January 6th, 2009
I'm currently in an argument over this... if you throw a ball directly
up in the air (ignore air resistance) what forces are acting on the
ball? More precisely, when the ball reaches the maximum height what
is the acceleration, I think it is still -9.8 m/sec^2 but my friend
says that it is 0, to me that is the same type of theory that if you
are passing a car you are going the same speed when you are next to
it, what I really need is some documentation either way as prove to
see who won the bet.mathisfun...
Your friend is correct. At the maximum height, the NET force
operating on the ball is zero - that is, the combined forces
of both the force applied to the ball when it was thrown, and
the force of gravity, which you correctly identify as 9.8 m/sec^2,
are, at that moment, balanced out and cancel each other out.
This same principle is used to expose astronauts to a weightless
environment, by flying a large plane as close to straight up
as is possible, and then cresting an arc as it begins to fly
downward. At the top of the arc, the astronauts can experience
weightlessness.
The formula for this is: Fnet = m * a where Fnet is net force,
m is mass, and a is acceleration. See this page on Newton's
second law of motion in this Glenbrook H.S. Physics discussion:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l3a.html
When there is more than one force applied to the object, the
results are calculated by using vectors, which are like arrows
with direction and with the amount of force indicated by the
length of the arrows. If the force applied to the ball in
throwing it upward were maintained at a constant level, the
vectors would be of equal length, with one pointing up, and
the other downward, and the object would exhibit the stationary
quality of inertia. This happens when you hold the ball, at
the same height, in the palm your hand. The force you are
applying to hold the ball up is equal to the force of gravity
pulling it down.
If you stop applying that force, your arm will fall, and the
ball will respond to the constant vector of gravity, and fall
to the ground.
When you throw the ball, your briefly apply a force which is
greater than gravity, but, since you cease applying it, the
upward acceleration of the ball is eventually overcome by the
constant acceleration of the force of gravity. At the uppermost
height, those forces are momentarily equal, and the NET force
applied to the ball at that moment is zero.
A discussion of vectors in the calculation of NET force appears
on the next page of the discussion cited above:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l2d.html#Practice
"Situation C" precisely represents the situation you're
describing, and if you click on the button to see the answer
it notes:
"The net force is zero Newtons. All the individual forces
balance each other (i.e., cancel each other out)."
I hope you didn't lose any money... : )
Please do not rate this answer until you are satisfied that
the answer cannot be improved upon by way of a dialog
established through the "Request for Clarification" process.
A user's guide on this topic is on skermit-ga's site, here:
http://www.christopherwu.net/google_answers/answer_guide.html#how_clarify
sublime1-ga
Additional information may be found from an exploration of
the links resulting from the Google searches outlined below.
Searches done, via Google:
"force equals"
://www.google.com/search?q=%22force+equals%22I'm still having troubles, from my understading there is no force
acting upward on the ball after it leaves your hand... maybe if I put
some numbers in I will be able to better understand your explination.
Say we have a 1 kg ball, now when throwing it up say we use a impulse
of 49 newton-seconds, and as soon as you let go of it it starts
slowing down at 9.8m/sec^2 meaning the net force is downward with
magnitude of 9.8 Newtons which implies the only acceleration is due to
gravity, and this force of 9.8 N stays constant for the 5 seconds for
the ball to reach the peak of it's path, and also on the return path
the Force is 9.8 N in the same direction, my question is what force is
added as an upward force against gravity for that point where the ball
is at its maximum, also where does it go for the ball to be able to
start moving again.mathisfun...
I apologize for missing a fine point in your question.
I will admit to having been rushing to complete the
answer in order to make an appointment on time.
I focused on the forces acting on the ball, and neglected
your focus on the acceleration on the ball at maximum
height. francf-ga has provided a very good explanation of
why the acceleration of the ball is -9.8 m/sec^2 for the
entire time it is in the air, once it leaves your hand,
which makes YOU correct in your assertion.
Since francf-ga's comment was added after your Request
for Clarification, it's not clear if it satisfied your
need for further explanation. It certainly satisfied
mine.
If you'd prefer that I remove my answer, I can ask the
editors to do so. Likewise, if there's something that's
not clear in francf-ga's explanation, I can try to
clarify that.
Let me know...
sublime1-gaI'm fine with giving you the payment and even a good rating, but if
you were paying attention to forces shouldn't and acceleration was
constant then the force acting on the ball would have been constant as
well? If you get back to me with either agreeing or disagreeing with
this (and if you disagree explain what other forces are in play
besides the impulse while throwing the ball) I have no problem giving
the small fee of $5, after all you did put probably that much time in
it.and as a side note, the funny thing is that it wasn't really a bet
between me and a friend but an argument between me and my physics
professor, I just didn't want to post that in case some would be
biased that the Dr. would naturally be correctmathisfun...
That's funny about it being between you and the prof!
And what you did to avoid bias makes perfect sense.
Yes, once the ball leaves your hand, having had a force
applied to it, the only force continuing to act on it
is the downward force of gravity, and, in that sense,
the ball appearing at rest at the top of the travel
is not the same as a ball at rest in your palm due
to a constant upward force opposing the constant
downward force of gravity.
The opposing vectors of force would only apply at
the very beginning when a longer vector (more force
than that of gravity) was (momentarily) pointing up
in direct opposition to the downward (and unchanging)
vector of gravity. Once the ball leaves the hand, there
is really only one vector in the picture, which gradually
overcomes the effect of the brief, initial upward vector.
Best regards...
sublime1-ga#If you have any other info about this subject , Please add it free.# |
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