Tension of two ropes?
816666
Let A = ten of rope at 28 deg angle from ceiling\n      B = ten of rope at 71 deg angle from ceiling\n \nSum forces in x direction\n\nA cos 28 = B cos 71\n \nsum forces in y direction\n\nA sin 28 + B sin 71 = 254.8\n \nSolving for A & B\n\nB = 254.8/ (tan28 x cos 71 + sin71)\n \nA = B cos 71/ cos28\n\nyou crunch the numbers.
377366
Depends how you measure it. \n\n1, Most people measure the length along the surface of the rope. \n2. Rope can stretch so if you stretch it taunt and measure it from end to end it will be longer,\n3. Rope is made of braid strands, So if you follow the twisted path of a strand, I that gives you the max length.
861484
1) The angle of the rope needs to be considered to calculate horizontal and vertical components of the clerk's force. \nFh = 188*cos(25) = 170 N This is what pulls the box.\nFv = 188*sin(25) = 79.5 N This actually lifts the box slightly.\n\nThe symbol for coefficient of kinetic friction is µk. A lower number makes it easier to move the box - if you spilled some olive oil on the floor µk would perhaps go to 0.1. N is normal force. If this was on a slope, normal would be different from weight, but the floor here is level so N = W? \n\nActually no. Because the rope pulls up on the box slightly, the normal force pressing the box to the floor is less than the box's full weight. The box's weight = m*g = 36.9 kg*9.8 m/s^2 = 362 N. The vertical component of the tension in the rope relieves 79.5 N of that so N = 362 - 79.5 N = 282 N. \n\nSo Ff = .450*282 N = 127 N is the force that friction fights with. The horizontal portion of the clerk's pull is 170 N. So there is 43 N left over to accelerate the box. Newton's 2nd law says\nF = m*a\n43 N = 36.9 kg*a\nso a = 1.17 m/s^2\n\n2) Here the force is directed down at an angle so the vertical component of the force will add to the normal force used in calculating the friction.\nFh = 435 N*cos(35) = 356 N\nFv = 435 N*sin(35) = 250 N\n\nN = the weight of the box + Fv = 259 N + 250 N = 509 N\n\na) Ff = µk*N = .57* 509 N = 290 N This is not enough to hold the box stationary so 356 - 290 N = 66 N are available to accelerate the box. Newton's 2nd law again:\nBut first we need the box's mass. W = m*g\nm = W/g = 259/9.8 N = 26.4 kg. Now\nF = m*a\n66 N = 26.4 kg*a\na = 66/26.4 m/s^2 = 2.5 m/s^2\n\nThis is getting complicated. To move 9 m will take t seconds:\nx = (1/2)*a*t^2 Solve for t.\n\nb) Ff = µk*N = .75* 509 N = 381 N This is more than the horizontal portion of the clerk's push, so the box will resist. The answer is 0.
30686
Let the length of the equal ropes be x ft\n    the length of the longer rope is (x+2)ft\n\ntherefore,\n\nx+x+x+(x+2) = 44\n4x = 44-2\n4x = 42\nx = 10.5
803924
(1) The tension in each of the two cables assuming the teacher sits exactly in the middle of the cables is (1/2)(9.8 N/kg) (115+4.0 kg) = 583.1 Newtons\n\n(2) The force of friction is equal to 40(9.8 N/kg)(0.06) = 23.5 Newtons\n\nHope this helps :-)
252925
The two simplest pulley systems demonstrate the principles of their operation.  First if there is a single pulley overhead and you pull down to lift a load, you must pull with the same force as the weight of the load because oonly one rope pulls up on the load.  100 pound load = 100 pound pull.\n\nSecond, if the load is below you and you pull up on a rope connected to your own level through a pulley, you pull with a force one half the weight of the load because two ropes are pulling now up on the load.  100 pound load = 50 pounds pull + 50 pounds on fixed rope.\n\nThe overhead pulley required you to pull the rope say 10 feet to lift the load 10 feet; but the pulley below you required you to pull the rope 20 feet to lift the load 10 feet.\n\nYou can calculate more complicated pulley systems either by counting the number of ropes that are under tension that actually lift the load and dividing the load by that number or you can determine the length of rope you must pull to raise the load a given distance and use that ratio to determine your force vs the load weight.  you can do this by making sketches rather than actually setting up real pulleys, ropes and loads.  Good luck.
