Okay, I saw this one previously posted, but the answer it kept leading me to was wrong.
A sports car of mass 1400kg (including the driver) crosses the rounded top of a hill (radius 88m) at 23 m/s. Find the normal force exerted by road on the car.
I thought that the normal force was mass times acceleration, just what the road is doing to the car. So I multiplied 9.8 times 1400, but that was wrong. What am i doing wrong? The same applies to a driver (weighing 75kg in the car)
normal force= apparent weight
= mg-mv^2/r
So, basically, the way I can look at problems like these is to subtract the two accelerations (normal's gravity and centripetal) then multiply them by the mass (F=ma)?
yes.
Why is the centripetal force subtracted from the force of gravity? Aren't both going in the same direction (since centripetal force is always in the direction of the center of the circle)
To find the normal force exerted by the road on the car, you need to take into account the forces acting on the car at the top of the hill. The normal force is not equal to the mass times acceleration in this case.
At the top of the hill, the car is experiencing both gravitational force pulling it downwards and centripetal force pulling it towards the center of the circular path. The normal force is the force exerted by the road perpendicular to the surface of the hill to counterbalance these forces and keep the car in equilibrium.
To calculate the normal force, you can start by considering the vertical forces acting on the car at the top of the hill. The gravitational force is given by the formula F_gravity = m * g, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the mass of the car, including the driver, is 1400 kg. Therefore, the gravitational force acting on the car is F_gravity = 1400 kg * 9.8 m/s^2.
Next, you need to consider the centripetal force acting on the car. At the top of the hill, the centripetal force is provided by the normal force. Using the formula for centripetal force, F_centr = m * v^2 / r, where m is the mass of the car, v is the velocity of the car, and r is the radius of the hill.
In this case, the mass of the car is still 1400 kg, the velocity of the car is 23 m/s, and the radius of the hill is 88 m. Therefore, the centripetal force acting on the car is F_centr = 1400 kg * (23 m/s)^2 / 88 m.
Now, since the normal force is the sum of these two forces (F_gravity and F_centr), you can calculate the normal force exerted by the road on the car. It is given by the equation:
Normal Force = F_gravity + F_centr
Substituting the values we obtained earlier, you can calculate the normal force.
Normal Force = (1400 kg * 9.8 m/s^2) + [(1400 kg * (23 m/s)^2) / 88 m]