Roller Coaster Car Force Calculation: Solving a Physics Homework Problem

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SUMMARY

The roller coaster car problem involves a mass of 310 kg traveling at a speed of 18 m/s around a horizontal curve with a radius of 40 m. The total force exerted on the car by the track must be calculated, taking into account both the weight of the car and the normal force acting at an angle. The correct approach involves drawing a free body diagram, resolving forces in the x and y directions, and applying Newton's second law to find the resultant force and its direction.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of centripetal acceleration and its calculation
  • Ability to draw and interpret free body diagrams
  • Familiarity with vector resolution of forces
NEXT STEPS
  • Calculate the normal force acting on the roller coaster car using the equation for centripetal force
  • Learn how to resolve forces in two dimensions using vector components
  • Study the effects of non-banked curves on roller coaster dynamics
  • Explore the concept of centripetal acceleration in circular motion
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Physics students, educators, and anyone interested in understanding the dynamics of roller coasters and circular motion in physics.

Agahnim
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hi i need help with a homework problem for physics: it goes like this:

A roller coaster car of mass 310 kg (including passengers) travels around a horizontal curve of radius 40 m. Its speed is 18 m/s. What is the magnitude and direction of the total force exerted on the car by the track?

_______N at _______° above the horizontal.

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The first answer I tried was calculating the normal force exerted by the track (which was simply mass * gravity, and 90 degrees above the horizontal). That was wrong. Then I tried giving the answer as the centripetal acceleration ((m * V^2)/ r), and the angle as 0 degrees above the horizontal. That was wrong too.

So how do I solve this?? Does anyone know, please help?!
 
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There are two forces acting on the car : (i) its weight acting downwards (ii) the normal reaction, acting at some angle (not vertically upwards) to be determined.

As a result of these forces, the acceleration of the car is v^2/r inwards.

Draw the free body diagram. Resolve all forces along the x, y directions. Apply Newton's second law to the forces and accelerations along these directions.

PS : Next time, post such questions in "Homework Help"
 
There are two forces acting on the car : (i) its weight acting downwards (ii) the normal reaction, acting at some angle (not vertically upwards) to be determined.

Hi, thanks for the probleb advice! and next time i'll post it in homework help.

But one thing I am not understanding: Wont the normal force be directly upwards and the weight be directly down, since the curve is not banked? Therefore the sum of forces in the y-direction will be ZERO right?

Now for the x-direction: The only thing acting on the car is centripetal acceleration since there's no friction. Therefore that acceleration will be in the x-direction. But since centri. accel. is not really a 'force' it can't be counted. so I am still confused.

P.S. If someone experienced can calculate the answer to this problem and let me know how he got it, i would like that!
 
As Gokul43201 stated, there are two forces acting on the car: its weight, acting downwards, and the force that the track exerts on the car, which acts in a direction that you must find. I would not call that latter force a "normal" force, since that assumes it acts in a particular direction: it's safe to assume that the car is somewhat "attached" to the track, not merely riding on top of it.

Yes, vertical forces must add to zero; horizontal forces must provide the needed centripetal force.

(I will move this to the homework help section.)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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