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?
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
Well, I have to say, your car really knows how to stick to the road! Now, let's calculate the force exerted on it by static friction.
The net inward force acting on the car during this turn is provided by the static friction between the car's tires and the road. This force is responsible for keeping the car moving in a circular path.
To calculate this force, we can use the equation:
F_friction = m * v^2 / r
where:
F_friction is the force exerted by static friction,
m is the mass of the car (1300 kg),
v is the velocity of the car (16.5 m/s), and
r is the radius of the turn (57.7 m).
Plugging in the values, we get:
F_friction = (1300 kg) * (16.5 m/s)^2 / 57.7 m
Now, let me crunch the numbers for you...
*drumroll*
The force exerted on the car by static friction is approximately 6056 N.
So, your car is really gripping the road tight! Just make sure it doesn't outshine you in the dance of physics!
To find the force exerted on the car by static friction, we will use the centripetal force formula. The centripetal force is the force that is required to keep an object moving in a circular path.
The formula for centripetal force is:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the car
v is the velocity of the car
r is the radius of the corner
In this case, the mass of the car (m) is 1300 kg, the velocity (v) is 16.5 m/s, and the radius (r) is 57.7 m. Plugging these values into the formula, we get:
F = (1300 kg * (16.5 m/s)^2) / 57.7 m
F = (1300 kg * 271.25 m^2/s^2) / 57.7 m
F = 352625 kg⋅m/s^2 / 57.7 m
F ≈ 6121.53 N
Therefore, the force exerted on the car by static friction is approximately 6121.53 Newtons.
thanks
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