Does anyone know how to do this problem about force???
954539
The net force acting on the bucket by Newton's Second law =\n3.9X2.8 = 10.92 N upwards.  Forces acting on the bucket are 3.9x9.8 = 38.22 N downwards by the earth and F, the force acted upon by the rope on the bucket which is upwards. So we have\n\nF - 38.22 = 10.92 or F = 38.22+10.92 =  49.14 N is the final answer.
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.
784877
How are they applied? Tangentially? In opposite direction? Or same?\nLet us assume oppo direc and on the border.\nThen net force=35N.\nTorque = r x F = I w\n=35x0.341= ''0.5'' m R^2\nR radius m mass\n ''Check if it is 0.5 only for a solid disk or cyliner about its axis.\n\nHope it helps
799229
a. Calculate the horizontal component of the 485 N.\nFh = 485 N*cos35\nCalculate the force of friction.\nFf = 0.57*319 N\nCalculate the net force.\nFn = Fh-Ff\nCalculate the mass of the box.\nW = m*g\nCalculate the acceleration.\nFn = m*a\nCalculate time to move s = 4.00 meters\ns = (1/2)*a*t^2\n\nb. Repeat with uk = 0.75
812688
1*6N right\n2*No Direction\n3*How much force and where\n4*???\n5*???
227728
Very simple. start out asking yourself what information do i know about the problem. Anything from masses and forces to symmetries of the problem. Next Ask yourself what additional information you need to solve the problem, values, formulas, etc. from there it should all fall out with ease.
975783
The force balance says that\n\nSFx = 0 = -2*g*sin(q) + m*v^2/r\nSFy = 0 = 2*m*g*cos(q) - m*g\n\nwhere m is the mass, v is the speed, g is the gravitational acceleration, q is the angle that the ropes makes with vertical, and r = L*sin(q) is the radius of the orbit. SFx and SFy represent the sum of the forces in the x- and y-directions, respectively.\n\nSFy indicates that\n\n2*m*g*cos(q) = m*g\n\nor\n\ncos(q) = 0.5\n\nHaving the angle, you can solve SFx for the velocity\n\n2*m*g*sin(q) = m*v^2/r\n\nor\n\nv^2 = 2*g*L*sin(q)^2\n\nGiven the velocity, the angular velocity is\n\nw = v/r.
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
1156730
force=mass*acceleration\nforce/area=pressure (usually measured in p.s.i. [pounds per square inch])
685243
Either the object is not immovable or the force is not irresistible.  Both things can not be true at the same time.
40451
well figure out the net force with F=ma and just subtract. like just say 93 x .26 is the net force, the force thats acually being used for motion. so you subtract this from 36 to see what force is opposing this. so the total equation is:\n36 - (93 x .26)
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.
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
655249
I know the following two ways:\n\n1) Force on a body  is directly proportional  to the product of it's mass and acceleration, F=K*m * a(k=constant).The units are so chosen so that K=1.\n Your's is also correct but it is limited to free falling bodies, where a=g=9.8 m/s^2.\n2)Force is directly proportional to the rate of change of momentum.\nF=k*(change in momentum,p)/t=K*p/t. Again K=1 for same reason..\nBoth are related as p=m(v-u)/t=m*a.\nv=final velocity, u= initial velocity.\nhope it helps.
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
118764
Newtons 2nd Law,\nForce = mass * acceleration\n\nIf you have two masses (m1 & m2) that accelerate at the same rate,\nF1 = m1*a --> a = F1 / m1\nF2 = m2*a --> a = F2 / m2\n\nset both acclerations equal to eachother, solve for the ratio,\nF1/m1 = F2/m2\nF1 = F2 * m1 / m2 --> F1/F2 = m1/m2\n\nSo the ratio of Force 1 (F1) to Force 2 (F2) must equal the ratio of mass 1 (m1) to mass 2 (m2) in order for both to accelerate equally.
494540
Haha whats with all these answers, if you know basic physics, One of Newtons laws of motion is that:\n\n"for an object that is stationary or travelling at constant velocity, no force acts."\n\nso you ask what is the force?\nthere is no resultant force, the driving force and resistance are equal and cancel each other out.\n\n\nAnd additionally the basic equation to find force is:\nF=ma (force = mass x accelleration), this is evidence that an accelleration needs to be taking place for a force to act.\n\nLets say you want the force for a 50kg object accellerating at 3ms-2. F=ma  ->  F=50x3 ->  F=150N.\n\nNow lets say you want the force for a 50kg object travelling at constant velocity, no accelleration.\nF=ma  ->  F=50x0  ->  F=0N\n\n\nThere you go, wasnt that easy