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Determine the normal force on the track

  • Thread starter Ry122
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  • #1
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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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  • #2


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
 
  • #3
565
2


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?
 
  • #4


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
 
  • #5
565
2


Doing what you said my answer is still wrong.
Centripetal force + weight force of car = 18405.44 + 4743.2 = 23148.64
 
  • #6


whoops!

f=(484)(18.5^2)/9
is also wrong..
it should be f=(484)(12.5^2)/9
 

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