Calculate Apparent Weight at Point S

In summary, the train has a weight of 2 2 times what it would have if there was no air resistance or friction. It has a speed of 14 m/s when it is at the point S.
  • #1
astrololo
200
3

Homework Statement


When the train has passed at the point C, the radii of curvature is 20 m and the apparent weight is 2 2 times bigger than the real weight. At the moment when the picture was taken, the train was at the point s, located 8 m higher than the hollow C. At s, the radii of curvature est 16m. Calculate the quotient between the apparent weight at the point S and the real weight. There is no friction and no air resistance.

http://imgur.com/R2A4pya

Homework Equations



Sum of force in y = (mv^2)/radii

Energy at s = energy s

k=mv^2/2

U=m*g*y

The Attempt at a Solution


I know what to do :

I calculate the speed at the point C. I get 62,6 m/s

I put up equations which respect that Es=Ec to find my speed at S

I do the sum of forces at S and find n and divide it by mg.

Now, the only point which I am not sure is what is the height the point S ? I is 48 m ?
 
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  • #2
astrololo said:
I calculate the speed at the point C. I get 62,6 m/s
How do you get that?
astrololo said:
Now, the only point which I am not sure is what is the height the point S ?
"The height of the point S" is an incomplete (meaningless) statement. When you measure the height of something, you must measure the height with respect to some reference point. So, from where do you want to measure the height of S?
 
  • #3
Nathanael said:
How do you get that?

"The height of the point S" is an incomplete (meaningless) statement. When you measure the height of something, you must measure the height with respect to some reference point. So, from where do you want to measure the height of S?
From 0 at the point C I guess
 
  • #4
astrololo said:
From 0 at the point C I guess
Ok, is there anything in the problem statement about that?

Also, double check how you found the speed, because I do not get 62 m/s.
 
  • #5
Nathanael said:
Ok, is there anything in the problem statement about that?

Also, double check how you found the speed, because I do not get 62 m/s.
Well, I set up n=2mg

Sum of forces y = mv^2/r

2mg-mg=mv^2/r

2g-g=v^2/20

which gives v = 14 m/s right ? I guess I did an error
 
  • #6
astrololo said:
which gives v = 14 m/s
Yes.
The problem statement does not say that the train merely coasts up the hill, but presumably that is the intent.
 
  • #7
haruspex said:
Yes.
The problem statement does not say that the train merely coasts up the hill, but presumably that is the intent.
Ok, but what is the height at the point s ? I am not sure, is it 48 m ? 2 times the radii + 8 m.
 
  • #8
astrololo said:
Ok, but what is the height at the point s ? I am not sure, is it 48 m ? 2 times the radii + 8 m.
astrololo said:
the point s, [is] located 8 m higher than [the hollow at] C
A clue is that it does not have enough KE to rise much more than 8m.
 
  • #9
haruspex said:
A clue is that it does not have enough KE to rise much more than 8m.
So it's 8 ?
 
  • #10
astrololo said:
So it's 8 ?
Yes, that is most likely what they meant.
(I'm not sure why they said "8 m higher than the hollow C" though... I don't know what the word hollow was supposed to mean here.)
 
  • #11
Nathanael said:
Yes, that is most likely what they meant.
(I'm not sure why they said "8 m higher than the hollow C" though... I don't know what the word hollow was supposed to mean here.)
It's a translation from another language lol But the final answer is supposed to be 3/4. Does it give this ?
 
  • #12
astrololo said:
It's a translation from another language lol But the final answer is supposed to be 3/4. Does it give this ?
Yes.
A tip: avoid using inexact numerical values calculated at intermediate steps. Ideally, keep everything in algebraic symbols until the final steps. This prevents the accumulation of rounding errors. In the present case e.g., you can avoid taking a square root to find the velocity, only to find that it's the square of the velocity that's needed.
 
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  • #13
haruspex said:
Yes.
A tip: avoid using inexact numerical values calculated at intermediate steps. Ideally, keep everything in algebraic symbols until the final steps. This prevents the accumulation of rounding errors. In the present case e.g., you can avoid taking a square root to find the velocity, only to find that it's the square of the velocity that's needed.
I obtained the following : quotient= (9,8m+(39,2m)/16)/9,8m=1.25

I feel like there is a minus missing
 
Last edited:
  • #14
astrololo said:
I obtained the following : quotient= (9,8m+(39,2m)/16)/9,8m=1.25

I feel like there is a minus missing
Your feeling is correct.
What are the forces acting on the train at S? What is the acceleration of the train at S? What equation relates the two?
 

1. What is apparent weight at point S?

Apparent weight at point S is the perceived weight of an object when it is placed on a scale at point S. It takes into account the gravitational force acting on the object as well as any other forces, such as normal force or buoyant force, that may be acting on the object.

2. How is the apparent weight at point S calculated?

The apparent weight at point S is calculated by subtracting the buoyant force and any other supporting forces from the actual weight of the object. This can be represented by the equation: Apparent Weight = Actual Weight - Supporting Forces.

3. What factors affect the apparent weight at point S?

The apparent weight at point S can be affected by several factors, including the mass of the object, the strength of the gravitational force, and the presence of any supporting forces such as buoyant force or normal force. It can also be affected by the location of point S in relation to the object.

4. Can the apparent weight at point S be negative?

Yes, the apparent weight at point S can be negative if the buoyant force and other supporting forces are greater than the actual weight of the object. This can occur when the object is immersed in a fluid, such as water, and experiences an upward buoyant force that is greater than its weight.

5. How is the apparent weight at point S different from the actual weight?

The apparent weight at point S is different from the actual weight because it takes into account any supporting forces that may be acting on the object. The actual weight only considers the gravitational force acting on the object. Therefore, the apparent weight may be greater or less than the actual weight, depending on the presence and strength of these supporting forces.

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