# Determine the normal force on the track

• Ry122
In summary, a roller-coaster car with a mass of 494 kg and a speed of 12.5 m/s at point A has a downward component of velocity of 0 and moves horizontally along a 9m circle. To determine the force exerted by the track on the car at this point, one must use the concept of vertical circular motion and consider the significant forces acting on the car: gravity and the support force exerted by the track. The support force must be greater than the weight of the car to provide the necessary centripetal force. Using the formula for centripetal force, the correct answer is 864.58 N.

#### Ry122

http://users.on.net/~rohanlal/roll.jpg [Broken]

A roller-coaster car has a mass of 494 kg when fully loaded with passengers.
If the car has a speed of 12.5 m/s at point A, what is the force exerted by the track on the car at this point?

To solve this don't i need to determine what the downward component of it's velocity is so i can determine the normal force on the track at point A? How can I do that when no angles are given?

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To solve this don't i need to determine what the downward component of it's velocity is so i can determine the normal force on the track at point A? How can I do that when no angles are given?

well the downward component of the velocity at point A is 0.
The cart at A moves at a tangent to the 9m circle, and as point A is directly at the bottom of the center of the 9m circle, the cart moves horizontally.

To solve this problem, you have to know "vertical circular motion" basic concepts.
At A (the bottom of the 9m circle) the significant forces acting are:
1. Gravity on the 494kg cart acting directly downwards
2. The support force which is exerted by the track on the cart acting directly upwards

force no. 2 is the one you must find.
HINT: force no. 2 must be greater that force no. 1 to provide the centripetal force of the cart so that it moves circularly at 9m radius.

centripetal force=mv2/r

heres my attmept
f=(484)(18.5^2)/4.5 = 36819.888 N
9.8 x 484 = 4743.2
at point A the centripetal force is less since the weight of the car is reducing the need for centripetal acceleration.
so 36819.888-4743.2 = 32067.688
this is incorrect, what am i doing wrong?

f=(484)(18.5^2)/4.5

The radius is 9 not 4.5

at point A the centripetal force is less since the weight of the car is reducing the need for centripetal acceleration.

Wrong. The weight of the cart acts downwards, whereas centripetal acceleration upwards (towards the center of the cart 9m circle). Hence, the heavier the cart, you need more centripetal force to move at the same circle compared to lighter cart.

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i tried to draw a diagram here for you. "0" is the cart
the arrow pointing downwards is the weight force.
the arrow pointing upwards is the support force (which you must find)
exerted on the cart by the track.
It has two components: one has 3 stripes, it is the reaction
force of the weight, acting with the same magnitude but opposite direction of
the weight. The one with 4 stripes is the centripetal force.
so, the force you must find is the sum of the weight and centripetal force
(in terms of magnitude only). Or, to be strict, it is the sum of the reaction force
(which, of course "has the same magnitude but opposite direction of the weight")
and the centripetal force Fc=(484)(18.5^2)/9

Doing what you said my answer is still wrong.
Centripetal force + weight force of car = 18405.44 + 4743.2 = 23148.64

whoops!

f=(484)(18.5^2)/9

is also wrong..
it should be f=(484)(12.5^2)/9

## 1. What is the normal force on a track?

The normal force on a track is the force that acts perpendicular to the surface of the track. It is the force that prevents objects from sinking into or falling through the track.

## 2. How is the normal force calculated?

The normal force can be calculated using the formula FN = mgcosθ, where FN is the normal force, m is the object's mass, g is the acceleration due to gravity, and θ is the angle between the object and the surface of the track.

## 3. What factors affect the normal force on a track?

The normal force on a track is affected by the mass of the object, the angle of the track, and the acceleration due to gravity. It is also influenced by any external forces acting on the object, such as friction or air resistance.

## 4. Why is the normal force important in determining an object's motion on a track?

The normal force is important because it is a reaction force that balances out the force of gravity. This allows the object to stay on the track and move in a controlled manner. Without the normal force, the object would either sink into the track or fly off of it.

## 5. How can the normal force be increased or decreased on a track?

The normal force can be increased by increasing the mass of the object, increasing the angle of the track, or increasing the acceleration due to gravity. It can be decreased by decreasing these factors or by adding external forces that counteract the normal force, such as air resistance or friction.