936912
Look at how the forces break down. For block 1\n\nSFx = m1*a = F*cos(q) - T1\n\nwhere SFx is the sum of the forces in the x-direction, a is the acceleration, and T1 is the tension from the string. For block 2\n\nSFy = m2*a = T2 - m2*g\n\nwhere SFy is the sum of the forces in the y-direction, a is the acceleration, and T2 is the tension from the string. Considering the pulley now\n\nSt = I*a/r = r*T1 - r*T2\n\nwhere St is the sum of the torque, I is the moment of inertia of the pulley, r is the radius of the pulley. Combining these three equations will let you solve for the acceleration.\n\nI*a/r^2 = T1 - T2\n\nT1 = F*cos(q) - m1*a\nT2 = m2*a + m2*g\n\nso we have\n\nI*a/r^2 = F*cos(q) - m1*a - m2*a - m2*g\n\nwhich rearranges to\n\n(m1 + m2 + I/r^2)*a = F*cos(q) - m2*g\n\nSolving for a yields\n\na = (F*cos(q) - m2*g)/(m1 + m2 + I/r^2)
401877
Well, I can't even imagine what you are trying to let us picture but, at the same time, I feel this question must have at least one answer so, here I go without having a clear picture in my mind and perhaps lesser knowledge than required to answer this question-\n\nLets first imagine that this fictitious rope is made up of x number of neutrons and y number of protons.\n\nNow, Mass of x number of neutrons = "x" multiplied by 1.00864 atomic mass units.\n\nAnd mass of y number of protons = "y" multiplied by 1.00727 atomic mass units.\n\nNow, total "expected" mass of the rope = "x" multiplied by 1.00864 atomic mass units + "y" multiplied by 1.00727 atomic mass units\n\nTo Convert total "expected" mass of the rope from a.m.u. to kilograms :-\nUse the relation- 1 a.m.u. = 1.66 * 10 raised to the power (-27) kg.\n\nNow, weighing the rope, we will find that the actual mass of the rope is lower than the expected mass. Let the diference between the Actual mass and the expected mass be [M], now as Einstein told us:\nE = mc^2\n\nWe can express this mass difference as:\nBinding energy = [M] c^2\n\nI think that to determine the physical strength of this rope, one shall have to divide the Binding energy obtained above by the total lengh of the rope.\n\nAfter doing so, you'll get the value of the physical force that is keeping the rope intact.
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...
943484
I expect that you have the data on stretch as a result of tension. \na) The object is just at rest. Tension, T = W = m*g.\n\nb) T = W + m*a\n\nc) Tmax = W + m*a
986137
If I'm not misinterpret your question , you mean that you use two pulleys, one is moving and one is stationary.\nThen the advantage of using two pulley is you can pull the rope from the load side (beneath), while if you have only one moving pulley then you have to pull the rope from above.\nIf you have motor to pull the rope then single moving pulley needs less rope.
832148
You could just use the old force = (G*M*m)/r^2\nWhat is missing from the diagram is that the small weights were held by a long thin wire.  The wire formed a weak torsional spring.  If there were some gravitational force, it would cause the weights to twist the wire until the torque of the twisted wire matched the force of gravitation (times the radius).  He could measure the amount of twist by shining a light at a mirror attached to the weights and seeing how far the reflection moved.
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.
864532
Set the tension of the left (T1) times the cos 53 degress equal to the tension of the right (T2) times the cos 37 degrees. Solve for either T1 or T2 in terms of teh other variable. Then set the sum of T1 times the sin 53 degees and T2 times the sin 37 degrees equal to 23 kg times 9.8 meters per second. Substitute the term from teh first equation into the second one, and solve for the variable that is left. Then just plug that variable into the first equation to get the other variable.
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!!!!
272672
Number of rope segments. only if the last segment rope us pulled upward
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