How Do You Calculate Forces on a Fairground Ride Ramp?

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SUMMARY

The discussion focuses on calculating forces acting on a fairground ride ramp, specifically addressing three key problems. The weight of the car, loaded with passengers, is 6.8 kN, leading to a parallel component of approximately 2.3 kN acting along the ramp at a 20-degree slope. The mass of the car is 690 kg, resulting in a deceleration of 3.3 m/s² due to this force. Finally, to ensure the car stops before reaching the end of the ramp, the minimum ramp length calculated is 73 meters, using the kinematic equation v² = u² + 2as.

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  • Understanding of basic physics concepts, including forces and motion.
  • Knowledge of trigonometric functions, specifically tangent for slope calculations.
  • Familiarity with kinematic equations for motion analysis.
  • Ability to apply Newton's second law (F=ma) in practical scenarios.
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  • Study the application of trigonometric functions in physics problems.
  • Learn about kinematic equations and their derivations in motion analysis.
  • Explore conservation of energy principles in mechanical systems.
  • Investigate real-world applications of forces on inclined planes.
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone involved in designing or analyzing fairground rides and similar systems.

Chris08
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Homework Statement


Hi i have a question I've been doing and need to know if i am completely wrong (or not) any help woul be really appreciated.

A fairground ride ends with the car moving up aramp at a slope of 20degrees to the horizontal

Homework Equations


a)The car carrying its maximum load of passengers has a total weight of 6.8kN.Show that the component of the weight acting parallel to the ramp is about 2.3kN.

b) The mass of the fully loaded car is 690kg. Show that the force in part a will deccelerate the car at about 3.3ms^2

c) The car enters the ramp at 22ms^1.Calculate the minimum length that the ramp must be in order for the car to stop before it reaches the end.Neglect the length of the car.

The Attempt at a Solution



Ok for a) i did tan20degrees =opp/adj so tan20degreesx6.8kn=adj =2.47kN (horizontal component).

b) i did parallel component = f=ma so a=2.3/690 = 3.3ms^2

c) i used v=u^2+2as = 0= (22ms^1)^2 +2(-3.3ms^2x s)
=484=-6.6xs so s =484/6.6 = 73m.

d) i am not to sure about this one can anybody help me with this please?

Thanks,Chris
 
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Part a) asks for the parallel component
c) would be solved easier with conservation of energy.
You did not post question d).
 

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