Physics balancing project! (regarding the center of gravity)?
969341
airplanes use the CG every time it gets on the runway.  Maybe something along those lines.  THere are a lot of info on it if you do google on it.
792058
you should visit the following link\nhttp://en.wikipedia.org/wiki/Centripetal_acceleration
870325
I suggest you build an anti-gravity lifter device that is able to float in the air without anything supporting it.\n\nhttp://www.amazing1.com/grav.htm
80144
Center of gravity of the displaced liquid is the center of bouyancy. The (resultant) weight of the vessel acts at the center of gravity(downwards) and the (resultant) bouyant force acts at the center of bouyancy upwards.\n\nIt is better to have the center of gravity lower than the center of boyancy. for stabality. The lower it is the more stable.\n\nThis is because, if the vessel is tilted, the weight and the bouyant force make a couple. This can be against the tilt or with the tilt. If center of bouyancy is lower than center of gravity, it will tilt more. And vice versa.\n\n YOu can find this out by drawing small diagram of the forces.\n\nAlso you can try this. \nTake a pencil. \nhold it vertical but little tilted. \nnow assume, the bottom end is center of bouyancy and top end is center of gravity.\n\nso, push up form the bottom and push down form the top. You will see that the pencil becomes horizontal. i.e you ship sinks.\n\nIf you assume the opposite - take the top to be the center of bouyancy. Pull the top up and the bottom down. The pencil straightens. YOur ship stabalises.
110744
Short answer: usam da inverse square law Jar Jar!\n\nLong answer:\n\nNewton simplified things by picturing all of a planets mass gathered in a single point in the center of the planet and the surface as a point on a certain distance from that center of mass. So the gravitationall pull (F) on a body (m) at a distance (r) from the mass (M) can be calculated using this equation:\n\nF=G*((M*m)/r^2)\n\nwhere G is the gravitational constant (6.67 * 10^-11 Nm^2kg^-2)\nEarths (equatorial) radius is 6.378 * 10^6 meters and its mass is\n5.974 * 10^24 kilogrammes\nSo lets test this on a dude weighing 100 kg:\n\n6.668 * 10^-11*((5.974*10^24 * 100)/4.068*10^13)\n==>\n6.668 * 10^-11*(5.974*10^26 /4.068*10^13)\n==>\n6.67 * 10^-11* 1.424*10^13 \n=\n979 N (what I got in my calculator)\n\nSo earth pulls on a guy, with the mass 100 kg, with the force of 979 Newtons. That means 9.79 N/kg, earths gravitational force (g). This should also mean that a mass dropped from a small height should accelerate at the rate of 9.79 m/s^2. It´s a simple experiment to do but where I am the value of g is actually 9.82. This is because I am closer to the earths center. Earth bulges out at the equator because of its rotation. I am farther north. But still a simple experiment can be done to validate this calculation. \nSo if the earths mass was 3 times greater than g shoud be 3 times greater. Right? You betcha. And I have already shown that moving closer to earths center increases g. Moving twice as far from the center should decrease g by half. Right? Wrong! Now what? We where so close. Oh man...\ng is reduced to a quarter if we increase the radius by 2. This is because of the inverse square law, a very important law in astronomy. It means that the apparent brightness of a light source is inversely proportional to the square of its distance. This also aplies to gravitational fields. Double the distance, divide by 4.\nSo lets put this all together. You should be able to do it yourself by now. I got 734 N or 75 % of the dudes weight on earth. And the reason he doesn´t weigh more than on earth is because of the inverse square law. Eventhough the new world is three times as massive we still moved twice as far away from the center. So take the weight of your astronaut, multiply by 3 and divide by 4.
1144696
gravity and air resistance.
611971
Mass is a constant; weight varies by gravitational pull.
533021
there are some solved physics problems on \nhttp://www.dctech.com/physics/help/\nbut I don't know if it will help you out.
955246
If you were right at the center then the symmetry of the situation would mean that you'd be weightless since the gravitational force would be balanced all around.  \n\nSimilar situation if you were just a little off the center.  Most of it would balance except for a little bit.  Think about starting at the earth's surface and moving to the center.  You'd start at your weight and become zero at the center.  Assuming continuity and noting the nice behaviour of gravity then it must decrease in a smooth fashion.  If you continued through your hole to the opposite side you'd weigh the same but in the opposite direction, always pointing towards the center.
938804
(1) Balance the forces\n\nSFx = -Tsin(q) + mv^2/r = 0\nSFy = Tcos(q) - mg = 0\n\nSFx and SFy represent the sum of the forces in the x and y directions, and T is the tension on the string. This leads to\n\nTsin(q) = m*v^2/r\nTcos(q) = mg\n\nDivding the first by the second yields\n\ntan(q) = v^2/(r*g)\n\nwhich is readily solved for the angle q.\n\n(2) The centripetal force is balanced by the gravitation attraction. This says that\n\nm*M*G/r^2 = m*v^2/r\n\nwhere m is the mass of the satellite, M is the mass of the Earth, v is the speed, r is the radius of the orbit, and G is the gravitational constant. Also, given the period (T) of orbit\n\n2*pi*r = v*T\n\nSubstituting for r and solving for v\n\nv^3 = 2*pi*M*G/T\n\n(3) At the top of the hill, in order to avoid losing contact the centripetal force must be balanced by the weight\n\nm*v^2/r = m*g\n\nsolving for v\n\nv = sqrt(g*r)\n\nAny faster and the centripetal force will exceed the gravitational force (weight) and the motorcycle will lose contact with the hill.
