homework help; torque?
964672
The answer is  - (0.635 - 0.324)* r * g = 2.9 Nm.\n\n\nYour equation for torque and other things are correct.\n\nFor the system to be in equilibrium the NET torque must be zero.\n\nSuppose that 0.635 kg mass rotates the pulley in clockwise direction.\n\nThen the other mass 0.324kg will rotate the pulley in the anti-clockwise direction.\n\nTorque produced by 0.635 kg about the axis is F x r = 0.635*r*g. (Let us take clockwise rotation  as positive)\n\nTorque produced by 0.324 kg about the axis is F x r = - 0.324*r*g. (anticlockwise direction)\n\nThe net torque about the axis is (0.635 - 0.324)* r * g = 0.311*r*g. \n\nIf there is no friction, the pulley will rotate in the clockwise direction by the action of this torque\n\nIf the frictional torque acts opposite to this torque, then the net torque will be zero and the system will be in equilibrium.\n\nTherefore the frictional torque is - 0.311*r*g. (anticlockwise direction)\n\n\nSubstitute the value for r = 0.95 m and g = 9.8m/s^2\n\nThe frictional torque is 2.9 Nm.
750470
The CG of the stick and the weight together is at the22.4 cm mark, because it is balanced when supported at this point.\n\nIf the CG of the stick were at x cm mark, the torque due to the stick at the 22.4 cm support\n\n134 g * (x - 22.4)\n\nThe toruque due to the 142.3 g weight at the support\n\n142.3g*(9 - 22.4)\n\nThe total torque must be 0:\n\n134 g * (x - 22.4) + 142.3g*(9 - 22.4) = 0\n>>\n134x -134*22.4 + 142.3(-13.4)=0\n>>\nx ~ 36.63 cm
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
838774
each person produces a torque of:\n\n4.1m x 180 = 738Nm\n\nSo total torque is 2 x 738.
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
841789
Assume s=275, v=43, r=0.21 & t is time from start to stop; then v=a*t & s=a*t^2/2, ‘a’ being deceleration. Therefore t=v/a, s=a*(v/a)^2/2; thus 2*s*a=v^2 or a=v^2/(2*s).\nAngular deceleration w’=a/r.\nNow I=0.175, so frictional torque T=I*w’. plug numbers in.
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try the formulas:\n\nforce = mass * aceleration\n\nwork = force * distance\n\n\n\n\nWell i dont speak english very well, but i hope i help u.\n\nKisses
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
69843
work = force times distance\nW=Fs
968030
Torque = Force * Distance(from fulcrum)\ntherefore\nTorque =(mass * acceleration) * D\nThe higher the torque, the greater the acceleration, but you'll need a duration (time) or distance travelled to determine a velocity.\n\nMore torque = more acceleration.
897942
You should only ask one question at a time.\n\nI will answer the first:\n\nConsider the forces on the box:\n\nGravity vertically downward\nfriction pulling backwards\nThe force of gravity parallel to the ramp \nminus the force of friction\n=m*a=m*1.22\nso\nm*sin(34)*g-f=m*1.22\nfriction can be expressed as\nf=m*u*cos(34)*g\nnote that mass divides out\nsin(34)*g-u*cos(34)*g=1.22\n\nu=.52\n\nj
754118
E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
238500
need to know the type of metal, then find the shear force & plug it into your equation.  don't forget to factor in each thread  (Basic solid mechanics)
889055
Power:\nP=dE/dt = dW/dt\n\nForce:\nf=ma\n\nEnergy:\ne=mc^2
790902
moment1 (torque) =0.500 x 20.0 = 10.0Nm\nmoment2 = F x (1.60-0.50) = 1.1F\nHere, F means the vertical component of the 35.0N force. [When you draw a rectangle whose diagonally opposite corners touch a line which denotes the force vector, you will see that it has a vertical component (length) and a horizontal component (breadth)] This is because moment is derived from F x d, where F is perpendicular to d.\n\nSO, after drawing a diagram, F = 35cos 30\nWith this, you can find the net torque, depending on which force is producing a torque in which direction (CW or CCW). \n\nHope this helps=)
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2t, where t = arctan(Electrostatic force/Weight);\nThe electrostatic force will act horizontal, and the weight vertical, and these two must compose to a force along the suspending thread.\n\nSubstitute the numerical values in and find the angle.
1831
Easiest convertor is just Google -- what a great tip someone showed me!\n\nJust go to google and type in\n  55 ft lbs in newton meters\nand up pops the answer:\n  55 (ft lbs) = 74.5699872 newton meters\n\nLikewise\n  65 (ft lbs) = 88.1281666 newton meters\n\nAnd\n  75 (ft lbs) = 101.686346 newton meters\n\nTo give you an idea of what this means, a standard mechanics torque wrench can apply up to 200 newton meters of turning force.