How Do You Calculate Forces on a Fairground Ride Ramp?

[Chris08]

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

 

[qspeechc]

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

 

[Moetasim]

I would first like to commend you for attempting to solve these force questions. It shows that you are actively engaging with the material and trying to understand it. However, there are a few things I would like to point out in your attempt at a solution.

For part a), you correctly calculated the horizontal component of the weight, but you did not show how you got the value of 2.3kN. To show this, you can simply multiply the weight (6.8kN) by the sine of the angle (20 degrees), as the weight is acting in the opposite direction of the ramp. This would give you a value of 2.33kN, which is approximately 2.3kN as stated in the problem.

For part b), you have correctly calculated the acceleration, but you have not shown how you got the value of 3.3ms^2. To show this, you can simply divide the horizontal component of the weight (2.3kN) by the mass of the car (690kg). This would give you a value of approximately 0.0033ms^2, which is equivalent to 3.3ms^2 as stated in the problem.

For part c), you have correctly used the equation v^2=u^2+2as, but you have made a small mistake in the units. The initial velocity (u) is given in units of meters per second (ms^-1), but you have squared it and then multiplied it by the acceleration (a) in units of meters per second squared (ms^-2). This would give you a value of meters per second cubed (ms^-3) instead of meters (m). To correct this, you can simply square the initial velocity and then divide by the acceleration, which would give you a value of approximately 148m.

For part d), I am assuming you are being asked to calculate the length of the ramp. In order to do this, you can use the equation s=ut+0.5at^2, where s is the distance (length of the ramp), u is the initial velocity, a is the acceleration (which we calculated to be 3.3ms^2), and t is the time it takes for the car to stop. We already know the initial velocity (22ms^-1) and the acceleration (3.3ms^2), so we just need to calculate