711024
Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)
935236
The answer depends on what point you use as the center.\nanyway,L=r(cross)p where r is the radius vector for the path, and p is the linear momentum. \n\nBut, to use this formula, we need to find the linear momentum of the bob. Since we know its mass, we need to find its velocity. There are   three forces of the string. tension, gravity, and centripetial force. since the object is not accelerating, the sum of these is 0. So, \nFt+Fc+Fg=0\n\nOr, Ft=-Fc-Fg\n\nIf y is the vertical axis and x is the horisontal axis, then the y compontent of the force of tension is\n\nFt*cos(angle)=-mg\n\nand the x component is\n\nFt*sin(angle)=-mv^2/r\n\ndividing the x component by the y component gives\n\ntan(angle)=v^2/rg\nso, v=sqrt(r*g*tan(angle))\nso, the linear momentum is p=mv=m*sqrt(r*g*tan(angle))\nwhere m is the mass of the bob, r is the length of the string, g is the acceleration from gravity, and angle is the angle of the string to the central axis. \n\nNow, you know the linear momentum, and you can find out the radius vector, so the angular momentum is easy enough to figure out.
225581
see if this helps.....\n\n\n\nAnalysis of a simple gravity pendulum\n \nFigure 2. Force diagram of a simple gravity pendulum.To begin, we shall make three assumptions about the simple pendulum\n\nThe rod/string/cable on which the bob is swinging is massless and always remains taut; \nThe bob is a point mass; \nMotion occurs in a 2 dimensional plane, i.e. pendulum does not swing in and out of the page. \nConsider Figure 2. The blue arrow is the gravitational force acting on the bob, violet arrows are that same force resolved into components parallel and perpendicular to the bob's instantaneous motion, the motion along the red axis, which is always perpendicular to the cable/rod. Newton's second law\n\n \nwhere F is the force acting on mass m, causing it to accelerate at a meters per second2. Because the bob is constrained to move on the green circular arc, there is no need to consider any force other than the one responsible for instantaneous acceleration parallel to the known path, the short violet arrow in our case\n\n \n \nLinear acceleration a along the red axis can be related to the change in angle θ by the arc length formula\n\n \n \n \nThis, however, is not the acceleration we seek because the gravitational force on the bob causes a decrease in angle θ. One accounts for that by placing a negative sign in front of a, thus:\n\n \n \n \nFigure 3. Trigonometry of a simple gravity pendulum.This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance h starting from rest, would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by\n\n \nchange in kinetic energy (body started from rest) is given by\n\n \nSince no energy is lost, those two must be equal\n\n \n \nUsing the arc length formula above, this equation can be re-written in favor of \n\n \nbut what is h? It is the vertical distance the pendulum fell. Consider Figure 3. If the pendulum starts its swing from some initial angle θ0, then y0, the vertical distance from the screw, is given by\n\n \nsimilarly, for y1, we have\n\n \nthen h is the difference of the two\n\n \nsubstituting this into the equation for  gives\n\n \nThis equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration\n\n \n \n \nsame as obtained through force analysis.\n\n[edit]\nSmall angle approximation\nThe problem with the equations developed in the previous section is that they are unintegrable. To shed some light on the behavior of the pendulum we shall make another approximation. Namely, we restrict the motion of the pendulum to a relatively small amplitude, that is, relatively small θ. How small? Small enough that the following approximation is true within some desirable tolerance\n\n \nif and only if\n\n \nSubstituting this approximation into (1) yields\n\n \nSolution to this equation is a well-known, and quite expected, oscillatory function\n\n \nwhere θ0 is the semi-amplitude of the oscillation, that is the maximum angle between the rod of the pendulum and the vertical.\n\nThe term  is a pulsation, which is equal to ,\n\nwhere T0 is the period of a complete oscillation (outward and return).\n\nSince ,\n\nthe period of a complete oscillation can be easily found, and we have obtained Huygens's law:\n\n \n \n[edit]\nArbitrary-amplitude period\nFor amplitudes beyond the small angle approximation, one can compute the exact period by inverting equation (2)\n\n \nand integrating over one complete cycle,\n\n \nor twice the half-cycle\n\n \nor 4 times the quarter-cycle\n\n \nwhich
83009
Center of gravity is not a point but is a line of action of force...\nSince a line is of infinite lenght there is no way you can think of the center of gravity of an object which does not pass through the object.\n\nhttp://www.schoolingkids.com/\n\nhttp://www.futureaccountant.com/
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
950064
Yes in the center the gravitational pull would be equal from all sides, freefall!