Rope-tension question?
832042
The vertical, horizontal components of all forces should balance.\n       Let the tensions in the ropes be T1 and T2 newtons.\n       T1*cos(90-71) + T2*cos(90-28) = 254.8\n       i.e., T1*sin71 + T2*sin28 = 254.8\n\n       For horizontal componets (which are opp. & equal)\n       T1*cos71 + T2*cos28 = 0\n\nSolve for T1, T2 please.
968006
The D=8" pulley is rotating at 120 rpm, so you can use its circumference (C) and its rotation speed (W) to find the length of rope (or whatever) that passes over it every minute:\n\nL = W*C = W*πD = 120*π*8 = 3015.9"\n \nNow, this must equal the length of rope passing over the smaller pulley (because its all 1 rope). So use the same equation as before, but with the smaller pulley's circumference (c) and rotation speed (w):  \n\nL = w*c = w*π*d\n\nSolve this for the diameter of the smaller pulley (d):\n\nd = L / (w*π) = (3015.9) / (150*π) = 6.4"\n\nThat's the long way.  It boils down to the (much simpler) expression:\n\nd*w = D*W\n\nFrom which you can get the smaller diameter:\n\nd = (D*W) / w = (8*120) / 150 = 6.4"\n\nBut you have to understand the first way in order to understand the easier way...
686255
your problem is statically determinate. so, use static equations to calculate the tensile force in the cable. maximum sag will occur at the center of the cable where the load should be placed.\n\nyoure asking for deflection and the force. it cannot be. the maximum allowable sag should be used to calculate the tensile force, either, calculate the sag using the allowable tensile force. this can be solve based on the area of the section and the allowable stress.\n\nI hope I was able to answer your question. better if you could send me the figure. my email is exaatto2000@yahoo.com or rbra_engrs@yahoo.com\n\nruel
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
973341
A)\n\nTo find tension in the wire, we'll calculate the moment about the base of the beam:\n\nweight of beam = 145 N  acting at l=1.7m/2=0.85 m\nweight of sign = 235 N acting at  l=1.7m\n\nonly the vertical component of force in the wire will be used \nbecause the horizontal component passes through the base\n\ntherefore Tvert=Tsin35 acting 1.35 m from the base\n\nso here's the Moment eq\n\n-235x1.7 (down) - 145x0.85 ( down ) + Tsin35 x 1.35 = 0\nT=(235x1.7 + 145x0.85)/(1.35xsin35)\n  =675.1N (up and to the right )\n\nB)  from above the T_vert=Tsin35 = 387.2 N (upwards)\n    and T_horiz=Tcos35=553.0 N (inward)\n\nNow we can find vert and hor forces at the base\n\nHorizontal====>       horizontal Force + 553 N (inward ) = 0\n\nhorizontal force = -553 N (wall is pushing outward ) \n\nVert====>Sign + beam + T_vert + vert_base_force = 0\n              -235 (downward ) -145 (downward ) + 387.2 (up) +vert_base_force=0 \nvert_base_force = 235 +145 - 387.2= -7.2 N ( down )
659102
Gimmy a rope to help you
837697
1.  hard description...but if the cables are on opposite sides of the "weight" than the answer is...\n\nwell, if the 30 degrees is to the vertical than tension is equal to  29/cos30 degrees\n\nif it is to the horizontal than it is 29/sin 30 degrees.\n\n2.  again depends on the instructions...if thre 30 degree cable is to the vertical its    29 x tan30   \nhorizontal is 29/tan30\n\nsorry my first one was wrong, had to edit it
838418
Take the mass 3M on the ramp.  The weight of the mass 3Mg will have a component 3Mg(cos(theta)) along the ramp; (theta is the angle the ramp makes with the vertical).\n\nThis will be the force acting on it along the ramp, downward.\n\nThe upward, along the ramp force on it will be Mg, the weight of the hanging mass.\n\nIf Mg > 3Mg cos(theta) the block will move up.  If Mg < 3Mg cos(theta), it will move down.  if the two are equal, it remains stationary.\n\nOr,\ncos(theta) < 1/3 , moves up\ncos(theta) = 1/3 , stationary\ncos(theta) > 1/3, moves down.
909402
Well, it's Wednesday morning.  I hope this answer still has some value to you.\n\nI started by summing the torques at the hinge.  There are three:\nT(light)=5*9.8*7.5*cos(37)\nT(cable)=-tension*6.31*sin(37)\nT(pole)=8*97.5/2)*cos(37)\n\nThese all sum to zero.\n\nI computed tension=140.5\n\nThe sum of the horizontal forces must be zero.\nThese are the tension in the cable and the horizontal reaction force at the hing.  So the horizontal component is 140.5\nThe vertical forces are the weights of the pole and the light offset by the vertical reaction force at the hinge\nthe vertical component is (8+5)*9.8\n=127.4\n\nI checked the answer by computing the magnitude of the reaction force two ways.  Using Pythagorean theorem \nsqrt(140.5^2+127.4^2)\n=189\n\nI then summed the forces parallel to the pole:\n5*9.8*sin(37)\n8*9.8*sin(37)\n140.5*cos(37)\n=189\n\nj
754405
Let L be the initial rope's length and R be the radius of the equator in metres.  Then \n\nL = 2piR\n\nAdding 1 to the length, we get:\n\nL+1 = 2pi(R+x)\n\nwhere x is the extra radius, i.e. the height at which the rope would float.  Combining these two equalities we get\n\n2piR + 1 = 2pi(R+x)\n2piR + 1 = 2piR + 2pi*x\n1 = 2pi*x\nx = 1/(2pi) = 0.159 or about 16 centimetres.\n\nThe cool part about this is, it's independent of the earth's radius!  The result is the same whether you start with the earth, a basketball or a marble.
866574
Mental picture for you.  Take a belt by the ends.  Now pull as hard as you can, you have now created tension in the belt.   Get it now?  a kind of stress that's built up.
427145
the vine doenst touch the ground so its harmonics with one fixed end, the answers are the the first and second harmonics\n\nthe tension is provided by his weight and the weight of the vine\n\nyou will proabably assume that he is hanging at the aboslute bottom of the vine and not grabbing anywhere in the middle
730943
tension force always act away from the point of contact. also tension is self adjusting (ie. tension will act only if there is there is tendency of motion like friction)
50723
its his weight * gravity. so the tension is: 75*9.8 which is about 735 N
708487
(A) centripetal force = weight of ball + Tension\n\n(mv^2)/r = mg + T\nT= (0.3x(4^2 )/0.72) - 0.3x9.81\nT = 3.72 N\n\n(B) centripetal force = Tension - Weight of ball\n\n(mv^2)/r =  T - mg\nT= (0.3x(4^2 )/0.72) = 0.3x9.81\nT= 9.61N
109486
This is an Static problem:\nThe forces in the wire are:\nIn the middle of the rope the mass of Jonathan: 50kg.\nRigth side: T1\nLeft side: T2\n\nThis is a Vectorial equation: T1+T2=mg\n\nSince Johnatan is in the middle of the wire T1=T2... \n2*T1=mg\nT1 has two components T1x and T1y (horizontal and vertical)\nThe vertical component of T1 is the one that act against weight.\nT1y=T1*Sin(10)\nThen:\n2*T1*Sin(10)=mg\n[using mks system g=9.8m/s^2]\nT1=mg/[2*Sin(10)]\nT1=50*9.8/[2*Sin(10)]\nT1= 1410 Newtons.\nThe tension at "this point" is 2T1 =2820 N.\nHope it helps
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.)