force that is pointing lateral of the object. In this case it is a car and it is the force of the friction of the tires to the road. So therefor Fcp=Fs for this question.
(Fs= static friction)
The diagram would look like this
↑( Net force)
🚖→ ( Fs)
↓(mg)
The question is only asking for ( Fs ) so that is all you need to find, which again is Fs= m ( v^2 / r )
I hope this helps any one that is confused like I was, Lol
The question stats that you do not skid so therefor you use Newtons second law formula ( Fcp = m ( v^2 / r ) ) only and the coefficient is not needed.
Also, Centripetal force is any force that is pointing lateral of the object. In this case it is a car and it is the force of the friction of the tires to the road. So therefor Fcp=Fs for this question.
(Fs= static friction)
The diagram would look like this
↑( Net force)
🚖→ ( Fs)
↓(mg)
The question is only asking for ( Fs ) so that is all you need to find, which again is Fs= m ( v^2 / r )
I hope this helps any one that is confused like I was, Lol