Falling in Water ```Name: Laura S Status: student Age: 17 Location: N/A Country: N/A Date: 1999 ``` Question: I am trying to conduct an experiment relating to the effect of gravity on things falling in water, namely a swimming pool. I know that on land things fall at a rate of 9.8 meters per second squared, yet I am trying to experimentally find a similar value for things falling in water. Is this possible to do? Replies: You may, but I don't think so. Gravity applies the same force to us in water as in air; the difference in the perceived force is because gravity is also pulling on the water as well. For, say, a block of metal to fall in water, an equal volume of water must rise. Thus, the total force on the metal will be equal to the weight of the metal block minus the weight of an equal volume of water. So, the weight of an object in water will not be a function of only the mass of the object, but also of its volume. The 9.8 m/s^2 gravitational acceleration for objects in a vacuum also translates to a weight of 9.8 N/kg. The weight is a function of the mass, and of the mass alone. Under water, you need to consider the force from the bouyancy of the water, which does not depend on the mass of the object, but on its volume. Richard Barrans Jr., Ph.D. Laura - I don't have your complete answer, but an important thought for your understanding. The 9.8 meters per second squared is the acceleration due to gravity near the surface of the earth in a vacuum. The force on an object is F=ma where a is the acceleration due to gravity (g). When falling through a fluid (e.g., air or water) the actual acceleration is the result of the combining of F(g) and the F(d)... the force of gravity compared to the force of the drag of the fluid. The latter is affected by the speed of the falling body. So, 9.8 is only true in a special case - a vacuum. To find the actual acceleration may take either someone with more insight in fluid dynamics than I or some experimentation on your part. I would expect though, that there exists a relationship between the viscosity of a fluid and the drag it creates on an object. Hope that helps. Larry Krengel The problem is that the 9.8m/s2 acceleration that you are referring to is due to the effect of gravity. This means that this acceleration can be calculated by an equation similar to the following: (G * m1 * m2)/(r*r) Where G is the gravitational constant, m1 is the mass of one of the objects (the earth), and m2 is the mass of the other object (the falling object), and r is the distance between the objects. When calculating the acceleration (F = ma), these equations cancel out the mass of the falling object, and that is why we say all objects fall at the same rate on earth. Also, since water, air, ice, or anything else is left out, this means the acceleration due to gravity on land or under water is the same. The one factor that does have an affect is the distance, or altitude of the objects. This effect can and has been measured. Since acceleration due to gravity is the same, why do objects fall at different rates under water? Well, these equation also ignore friction. There is a point at which, the drag (or air friction) causes falling objects to stop accelerating. This is known as terminal velocity. Different fluids, in this case, air and water, affect falling objects differently. What you are observing is the difference between air and water, not differences due to gravity. This also explains why you are having difficulty in measuring the acceleration for different objects. Unlike gravity, which is the same for all objects, drag is much, much more complex, and depends very highly on the characteristics of the fluid (including temperature, currents, etc) and the object being affected, including shape, mass, surface smoothness, direction, and velocity. This means the effects are different for each object, and are even different for the same object on different drops. For example a feather never drops exactly the same way twice--because even very minor differences in its starting position change how it interacts with air and cause it to fall differently. That said, the equation for the drag force is: D = 1/2 CpAv*v Where D is the drag force, C is the drag coefficient for the object (usually from 0.5 to 1.0), p is the density of the fluid, A is the cross-sectional area of the falling object, v is the velocity of the falling object. Notice that the cross-sectional area is used. This means for any object not a sphere, the drag will change as the orientation of the object changes. I hope this helps explain your situation a bit. Unfortunately, as you can see from the equation for drag, it is a pretty complex problem. The acceleration of an object affects its velocity and is determined by the forces acting on the object, and in turn the forces are determined by the velocity, this problem is difficult to solve, and is a system of differential equations. Thanks, --Eric Tolman Laura, In air things ACCELERATE at 9.8 meters per second squared. Actually they only accelerate at that rate for a short period of time until the forces due to wind resistance start to counteract the forces due to gravity. In water you'll have a similar situation. Things will initially accelerate at 9.8 meters per second squared but, because of the greater viscosity of water (compared to air), the water resistance will partially counteract the gravitational force with much smaller speeds. A heavy, streamlined object will fall much faster than, for example, a towel thrown into a pool. In addition, the gravitational force acting on an object in a pool is partially counteracted by the buoyancy effect of the displaced water. In the extreme case the object doesn't fall at all but floats on top of the water. One thing you could try would be to get a set of balls of the same size but different weights and see what their terminal velocities are in water as opposed to in air. Have fun! Greg Bradburn Yes it is. Falling in water is not qualitatively different from falling in air, but the balance of the forces acting on the falling object is usually quite different. There are three forces you need to pay attention to: gravity (how strong is the attraction between the object and the earth), the buoyant force (what does the object weigh when it is at rest in the water/air, compared to what it would weigh if it were not immersed in water/air), friction (what is the force required to push water/air out of the