Help in finding the magnitude of a force

[tmn50]

Homework Statement

a skater moves from point A to C in a circulaire path as shown in the picture

C is the end of the circular path
find R the force that the ground exerts on the skater at point C using Newten's second law
there is no friction nor air resistance

Homework Equations

p-R=ma

The Attempt at a Solution

well I did this
p-R=0
so that P=R
am I right ? cause I think in point "C" there is no acceleration

 

[Delphi51]

This is a circular motion problem. Look up the formulas for the centripetal force required to hold something in circular motion.

The problem is not complete; you'll need things like the radius of the circle, the speed and mass of the skater.

 

[tmn50]

This is a circular motion problem. Look up the formulas for the centripetal force required to hold something in circular motion.

The problem is not complete; you'll need things like the radius of the circle, the speed and mass of the skater.

I only want to know if there is a perperndicular acceleration in "C" I think that it is 0 because the skater would go up if there was any

 

[Delphi51]

Upward? I thought the skater was on flat ice. Does the diagram mean that the skater goes down a ramp and then upward on a circularly curved ramp?
If so, the force exerted by the ice at point C is certainly upward. It must cause the skater to accelerate upward with the centripetal force necessary to maintain the circular motion.

 

[tmn50]

no the skater starts from point "A" with no speed and accelerates downward the uestion is when it arrives at point "c" does it have any acceleration ?

 

[Delphi51]

no the skater starts from point "A" with no speed and accelerates downward

I'm still not clear. Is the motion in the horizontal plane on a flat ice arena?
Or in a vertical plane as "accelerates downward" implies?
Despite your "no" I think you mean the latter. At C it has stopped accelerating down the ramp. And started in circular motion, so it has centripetal acceleration and there must be a force toward the center of the circular path. What about gravity?

 

[tmn50]

maybe my drawing wasn't clear here's a new one
or let's put the question in another way
define the acceleration of the body in "C3

 

[Delphi51]

Oops, I missed the fact that it starts on the circular part, at A.
You still haven't said whether the diagram is vertical or horizontal, so I'm still guessing vertical.

I see the confusion now. Is it still in circular motion or is it on a flat part? Not very clear, is it? I would say that it is still in circular motion but others would go the other way. Work it out both ways?

Whether or not you have centripetal acceleration and force, you still have gravity pulling the skater down and some force must oppose that because the skater is certainly not accelerating downward.

 

[tmn50]

Oops, I missed the fact that it starts on the circular part, at A.
You still haven't said whether the diagram is vertical or horizontal, so I'm still guessing vertical.

I see the confusion now. Is it still in circular motion or is it on a flat part? Not very clear, is it? I would say that it is still in circular motion but others would go the other way. Work it out both ways?

Whether or not you have centripetal acceleration and force, you still have gravity pulling the skater down and some force must oppose that because the skater is certainly not accelerating downward.

what do you mean by horizontal and vertical ?
and yes the gravity is the force pulling the skater
and the original question is to define the magnitude of the force acting on the skater by the ground
I said P-R=0 so P=R
but if there is a vertical acceleration towards the centre of the circle the equation would be P-R= ma so R = P -ma
so which one is right ?
and for the vertical and horizontal stuff
the ground is the buttom of the diagram

 

[Delphi51]

Okay, we're on the right path! I'm not sure what you mean by P and R, though. I do think the ma force should be included but I'm not sure about it. Were you given any information that would allow the centripetal force to be calculated? If not, then I would change my mind and ignore it. The question would be awfully trivial that way, though: simply F = mg.

 

[tmn50]

yup we have the speed at "C" also ma isn't a force "m" is the mass of the body and "a" is the perpendicular acceleration at point "c" (the pink arrow in the last diagram) I'm in a hurry mate

 

[Delphi51]

ma = F is definitely a force! Centripetal force.

 

[tmn50]

are you sure you know what's the question ?

 

[Delphi51]

find R the force that the ground exerts on the skater at point C using Newten's second law

Good point; always a good idea to read the question again! "Newton's second law" is F = ma, so that clears up any doubt we had about whether there is a centripetal force - there is and it is F = ma where "a" is the centripetal acceleration.

I also see what your R force is. Still don't know what the p in "p-R=ma" is.

 

[tmn50]

Good point; always a good idea to read the question again! "Newton's second law" is F = ma, so that clears up any doubt we had about whether there is a centripetal force - there is and it is F = ma where "a" is the centripetal acceleration.

I also see what your R force is. Still don't know what the p in "p-R=ma" is.

p=mg where m is the mass and g is 9.81m/s^2
and if there were a centripetal force there the body would continue his circular motion it wouldn't follow a straight line there

 

[Delphi51]

Okay, so mg - R(ground force up) = ma
where a is the centripetal acceleration.
Finished? Put in the numbers?

