Roller Coaster Car Force Calculation: Solving a Physics Homework Problem

  • Thread starter Thread starter Agahnim
  • Start date Start date
AI Thread Summary
To solve the roller coaster car force problem, the total force exerted by the track must be calculated by considering both the weight of the car and the centripetal force required for circular motion. The weight acts downward, while the normal force from the track acts at an angle, not vertically. The vertical forces must balance, resulting in zero net force in the vertical direction, while the horizontal forces must provide the necessary centripetal acceleration. A free body diagram and application of Newton's second law are essential for resolving forces in both the x and y directions. Understanding these concepts will lead to the correct calculation of the total force and its direction.
Agahnim
Messages
6
Reaction score
0
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.

--------------------------------------------------------------------

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?!
 
Physics news on Phys.org
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.)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top