way as the object moves). Friction depends on the speed of the object relative to the stuff it's moving through; the other two forces are approximately constant over the distances you're interested in. Another project that might be fun and instructive is to compare the rates at which various objects (e.g., baseballs, balloons, lead weights) fall though air and through water. Tim Mooney you might have a little trouble with this. Your idea is to take experimental data, and then try to match to theory -- good physics. You just have to try to get as good a theory as you can. The object will still be pulled on by the earth's gravity, trying to accelerate the object at your 9.8m/sec/sec. But other things will be going on, other forces will be at work. Will they be properly modelled by simply changing the 9.8 number to another (lower) number, OR, do you need to add other terms to the equations???? For example, with just gravity, distance fallen = 1/2 acceleration * time * time. What is the equation for distance fallen in water? This equation predicts that distance fallen will at least follow a square law for time. If you graph out what actually happens, does it follow this type a square law? You can cheat here by adjusting (accleration) to match your experimental results -- this is how you would afterall MEASURE the new accleration. BUT, it may be that no matter what you do, you cannot make this simple equation "fit", with the form of time * time. You might want to put in a new term, say d = 1/2 a t*t + (new constant) * t or something else, you figure out. Now you have a couple ways to make the theory "fit". You also have to come at this from the other direction, pure "theory", as decide what kind of new terms you want to allow. For example it might be hard to justify a term here that goes as time to the 20th power -- too weird a term. Maybe this term does not really depend so much on "time" as on "speed" or position, or something else. Another thing to think about at least is if you changed the water into something more gooey, like syrup. This should not change the form of the equation, but your new constants out front might get adjusted. This interplay is pretty typical in physics. Your experiment may be simple on the surface, but it does a good job of making you think about this stuff. good luck. Steve Ross For exceedingly dense objects the answer would still be 9.8 m/^2, while for other objects the question itself does not make sense. 9.8 m/s^2 is the acceleration due to gravity at the Earth's surface *in a vacuum*. When objects fall through some other media, for example air or water, then in addition to the force of gravity there is also a drag (friction force) on the body from the air/fluid, because the object must push aside the air/fluid in order to fall. The drag is proportional to the downward velocity of the object: when the object is at rest, there is no drag. When it starts moving, the drag grows, and keeps on growing as long as the object accelerates. The drag cancels some of the force due to gravity, so that the object accelerates more slowly as time goes on. And, since the force due to gravity is constant, eventually the drag becomes large enough to completely cancel the force due to gravity, and the object stops accelerating entirely. It then falls at a constant velocity called the terminal velocity. In air terminal velocity for something the size of a human being is about 180 MPH, so a falling human being will accelerate only up to that speed, i.e. for about 80 m/s / 10 m/s^2 = 8 seconds. After 8 seconds he will fall at a constant velocity and no longer feel weightless. The drag on a compact object is proportional to the size of the object, the velocity, the density of the medium, and the viscosity (stickiness) of the medium: F_drag ~ R p_fluid n_fluid v_object The gravitational force is proportional to the mass, which is in turn proportional to the density times radius cubed: F_gravity ~ p_object R^3 Terminal velocity is where drag equals gravity: F_drag = F_gravity or R p_fluid n_fluid v_terminal = p_object R^3 Thus terminal velocity is proportional to the ratio of the density of the object to the density of the medium, the square of the size of the object, and inversely to the viscosity of the object: ``` v_terminal ~ (p_object/p_fluid) R^2 ---------------------- n_fluid ``` For dense objects in tenuous media, like rocks falling through air, p_object >> p_fluid, and terminal velocities are high. For rocks falling through water p_object is much closer to p_fluid, which means terminal velocities are lower. For rocks falling through viscous, sticky stuff like molasses or heavy oil n_fluid is bigger, and you have even lower terminal velocities. Bigger objects have bigger terminal velocities, so if you are in an airplane that suddenly loses its wings it really is better to jump out, like Bugs Bunny, because the plane fuselage will hit a lot harder than you will. And even though humans and spiders have the same density, roughly, a spider but not a human can jump out of a tall tree and live, because the spider's terminal velocity is much lower. I think even very large and dense objects, like ocean liners filled with water, have terminal velocities in water that are only 30 MPH or so. We know this is the case since sunken wrecks are not all smashed to smithereens by their impact with the sea floor. So they accelerate at 9.8 m/s^2 for only half a second or so. But if you got something very compact and exceedingly dense, a ball of neutronium or something, it might accelerate for a few seconds. Grayce Dear Laura, Yes, it is possible to find the acceleration due to gravity of the Earth in water, and indeed, you already have. The acceleration due to gravity of the Earth is still 9.8 meters per second squared, as long as you are near the surface of the Earth, regardless of what the object is falling through. What makes it seem like gravity accelerates objects slower through water is resistance. In air, there is some resistance, though it is usually not noticable until you are moving really fast, for example skydiving, or you have a large surface area perpendicular to your downward path, i.e. a parachute. Water has a much greater resistance and therefore exerts greater pressure opposite the force of gravity, slowing the object more than if the object were falling through air. Click here to return to the Physics Archives

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