The vector sum of three co-planar forces?
925243
e is the correct answer\n\ncoplaner means all in the same plane. the magnitude of the three can result in a, b, or c.\n\nd is impossible
610533
answer is 350newton\nsince the system is in equilibrium, we have,\n400*1.4=W(3-1.4)\nthat is W=400*1.4/1.6=350\nand the force on the pivot would be 400-350=50newton
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
834482
The answer is C. Although B seems to be a possible answer at first, there could still be an n-number of forces acting on the object such that the sum of all the force vectors is zero, thus making C the more complete answer. One of Newton's laws says that F=m*a (force on an object equals its mass times its acceleration), or a force of 1 Newton on a 1kg object will make it accelerate at a rate of 1m/s^2 in the same direction of the force.
634747
What you want is express \n\nv = a*v1 + b*v2 + c*v3 \n\nWhere a, b and c are, of course, scalars. The coordinate vector will then be (a, b, c).\n\nNow how do we do this: \n\n(2, -1, 3) = a(1, 0, 0) + b(2, 2, 0) + c(3, 3, 3) \n\nEither you just look at it and try to guess what a b and c can be, or you write this up as a set of 3 equations in a,b and c: \n\n2 = 1a + 2b + 3c\n-1 = 0a + 2b + 3c\n3 = -0a + 0b + 3c \n\n...and then solve it: \n\nFrom the last equation we see that c = 1, then using the second equation, -1 = 2b + 3  <=>    -4 = 2b  <=>  -2 = b\n\nAnd substituting into the first, \n\n2 = 1a + 2(-2) + 3(1) = a - 4 + 3 = a - 1 \n\nthus 3 = a. \n\nSo the coordinate vector is (3, -2, 1) and with these numbers the criteria are (supposedly) satisfied. :)
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
740929
Without a little more context I can't say for sure but, most likely, it's just another way of asking for the vector sum. That's the one vector that is equivalent to adding up the directions and magnitudes of all other vectors in the problem.
829822
The vector potential is defined only to within the gradient of a scalar. That is, you can add the gradient of any scalar field to the answer below and get an equally valid expression for the vector potential.\n\nIf B is the constant (vector) value of the magnetic field then the vector potential is:\n\nA(r) = -r x (B/2)\n\nwhere r is the the position vector and x means cross product. In Cartesian coordinates in which the magnetic field is pointing in the direction of the z axis, the three components of A are:\n\nA_x = -(B/2)y\nA_y = (B/2)x\nA_z = 0\n\nwhere B here is a constant number.
948619
The x direction net force is the sum of the left and right forces.  Let the right force be positive and the left negative.  The y direction force is the sum of the upward and downward forces.  Let up be positive.  \n\nWhen you have both of these (each of which is considered a vector), you can add them together to get a total force vector.  Then use the Pythagorean theorem to get the magnitude, and arcsin, arccos, or arctan to get the angle.
35459
You need to apply vectors.
968370
If the origin of the coordinate system is placed at the middle of the two particles positions are -r/2 and r/2. Two charges of the same sign will repell each other so the force has a positive sign and it's magnitude is given by (C Q^2) / (r^2) Where C is the Coulomb constant.
1043313
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
966154
The magnitude of a force is measured in newtons(N)
653555
Not considering the spheric harmonic of high orders I would have to tell you that my physics spirit tell the surface area  is what counts and there you could have k0.2Pi.R=2Pi <=>k0=1/R\nk1=1/(2R), k2=1/(distance of the 3 points more away from each other)........................
835150
Consider two vectors A and B inclined at angle a.\n\nThe component of A in the direction of B is the vector whose magnitude is A cos a  \n\n The component of A in the direction perpendicular to B is the vector whose magnitude is A sin a.\n\nFor the resultant of the two vectors to be zero, there should not be any vector in the perpendicular direction.\n\nTherefore A sin a must be zero. That is the angle a must be zero or 180 degree.\n\nThus the two vectors must be in a straight line.\n\nWhen a = 0 or 180, cos a = plus one or minus one.\n\nTherefore either A must be in the direction of B or opposite to B as for as the direction is concerned.\n\nNow for the magnitude of the resultant to be zero, A must be opposite to zero.\n\nThus two vectors will have a zero resultant only if they are opposite and equal.\n\nThe resultant of three vectors will be zero if they form a triangle.\n\nWe choose one vector and place another vector with its tail coinciding with the head of the first vector.\n\nWe can repeat if for many vectors. If finally the head of the last vector coincides with the tail of the first vector, we get a closed polygon. In that case the resultant of all vectors is zero.\n\nIf three vectors of equal magnitude form an equilateral triangle in order, then their resultant is zero.\n\nIf four vectors of equal magnitude form a parallelogram in order, then their resultant is zero.
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
795081
I think you want a vector that's perpendicular to a plane passing through 3 given points (each with 3 coordinates), say a = (a1, a2, a3), b = ..., c = ....\n\nTake one of these points and find the two vectors joining it to the other two, that is, the vectors v = a - c and w = b - c. Now take the cross product v x w. It is a vector perpendicular to both v and w, so it is perpendicular to the plane. The formula for the cross product in terms of the coordinates (of v and w) is:\n\nv x w = [(v2w3 - v3w2), (v3w1 - v1w3), (v1w2 - v2w1)] = ...\n\n(substitute v = a - c and w = b - c).