2 forces, f sub 1 & f sub 2 are given. The angle between them is 57. Find the magnitude of the resultant?
954885
First lets assign the direction of f(1) as being at 0 degrees, then f(2) is at 57 degrees.\n\nResolve both forces in x, and y\n\nF(x) = f(1) + f(2) cos (57)\n\nF(y) = f(2) sin (57)             f(1) is at 0 degrees so has no y component.\n\nThe resultant force F = sqrt ( F(x)^2 + F(x)^2)\n\nThe angle of the resultant force \n\nArctan( F(y)/ F(x))\n\nYou do the number crunching
257076
dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
391655
force = constant * distance\nF=kx\nforce = mass* acceleration\nF=ma\n\nx=F/k\nx=m*a/k = 4.0*9.8/800= about .05m, or 5cm
1055166
Let v1=C-B , v2=A-C , v3=B-A be the three displacement vectors making up the sides of the triangle. Define f1,f2,f3 to be the three forces X,Y,Z each rotated 90 degrees counterclockwise. Then f1 is parallel to v1, and so on. That is:\n\nf1 = s1*v1\nf2 = s2*v2\nf3 = s3*v3\n\nwhere s1,s2,s3 are three scalars. It's only necessary to show that s1=s2=s3 to prove that the three forces are proportional to the three opposite side lengths (|f1|/|v1|=|f2|/|v2|=|f3|/|v3|). This is easily done as follows:\n\nSince the forces are in equilibrium (f1 + f2 + f3 = 0):\n\ns1*v1 + s2*v2 + s3*v3 = 0\n\nMultiplying (v1+v2+v3)=0 by s1 and subtracting from the above equation:\n\n(s2-s1)*v2+(s3-s1)*v3 = 0\n\nSince v2 and v3 are linearly independent both coefficients must be zero\n\n-->\n\ns1 = s2 = s3\nQED.
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
233618
1 mass\n2 magnitude of separation
948754
we juss need to take vector sum of the two forces. assuming the wind is directly in horizontal direction....... sqrt(9.25sq+1.05sq)=9.3N
774792
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
989843
net force IN THE DIRECTION OF DISPLACEMENTis (1.7*10^6 * cos 14)*2 \n= 3.3* 10 ^ 6 N\nwork done = 3.3*10^6*860=2.838 * 10^9 Joules
784797
Suppose there is load and two men are pulling it in one direction each one giving a force of 5N.\n\nIn the language of vector, two vectors of magnitude 5 are acting in the same direction,\n\nThe resultant is a vector of magnitude 10N in the same direction. \n\nIf they are pulling it in opposite direction the load will not move.\n\nIn this case the two vectors of equal magnitude are acting in opposite direction.\n\nThe resultant of these two vectors is zero.
804380
Torque = F x R, F = (2,0,-3) and R = (0, 0.5, -2)\n\nTherefore Torque = (2,0,-3) x (0, 0.5, -2)\n\n(a) Relative to the origin:\n\nT = [0x(-2) - (-3)x(0.5)] î, [-3x0 - 2x(-2)] ĵ, [2x(0.5) - 0x0] k\n\n   = (1.5, 4.0, 1) N m\n\n(b) Relative to (2.0, 0.0, -3.0), the moment arm becomes\n\nR = (2.0, 0.5, -5.0), then Torque = (2,0,-3) x (2.0, 0.5, -5.0)\n\nT =  [0x(-5) - (-3)x(0.5)] î, [-3x2 - 2x(-5)] ĵ, [2x(0.5) - 0x2] k\n\n   = (1.5, 4.0, 1) N m\n\nThe same as before!!!
519602
Just draw the vectors\nMake sure you understand at what nagles they are\nr1 and r2 are inclined at 45 deg\nYou have asked only for magnitude\nJust apply the formula sqrt(4^2+2^2+2*4*2*cos 45)\nsqrt(16+4+16/sqrt(2))=5.595\nFor the difference part.Just reverse the sign for cos term\nyou get |r1-r2|=2.9472
570716
ok my mechanics is rusty but I beleive the answer goes like...\n\nResolve (down the slope) m.g.cos(90-theta) - F.cos(theta)=0\n\nWhere F is the force exerted on the stand.\n\nF = m.a and cos(90-theta)=sin(theta) so...\n\nm.g.sin(theta) - m.a.cos(theta) = 0 (divide by m)\n\ng.sin(theta) - a.cos(theta)= 0  (rearrange to get)\n\na = g.sin(theta)/cos(theta)  \n(sin(theta)/Cos(theta)= tan(theta))\n\na = g.tan(theta)\n\nHope this helps although without the aid of a diagram I don't thinks its too great!!!
774120
Sure,\nYou are adding two vectors:\n\nV1 has magnitue 0.45 m/s and in the W direction (head of vector faces W) \n\nThe 2nd vector is has magnitude 0.35 m/s and 30 degrees E of N (or 60 degrees from +ve x-axis) and its tail is added to the head of the first). So the angle betwen the two vectors is 60 degrees.\n\nUse the SAS rule for finding the magnitude of the resultant vector\n\nc = sqrt(a^2+b^2 -2ab cos(theta))\n\nVr = sqrt((0.45)^2 +(0.35)^2 - 2*0.45*0.35 cos60)\n   = 0.409 m/sec \n\nThe direction of Vr is from the tail of the first vector to the head of the 2nd.\n\nTo determine the angle, A, bewteen Vr and the first vector use the sine law.\n\n0.409/sin60 = 0.35/sinA\n\nSinA = 0.74 and A= 47.8 degrees\n\nSo the direction of the craft is 90-47.8 = 42.2 degrees W of N
260952
R^2 = 45^2 +52^2 + 2x45x52cos25\n.......= 8970\nR......=  root of 8970\n.......= 94.71 pound.\n\ntan A = 45sin 25/( 52+ 45 cos 25)\n.........= 19.01/92.78\n........=0.2049\nA.....= 11.57 degree from 52 pound\n\n94.71 pound .acts at an angle of 11.57 degree from 52 pound
35588
They're vectors, so take into account the direction. Thankfully, they're all in the same dimension, so you can make north positive and south negative (or vice versa, if you prefer), then add. -1 + 3 - 6 + 2 = -2, that is, 2 km south.\nIf they were in different directions, you'd do the same thing, adding horizontal and vertical separately, then using the Pythagorean theorem to find the resultant vector.
1036294
letter b is the answer.. because the weight of the object is approx. 627.20kg multiplied by 9.8(acceleration due to gravity) so the product is the force..