Does weight affect the number of G forces applied?
541878
No.  The G-force is multiplied by the mass.  A 500 lb mass will feel like a 1000 lb while experiencing 2G.
894777
Kinesiology II\n Physics of Motion\nI.  Force \nA.  Definition\n1.  Force = mass * acceleration\n2.  Mass is a fundamental dimension, units g, kg\n3.  Acceleration is change in velocity, units m/sec/sec or m/s2\n4.  “Weight” is a force, units pounds or Newtons\n- Pounds = mass * acceleration of gravity\n- Why your weight, but not your mass, is different on the moon\n- Gravity on Earth = 9.8 m/sec2 \n5.  Example:  A person with a mass of 70 kg weighs:\n                        70*9.8 = 686 kg*m/sec2 or 686 N\n \nB.  Aspects\n1.  Force has a point of application or origin\nExample:  location where muscle attaches to bone\n2.  Force has a line of application\nFollows the tendon\n3.  A rotational force is called torque\n4.  Force has direction\nCan be indicated in two (x,y) or three (x,y,z) dimensional space\n5.  Force has magnitude\nMagnitude is measured in Newtons\n \nII.  Rotational Force \n1.  Force (mass*acceleration) may cause rotation about an axis\n2.  Moment of force (=Torque) is equal to the force times the lever arm – the perpendicular, straight line distance between the axis of rotation and the force\n3.  Moment of force (torque) = force* perpendicular distance \nUnits:  N*m\n4.  Sign convention:  \nClockwise moment = negative\nCounterclockwise moment = positive\n \nIII.  Levers – Location of effort force, load or resistance force, and rotational axis\nA.  Three types of levers\n1.  First class\n2.  Second class\n3.  Third class\n4.  Differences between the three have to do with the relative locations of the effort, load, and the axis of rotation\n \nB.  First Class Lever\n1.  “EOR”\nO = axis or pivot point\nE = effort force (as applied by muscle/tendon)\nR = resistance (load) force (outside force such as barbell)\n2.  The effort force and the load force are located on opposite sides of the axis of rotation\n3.  Example:  seesaw\n4.  Anatomical example:  Triceps muscle – elbow extensor\nInsertion:  Olecranon process of ulna\n5.  First class levers are rare in human body\n \nC.  Second Class Lever\n1.  “ORE”\n2.  The resistance (load) force and the effort force are on the same side of the axis of rotation\n3.  The effort lever arm is GREATER than the resistance lever arm\n4.  Example:  wheel barrow\n5.  Anatomical example:  brachioradialis – elbow flexor\nInsertion:  Styloid process of radius (distal – past the center of mass)\n6.  Second class levers are rare in the human body\n \nD.  Third Class Lever\n1.  “OER”\n2.  Effort and resistance (load) forces are on the same side of the axis of rotation\n3.  The effort lever arm is SMALLER than the resistance lever arm\n4.  Anatomical example:  Biceps brachii – elbow flexor\nInsertion:  Tuberosity of radius (proximal)\n5.  Anatomical example 2:  Quadriceps group – knee extensors\nInsertion:  Patella, tibia \n6.  Third class levers are common in the human body\n \nIV.  Mechanical Advantage\n     A.  Mechanical Advantage defined as:\nEffort Lever Arm/Resistance Lever Arm\n1.  A second class lever will always have a mechanical advantage greater than one because the effort lever arm is always GREATER than the resistance lever arm\n2.  A third class lever will always have a mechanical advantage less than one because the effort lever arm is always LESS than the resistance lever arm\n3.  A first class lever can have a mechanical advantage less than, equal to, OR greater than one, depending on the locations of the effort force and resistance force versus the axis of rotation\n \nB.  Mechanical Advantage VS. Speed\n1.  The mechanical advantage of a third class lever – common in the body – is poor – always less than one\n2.  However, the SPEED of rotation created by a third class lever is high\n \n\n3.  Because the origin of the resistance force is located farther from the axis than the origin of the effort force, it must travel a greater distance in the same time\n4.  Greater distance per unit time = greater speed\n5.  Good for throwing objects, kic
227729
The units of a gravitational field are Newtons per kilogram (Force per mass). It shows that the more kiograms an object has, the more force the gravitational field is applied to that object. Also, a gravitational field's strength varies inversely with the square of the distance from the object causing the field. That means the closer an object is in the field, the more stronger the field.\n\nThe same goes for electric fields, except the units are Newtons per Couloumb (Force per charge).
846277
It decreases to 1/4 of the original force.  This is because the force of gravity is inversely proportional to the distance between the two bodies.  Thus, increasing distance by a factor of n reduces force to 1/n^2.
