Resultant Force for circular motion on a banked track

In summary, we are trying to find the magnitude of the resultant force on a 2800 kg car and driver traveling at 77 m/s around a circular track with a radius of 71.9 m, an angle of 22 degrees with the horizontal, and an acceleration of gravity of 9.8 m/s^2. The formula for centripetal force is mv^2/r, which in this case equals 230892.9068N. However, the normal force and force of friction do not create enough centripetal force, so we must also consider the horizontal components of these forces, which adds up to a resultant force of 11889.6N. Therefore, the magnitude of the
  • #1
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[SOLVED] Resultant Force for circular motion on a banked track

Homework Statement


A car travels 77 m/s around a circular track of radius 71.9
mass of car= 2800 kg
coefficient of friction = .1
angle of track with horizontal = 22 degrees
acceleration of gravity = 9.8 m/s^2
What is the magnitue pf the resultant force on the 2800 kg driver and his car?


Homework Equations


normal force= mgcos(angle)
force of friction = (normal force)(coefficient of friction)


The Attempt at a Solution


i know that the force of friction and normal force do not create the centripetal force necessary to cause uniform circular motion as
Fn= 2800(9.8)cos(22)=25441.9N
Ff= 25441.9(.1)=2544.19N
and the necessary centripetal force= mv^2/r= 2800(77)^2/71.9= 230892.9068N

so i tried making the resultant= the horizontal component of Ff+horizontal component of Fn= cos(22)(2544.19)+sin(22)(25441.9)= 11889.6N which was not the right answer
 
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  • #2
What is the magnitude pf the resultant force on the 2800 kg driver and his car?

You may ignore friction. At any moment the force will be the resultant of the centripetal force and gravity. The question isn't about whether the car will stick.
 
  • #3
never mind, the answer was just centripetal force
 

FAQ: Resultant Force for circular motion on a banked track

What is the resultant force for circular motion on a banked track?

The resultant force for circular motion on a banked track is the net force acting on an object moving in a circular path along the banked track. It is the combination of the centripetal force, which is directed towards the center of the circle, and the horizontal component of the normal force, which is directed towards the center of the circle.

How is the resultant force calculated for circular motion on a banked track?

The resultant force can be calculated using the equation F = mv^2/r, where F is the resultant force, m is the mass of the object, v is the velocity of the object, and r is the radius of the circular path. This equation takes into account the centripetal force and the horizontal component of the normal force.

What is the role of the banked angle in determining the resultant force?

The banked angle, also known as the angle of inclination, plays a crucial role in determining the resultant force for circular motion on a banked track. It affects the magnitude of the horizontal component of the normal force, which in turn affects the magnitude of the resultant force. The steeper the banked angle, the larger the horizontal component of the normal force and the resultant force.

Does the resultant force for circular motion on a banked track always point towards the center of the circle?

Yes, the resultant force for circular motion on a banked track always points towards the center of the circle. This is because the centripetal force and the horizontal component of the normal force are both directed towards the center of the circle, resulting in a net force that is also directed towards the center of the circle.

How does the speed of the object affect the resultant force for circular motion on a banked track?

The speed of the object affects the magnitude of the resultant force for circular motion on a banked track. As the speed increases, the magnitude of the resultant force also increases, as seen in the equation F = mv^2/r. However, the direction of the resultant force remains the same, always pointing towards the center of the circle.

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