by Ron Kurtus
You can read them to further your understanding of the subject.
|Work Against Gravity||Applied force and force of gravity||India|
|Gravity||Throwing objects from different heights||USA|
|Falling objects||Terminal velocity of falling object||USA|
|Center of Gravity||Center of gravity for a one-axle trailer||South Africa|
|Artificial Gravity||Disagree with artificial gravvity||Australia|
|Overview of Gravity||Meaning of gravity constant "g"||Pakistan|
|Center of Gravity||Determine the center of gravity with plumb line||Ghana|
|Center of Gravity||Physics problem||USA|
|General||Is gravitation force in space?||Pakistan|
|Velocity Equations for Objects Projected Upward||How long does object remain at its peak?||Australia|
Applied force and force of gravity
Topic: Work Against Gravity
January 17, 2015
work done against gravity =mgh in which mg is the force acting on the object and h is the vertical displacement occured due to that applied force. but force of graviy is also mg . so my question is that how will the displacement occur if the applied force and force of gravity are equal
Reetinder Singh - India
The applied force must be greater than the force of gravity for the object. Otherwise the object will not be lifted higher.
When you hold an object, the applied force equals the force of gravity. When your applied for is greater than gravity, you lift the object and do some work.
Throwing objects from different heights
January 12, 2015
If you throw two objects same size and speed from different heights would they hit the ground at the same time?
I assume the answer is no, because gravity does not effect horizontal direction.
Nichole - USA
If you drop the objects or throw then downward from different heights, they would hit the ground at different times, according to the height. The same is true if you threw them in a horizontal direction.
See Gravity Time Equations for Falling Objects for more information.
Terminal velocity of falling object
Topic: Falling objects
January 12, 2015
Let some object hang by a string. So no motion since weight was counter balanced no net force. Now string is cut. Object is falling under gravity and experiences air resistance. Terminal velocity is reached when weight is equal to the air resistance. So no net force and no net acceleration after terminal velocity. But why does the object continue to fall if there is no net force acting on it, why doesn't it just hang in mid air dead stopped ? Is this motion due to inertia ? If it falls a distance of 'x' at terminal velocity what is the work done in terms of W = F x d ?
MD - USA
The force of gravity accelerates a free-falling object. Air resistance is a function of the velocity and it slows down the falling object.
At what is called the terminal velocity, the force caused by the air resistance equals the force of gravity. Thus the object will no longer accelerate. However, the object is traveling at the terminal velocity and has momentum according to that velocity, due to the Law of Inertia. Thus, it will continue to move at the terminal velocity.
Also, since the force on the object is the force of gravity minus the air resistance force, resulting in a no net force. Thus, the equation W = F x d cannot be used. However, the work on a falling object can also be defined as the difference in potential energy times the distance. Since PE = mgh, where "h" is some initial height, the work done at a distance "d" when moving at the terminal velocity is P = mgd.
See Work Done by Gravity Against Inertia and Air Resistance.
Center of gravity for a one-axle trailer
Topic: Center of Gravity
January 1, 2015
Please could you help with formulae to calculate the horizontal & vertical centre of gravity of a single axle trailer/caravan with the only measurements are mass, length & width of trailer/caravan & distances from front/rear to axle & axle to front/rear - eg
Mass - 800kg
Total length - 5m
Total width - 2m
Length front to axle - 3m &
Length axle to rear - 2m
Only thing I possibly fathomed was
Xcg = (mass X length)/mass & Ycg = (mass X width)/mass
I am more than likely far off the mark.
Ian - South Africa
There seems to be information missing in your problem.
Assuming the trailer is rectangular and has equally distributed weight, the CG would be at 1/2 the length and 1/2 the height.
If the trailer has a single axle, you would want to move the CG to be over that axle. That means redistributing the weight (or mass) such that 2/5 is on the side from the front to the axle and 3/5 of the weight is on the shorter side to the rear.
Disagree with artificial gravvity
Topic: Artificial Gravity
December 6, 2014
I came across some incorrect information on your website and thought I should bring it to your attention. The page is the one dealing with Artificial Gravity
The topic is
Generating or Simulating Gravity in a Zero-G Environment:
In the movies, or even on some websites, we are led to believe that gravity can be created in space (in a zero-gravity environment), by spinning the chamber/ room so fast, that the people inside it are forced against the outside of floor of the spinning room (as per the figure in your article). This, however, will not work.
We've all been on the spinning ride at the carnival when we were young, and many of us have seen the spinning bucket experiment at some point, so most people just assume that the act of spinning in and of itself will create a gravitational force, exerting rimwards from the hub of the spinning room, aka centrifugal force. This can only work if there is a gravitational force already acting upon things in said room. As rotational force is applied and increased, gravitational force starts to compete with the rotational force, until "gravity" is applied -at most- at a 90 degree offset to its normal downward pull. I say again, this kind of gravitational simulation only works because it is being done in an environment where gravity already exists and is being exerted.