368233
N is Newtons, a unit that measures force. For every 1kg of something, if it accelerates at 1 meter per second per second (m/s/s or meters per seconds squared), than there is a force of 1N acting on it.\n\nForce is calculated by multiplying the mass (not weight) with the acceleration. For example, if you are 75kg, than there is a force of 75kg X 9.8 m/s/s (acceleration due to gravity: roughly constant on Earth) = 735N pulling you down. BUT, there is a reaction force, in this case the NORMAL, pushing you back up at the same time and force. Therefore, the force pulling you down, 735N, take away the NORMAL, also 735N, and you will have a NET FORCE of 0N because you are stationary. \n\nSo in order to resist something pulling at 14N (ignoring air resistance, friction, etc), it needs to pull back 14N.
337761
more than likely multiply the number by 10 so 5 = 50 lbs.  However if you are using a machine with multiple cables, ex. like a high cables and low cables both conected to the weights through pulleys.  It is possible the pulleys have certain amounts of resistance added because certain muscles are stronger than others, so the stronger muscles used in exercising with that cable need more weight.
514076
It's true - gravity pulls harder (i.e. more force) on a more massive (i.e. heavier) object. BUT that object is harder to accelerate - it has more interia. The two effects cancel each other out.\n\nClearly you don't like equations, but putting it in numbers can help:\n\nIf a 1kg object is pulled towards the earth, what will be the force, and what will be the acceleration?\nForce = mass * earth's gravitational field strength\nso Force = 1kg * 9.81N/kg\nForce = 9.81N (or 10N, roughly)\n\nAcceleration = Force / Mass\nSo Acc'n = 9.81N / 1kg = 9.81 m/s^2\n\nNow consider the same for a 10kg object:\n\nForce is now 98.1N (or 100N, roughly)\nSo Acc'n = 98.1N / 10kg = 9.81 m/s^2\n\nSo the acceleration is the same.
57362
Weight is a unit measure of gravitational force.
28057
Weight is the force on an object caused by gravity. It is usually measured in pounds in the English system and newtons in the metric system. To find the weight (in newtons), multiply the mass in kilograms (44) by the acceleration of gravity (9.81).\n\nP.S. Newtons do measure weight since weight is a force.\n\nP.P.S. Don't listen to the guy below me. The acceleration of gravity is not 10. It is 9.81 (assuming you are on the Earth). The weight would be 44*9.81.\n\nhttp://en.wikipedia.org/wiki/Gravity#The_Earth.27s_gravity
397135
The weight of the object is irrelevant unless one is taking into account air resistance. This is because the net force on the object is the force of gravity, so we have ma = -mg, or a = -g, which is all we need to know for the vertical kinematics equations. The only thing we need to know for the horizontal equation is the initial horizontal speed of the package.\nIf, on the other hand, one was concerned with air resistance, one would also need to know approximate air density and cross-sectional area of the object, as well as be able to numerically solve a nonlinear ordinary differential equation (the frictional force is proportional to some function of velocity) so I doubt this is what you wanted.
119697
It depends if it is acting as a Force or a Pressure (a distributed Force). If it is a force, it has units of N (Newtons, or kg m/s²) and if it is a pressure it has units of Pa (Pascals, or N/m² which is the same as kg/(m s²))
44051
The other posters are correct.  The mass of the object is unchanged.  but, the force of gravity is different on mars.  If you were standing on mars holding the object, mars would be exerting less gravity on the object than earth and thus the object would feel lighter.\n\nNewton determined that the force acting on an object is F=-GMr/r3 \n\nWhere G is the universal gravitation constant = 6.672 e-11 m^3 kg^-1 s^-2 and r is the radius (from the center of the planet) and M is the mass of the planet.  Mars is about one tenth the weight of earth, but it is also much smaller, thus you are closer to the center of mars.  The net effect is that an object feels 0.38 as heavy on mars as it does on earth.
247359
Mass interacting with gravity
618210
Not due to gravitational pull?!
452980
Force here in the USA is measured in Foot-pounds.\n\nEuopean measurement of force is either in Newtons or Kg-meter.\nThere is 2.2Lb per 1kg which equals 9.8N which equals 1Kp
1095815
Mass is the amount of matter present in a body and is an intrinsic property of the body. Mass of an object remains the same always at any place.\n\nWeight on the other hand is the force which a given mass feels due to the gravity at its place. Weight is measured in units of Force like Newton (which is the SI unit of Force).\nIf your mass is 60 kgs then your weight is approximately 60 x 10 = 600 Newtons. This is because\n\nForce = mass x acceleration (From Newton's second Law)\n\nThus, weight = mass x acceleration due to gravity
474542
Gravity does not change the mass of an object, only its weight.\n\nIf you travel away from the Earth, or any concentration of gravity, you will weigh less, but your mass is unchanged.