what is the formula for force?
539746
F = ma\n\nF - Force\nm - mass\na - acceleration
559516
Force of gravitaional attraction is given by:\n\nF=G*m1*m2/(r^2)\n\nwhere:\nF=force\nG=universal gravitational constant\nm1=mass of first object\nm2=mass of second object\nr=distance between objects\n\nr must be infinitely large before force of attraction becomes zero.
83668
There are several definitions of the word "force," but since your question is in the Other Science section, this should be the one you're looking for ---\n\nForce is a vector quantity; its size is measured in Newtons. One Newton will make a body of mass 1kg accelerate from rest to 1 metre per second, in a time of one second.
876548
Newton's Laws of Motion:\n\nI. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.  \n\nII. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.  \n\nIII. For every action there is an equal and opposite reaction.
893876
Simply keep in mind that summation of forces equal to zero.\n\nSum (Fx)=0\nSum (Fy)=0\nlet th\nF1- force o string supporting W1 (weight on the left)\nF2- force o string supporting W2 (weight on the right)\nF3- force o string between the weights\n\nFx=0=F1cos(54) - F2cos(68) +F3cos (13)\nFy=0= W1+W2-F1sin(54) - F2sin(68) -F3sin (13)\n\nwe have 2 equations and three unknowns.\nSince the dimensions or length of the ropes ar enot given we can't employ moments.\n\nlet's use superposition and solve one side at a time\n\nW1=F1sin(54) then F1=W1/sin(54)=37.73/sin(54)=46.6N\nsimilarly\nF2=W2/sin(68)=90.944/sin(68)=98.08\n\nsince \nF1cos(54) - F2cos(68) +F3cos (13)\n\nthen  F3=(F1cos(54) - F2cos(68))/cos (13)\n\nF3=(46.6cos (54)-98.08cos(68))/cos(13)= - 9.6 N\n\n(Negative sign means that the originally assumed 'positive' direction to the left was wrong. F3 is pointing to the right.)
6524
Newton's second law:\nFirstly, this law states that if you do place a force on an object, it will accelerate, i.e., change its velocity, and it will change its velocity in the direction of the force.\nSecondly, this acceleration is directly proportional to the force. For example, if you are pushing on an object, causing it to accelerate, and then you push, say, three times harder, the acceleration will be three times greater.\nThirdly, this acceleration is inversely proportional to the mass of the object. For example, if you are pushing equally on two objects, and one of the objects has five times more mass than the other, it will accelerate at one fifth the acceleration of the other.
1156730
force=mass*acceleration\nforce/area=pressure (usually measured in p.s.i. [pounds per square inch])
4161
Force is a push or pull that one object exerts on another. ex. when you play hand ball your hand exerts a force on the ball. The ball also exert a force on the your hand, the effect that you feel. Forces always exist in a pairs nothing exerts force without having a force exerted on it.
912301
Well, for centripetal force:\n\nForce = (mass) * (velocity squared) / radius\n\nand from there?  I'm lost.  I've tried plugging in all your information and it feels like we're missing a known variable.  My first assumption was to put in 7.18 as the force, but then we don' t the mass.   And if we plug in (7.17*mass), we're left with force as a variable.   \n\nOye, good luck.  It's been 10 years since I've taken physics.
544891
The formula is actually divided up into two cases:\nForce friction = Force Applied when Force Applied < mu Static x Force Normal (i.e. nothing'll move below this threshold)\nForce friction = mu dynamic x Force Normal otherwise.\n\nThere is no formula for the two mus; they are measured experimentally.
755944
Net force is required to change the motion of an object, or accelerate the object. Often forces are balanced, so net force = 0, so motion does not change.\n\nWork is in the same units as energy, not force. Work is related to force by \nW = F*d*cos(theta) where F is the applied force, d is the distance through which the force has been applied, and theta is the angle (if any) between the force and the distance. \n\nFor example if a 100 Newton vertical force is applied to lift an object to a table 5 meters above the starting point, 500 Newton*meters, or 500 joules, of work have been done.
889055
Power:\nP=dE/dt = dW/dt\n\nForce:\nf=ma\n\nEnergy:\ne=mc^2
217280
According to Newton's second law,\nF = m*a \nF = Force\nm = mass\na = acceleration\n\nnow, also,\n\na = (v-u)/t \nv = final velocity\nu = initial velocity\nt = time\n\nput this expression into first equation, you get\n\nF = m (v - u)/t\n= (mv - mu)/t\n\nalso, momentum = mass * velocity\n\nso,\nF = (final momentum - initial momentum)/t\n\nwe can also say that\n\nF = Rate of change of momentum
853308
The net force on the man is the sum of the force of gravity and the force on the ground. You've already calculated the net force, so simply subtract the man's weight from it. Remember that gravity is pulling him down, so you'll be subtracting a negative number (i.e. the final result should be greater, not less, than the net force).
589619
You need to find the spring constant, k, which should be in units of force (e.g. meters per Newton or m/N) per unit  length (m).  In this case:\n\nforce (N) = mass, m (kg) times acceleration, a (m/s^2)\nm = 0.2 kg\na = g = 9.8 m/s^2\n\nf = ma = (0.2 kg)(9.8 m/s^2) = 1.96 N\n\nk = force, F (N) divided by elongation, delta L (m) = F / delta L\n{The actual equation has a negative in front. Don't worry about the negatives in this exercise. Since force is in the opposite direction of the extension, one of them would be negative, and therefore would cancel out the leading neagtive sign.}\n\ndelta L = 0.1048 - 0.1037 = 0.0011 m \n\nk = 1.96 N / 0.0011 m = 1781.81 N/m\n\nFn = 0.5(3.14)sqrt[1781.81 / 0.2] = 148.26 Hz
738706
Both questions use the formula F=ma (Force = Mass * Acceleration):\n\n1) F = ma\n1310 = 214*a\na = 1310/214 = 6.1215 m/s^2\n\n2) F = ma\n3.6 = m*9.8\nm = 3.6/9.8 = 0.367 kg
72896
Newtons is the unit for measure of force