In a zero-gravity environment, a spaceship/ space station could have a room spinning at bone breaking velocities, but it cannot create gravity from nothing. An astronaut floating in the middle of the spinning room would continue to float in the middle of the spinning room, regardless of how fast the room spins, and no amount of careful maneuvering would create a gravitational force gluing him onto the floor of that room. Newtons first law of motion: An object in motion stays in motion unless acted upon by an outside force. Similarly, an object not-in-motion, will continue to remain not-in-motion, unless acted upon by an outside force. The only way a spinning room can simulate gravity in a zero-gravity environment, is if there are 'walls' that extend outwards in significant length and width, from the centre of the spinning room, just like spokes on a bike (but significantly wider than spokes, of course). These 'walls' will in fact become the 'floor'. As the room starts to spin, an astronaut floating in the centre of the room will soon find himself struck by one of the spinning 'walls'. As this 'wall' continues to push against the astronaut, it will apply force against him. The faster the room spins, the more force the wall will apply against the astronaut. When moving at a speed -which I've not yet worked out- the force applied by the 'wall' against the astronaut will be equivalent to the force normally applied by gravity, and the astronaut can then start using this 'wall' as his new 'floor'.
Thus a circular room is unnecessary, as the thing traditionally thought of as the floor, would actually become the wall, and what would traditionally be a wall, would become (when spinning) the floor. So, beyond the desire for the extra structural stability that might come from spinning a circular room vs. spinning a square room, the shape of the spinning room is entirely unimportant. So long as there are long and wide spokes that radiate outwards from the centre of the rotational force, then gravity can be simulated as these spokes become the floor.
So, the figure on the page is incorrect. Instead, if the circle were divided into 4 equal quarters (or it could be any denomination, really, I just thought 4 quarters would be easy), with the lines radiating straight outwards from the centre, to the circumference. The astronaut would stand on any one of these lines and would exert force against it equal to the force it exerts against him (Newtonian physics again). Eventually, when spinning fast enough, these spokes would simulate normal gravity for the astronaut and would become the new floor. If that doesn't make sense, think instead of a windmill. If you could take a windmill into space and have it spin there, then someone standing on one of the blades would feel that gravity was being simulated.
I hope this makes sense. I hope I've done a good enough job at translating my thoughts into words. I expect that this should all make sense to you, though. This is one of those things that I expect we all take for granted at some point, but as soon as we see the reality of it, it seems so obvious, that we can't believe that we ever thought otherwise.
Raz - Australia
Thanks for your comments an analysis of Artificial Gravity. I tried to be very careful in stating that motions such as acceleration and rotation (centrifugal force) only simulate the effect of gravity. They do not create gravity.
Centrifugal force is a function of inertia, as is the force created from acceleration, and does not require gravitational force to apply.
The spinning structure, with the curved wall acting as the floor would not have to spin at break-neck speed. The example I provided in the lesson stated that the rotation would be less than 5 revolutions per minute.
Anyway, it is good that you put some thought into the theory.
Meaning of gravity constant "g"
Topic: Overview of Gravity
November 19, 2014
My question is about physical meaning og g=9.8 value.
Does it mean that an object having initial velocity as zero will cover a 9.8 meter distance in 1 second, falling under gravity near earth?
If yes then what if initial velocity of an object is not zero? Let say it 5 m/s then how much distance it will cover in 1 second falling under gravity?
Naveed - Pakistan
"g" represents the acceleration of an object due to the force of gravity. g = 9.8 meters per second squared. It is correct that a dropped object will fall 9.8 meters in one second. The equation for distance with respect to time is y = (gt^2)/2. Thus y = 9.8*2*2/2 = 18.6 meters. See Gravity Displacement Equations for Falling Objects for more information.
If the initial velocity is not zero, it depends whether the object was thrown up or down. See Objects Projected Downward and Objects Projected Upward.
Determine the center of gravity with plumb line
Topic: Center of Gravity
November 9, 2014
describe an experiment how to determine the center of gravity of an irregular shape using the plumb line method
eric - Ghana
When an object is suspended so that it can move freely, its center of gravity is always directly below the point of suspension.
You can find the center of gravity in an object experimentally by hanging it from several points and using a plumb line to mark the vertical line. The intersection of two or more vertical lines from the plumb line is the center of gravity for the object.
For an irregular three-dimensional object, you need to carefully mark the path from the plumb line.
Topic: Center of Gravity
August 11, 2014
Will you answer a physics problem for me with regards to center of gravity and torque? It is a work problem form a college physics text on finding the mass that is distributed on the wheels of a car after another load is added to the vehicle. The answer is supplied by the book However, I simply can't get my calculations to match those of the authors.
Phillip - USA
I really can't tell unless I see the exact problem. One common error is not using the correct units.
Is gravitation force in space?
May 7, 2014
is there gravitational force in space? if it is there then explain me how it works?
saba - Pakistan
Gravitational force fields move outward through space from all bodies of mass. For example, we observe the force from the Sun and the Moon in the changing tides. See Influence of Gravitation in the Universe for more information.
How long does object remain at its peak?
Topic: Velocity Equations for Objects Projected Upward
April 18, 2014
When an object is thrown upward, what is the duration of time at the peak of its travel, ie when v=0? Or how long is it "suspended" between the upward force and gravity? Is it actually stopped mid-air for a finite time, and can that time be measured? Is that time the same regardless of mass and force? Thanks for your help! Martin.
martin - Australia
That is a good question. In theory, the object should change directions instantaneously upon reaching the maximum height, such that the velocity is zero at that point. It is just like holding an object from a height and dropping it. The object would immediately accelerate from v = 0.
However intuitively, it would seem like there would be a pause before changing directions. The instantaneous change in direction would mean an infinite acceleration.
Another way to look at it is when an object is projected upward at an angle, such that its path is in a curve. Reaching the maximum height and curving downward would only mean a continuous change in velocity.
I hope that helps.
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