 

[yster]

p=mg where m is the mass and g is 9.81m/s^2
and if there were a centripetal force there the body would continue his circular motion it wouldn't follow a straight line there

you need to remember that at that point the body is in a transition from circular motion to linear motion. I also believe it's a centripetal force. If what was suggested above doesn't work. You can try determining the centrifugal force exerted by the body after the 90 degree circular motion. It should be equal to the force exerted by the ground

 

[tmn50]

can you explain why there is a vertical acceleration although if ther path after the circular one were flat the body will go on a straight path with a constant speed ?
so that acceleration didn't even affect the body (the speed nor the direction) how does it exist than ?

 

[Delphi51]

That's a tricky point, all right. If it is transitioning from a circular path to the ramp with only a single point where the tangent of the path is horizontal, then the answer to your question is that the centripetal force is still holding the skater on the circular path, changing the motion from slightly down/left to horizontal. In the next instant the acceleration will be upward and to the right as the path becomes a ramp. The diagram does not make it clear whether or not there is a horizontal part of the path of finite length. Remember the question asks you to use the acceleration to calculate the force, so likely the intent is for you to use the centripetal force.

 

[yster]

if you understand what Delphi just said and what i said. I believe it will be easier to come to a solution. When the skater is at point c the is no more vertical acceleration. This means that the resultant force is 0. It does not mean that there is no force acting on the body. When i look at the problem the only force that the body can exert is a centrifugal force. The ground is merely balancing/reacting to the force. I think you should add the weight to this force.

 

[tmn50]

ok forget that problem imagine this situation with me ok ?
a skater moves in that 1/4 circle than it moves to a flat surface there is no friction in it
in the transition point (the end of the circular path and the start of the flat area )
does this body have a vertical acceleration if so why doesn't it change the speed nor the direction of the body ? expalin this to me avoid the first problem i want the truth not according to a question

 

[Delphi51]

Excellent thought! Your new situation is very clear. I would say there is still a centripetal acceleration (upward) at the point where it changes from circular to horizontal motion. A nanometer out onto the flat part, the acceleration is zero.

 

[tmn50]

I wouldn't say that how can a skater's feet stand such a force this sport would have been banned if it was so

 

[yster]

No there is no more vertical acceleration at that point.

The RESULTANT vertical force is 0.
That is why the vertical displacement and speed do not change from that point onwards.
The skater is most likely to feel himself being pulled down at that point as a result of the centrifugal force that his body is exerting/inertia, however you choose to view the situation. The ground just reacts to this force

 

[tmn50]

can you explain little further please?

 

[yster]

I am trying but I'm having difficulty getting through to you.
Read all three of Newtons laws
http://en.wikipedia.org/wiki/Newton's_laws_of_motion
What I am saying is a combination of all three of them. Let me just think a little more about to break it down so you see how each one applies in this case.

 

[Delphi51]

I wouldn't say that how can a skater's feet stand such a force this sport would have been banned if it was so

Have you calculated the force?
I think we agree there is a centripetal acceleration and force for the circular path up to the point C. If you accept the force a nanometer before C, why does it suddenly become so unreasonable to have it for one more infinitesimally small distance?

 

[tmn50]

what about the first poin of the circular path is there any vertical acceleration ?
the body started with no speed
and I don't know there is something which make no sense in this when i calculated the force it was 2400 N (nothing wrong ) while mg = 800 N

 

[Delphi51]

At the top of the circular path, velocity = 0 so no centripetal acceleration.
The skater does accelerate downward as if in free fall. However, a nm down the path there is some velocity and a tiny bit of centripetal acceleration to the left. As the skater continues to fall, velocity and centripetal acceleration increase.

Your 2400 N calc is for centripetal force plus gravitational force, right?
I'm finding it interesting that the centripetal force is twice the gravitational force regardless of the height = radius. Since potential energy is converted entirely to kinetic energy as it falls,
PE at top = KE at bottom
mgR = ½mv²
v² = 2gR
Putting this into Fc = mv²/R
gives Fc = 2mg.
The total force pushing up on the skater at the bottom of the path is three times gravity regardless of the radius. Most interesting! Skateboarders are experiencing 2 G's of acceleration but feeling 3 times their normal weight at the bottom of a bowl.

 

[tmn50]

thanks man I really apreciate it

 

[Delphi51]

Most welcome. Quite an interesting problem, too.

 

[yster]

thanks a lot Delphi51. I don't think would have managed to make it that clear.

 

[tmn50]

I don't mean to be a bother but i have a problem with an electrochemical cell
here is the link if you would like to give me a hand
https://www.physicsforums.com/showthread.php?p=3317518#post3317518"