Calculating Force on a Roller Coaster: 21kg Boy, 6.7m/s Speed, 14m Radius

[Rwarno]

A(n) 21 kg boy rides a roller coaster.
The acceleration of gravity is 9.8 m/s
With what force does he press against the
seat when the car moving at 6.7 m/s goes over
a crest whose radius of curvature is 14 m?
Answer in units of N
2. F=ma a=(v²⁾/r

Fn= ma-mg

The Attempt at a Solution

Fn= 21kg(6.7m/s²/14m) - 21kg(9.8m/s²⁾
Fn = 67.335N - 205.8 = -138.465

(my homework database says this is wrong, so i tried...)

Fn = ma + mg
Fn= 67.335 + 205.8 = 273.135

(my homework database says this is also wrong what am i missing or doing wrong?)

 

[berkeman]

A(n) 21 kg boy rides a roller coaster.
The acceleration of gravity is 9.8 m/s
With what force does he press against the
seat when the car moving at 6.7 m/s goes over
a crest whose radius of curvature is 14 m?
Answer in units of N



2. F=ma a=(v²⁾/r

Fn= ma-mg

The Attempt at a Solution

Fn= 21kg(6.7m/s²/14m) - 21kg(9.8m/s²⁾
Fn = 67.335N - 205.8 = -138.465

(my homework database says this is wrong, so i tried...)

Fn = ma + mg
Fn= 67.335 + 205.8 = 273.135

(my homework database says this is also wrong what am i missing or doing wrong?)

It looks like you are subtracting the two forces? If he's going over a bump on a crest, that increases his apparent weight, it doesn't decrease it...

 

[Doc Al]

Fn= ma-mg

Fix this. Hint: The normal force and the weight are in opposite directions. What's the direction of the acceleration?

 

[oo7 - 2oo]

I think it -Fn = Fc - Fg

* = (ma - mg) / -1

 

[Rwarno]

I tried to add them and the answer was also incorrect. I did it a little backwards(possibly?) and came up with 138.465, Is that correct?

 

[Doc Al]

I tried to add them and the answer was also incorrect.

Apply ΣF = ma, making sure that everything has the correct sign.

 

[Rwarno]

I tried F=ma in all forms:
the given acceleration : F = 21kg(9.8m/s2) = 205.8 which was said to be wrong
and applying the acceleration using 21kg x v = (6.7m/s)squared divided by r = 11m
which equals 67.3335, which is also incorrect
then i subtracted mg = 21kg(9.8) which was also incorrect
then i added mg, which was also incorrect

 

[berkeman]

I tried F=ma in all forms:
the given acceleration : F = 21kg(9.8m/s2) = 205.8 which was said to be wrong
and applying the acceleration using 21kg x v = (6.7m/s)squared divided by r = 11m
which equals 67.3335, which is also incorrect
then i subtracted mg = 21kg(9.8) which was also incorrect
then i added mg, which was also incorrect

The crest radius is 14m, not 11m.

And please add the weight and the extra force, there is no reason to subtract...

 

[Rwarno]

i meant to put 14, that's still the right calculation
and i tried it when i added the weight and it was still wrong

and what is the extra force?

 

[Doc Al]

i meant to put 14, that's still the right calculation
and i tried it when i added the weight and it was still wrong

and what is the extra force?

Only two forces act, the weight and the normal force. Again, write ΣF = ma, making sure that everything has the correct sign. (Use symbols, not numbers.)