working principle of crane?
449839
What type of crane are you interested in learning about?
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
1148481
it is a type of Machinery.
755704
How Does the Windmill Work?\n\nWind Energy\nWind is an indirect source of solar energy. It is caused by the uneven heating of the earth by the sun and the resulting redistribution of air to equalize energy in the atmosphere moving air from areas of high pressure to areas of lower pressure. The energy of the wind is transferred to the windmill’s sails where it is harnessed to perform work.\n\nThe cap of the windmill is turned with the winch attached to the bottom of the tail pole until the sails face the direction of the wind. Depending on the wind speed, wind boards are installed on the leading edge of the sails and canvas sail-cloths are unfurled across the sail-frames. The Miller climbs the sail-frame using the sail-bars like rungs of a ladder attaching the canvas to the cleats on the whip. High winds require less open sail than low wind. The Miller determines the amount of sail surface required based on wind speed and partially furls (reefs) the sails as required. The sails catch the wind and turn the wind-shaft.\n\nThe wind-shaft which is connected to the brake-wheel can be stopped with the brake which consists of a band of friction blocks encircling the brake-wheel attached to the brake lever and controlled via a rope near the tail pole attached to the brake lever. The turning brake-wheel turns the wallower connected to the upright shaft on the fifth level. A wood wheel on the fourth level is used to adjust any irregular spinning of the upright shaft.\n\nThe upright shaft is connected to a large gear called the spur-wheel on the third level. Power to run many things in the windmill is drawn from this wheel. Two stone-nuts turned by the spur-wheel drive quant shafts in turn rotating the runner stones; the top stones of the grinding stone pairs. The lower stone of the grinder is the bedstone; it remains stationary. The spur-wheel also turns a third lantern gear (trundle) to drive power transfer belts, pulleys and shafts used to operate the grain elevators, sifters, grain separator, corn sheller and a vacuum system.\n\nHow is Grain Ground into Flour?\nGrain and flour move through the system using grain elevators and gravity. Grain is carried into the windmill and poured into first level hoppers. From there it is cleaned in the basement with the corn sheller and grain separator. Grain elevators carry cleaned grain to the third level hoppers which hold it until it is dropped into the second level grinders. Corn & oats are ground in the west grinder and dropped into a ceiling mounted sifter on the first level. Wheat & rye are ground in the east grinder. The flour drops into an elevator at the ceiling of the first level, is carried back up to a hopper on the third level and dropped into the sifter on the second level. The sifted flour is stored in hoppers suspended from the ceiling on the first level.\n\nInteresting Notes\nThe windmill contains two scales; a floor scale in the windmill and a wagon scale outside the north end of the windmill. The scale inside the windmill can weigh individual bags of grain. The scale outside is used for weighing loaded and unloaded wagons.
981852
You can solve this problem by resolving the forces along the ramp for part a) and normal to the ramp for part b).\n\na) The net force along the ramp must be zero for the crate to move at constant velocity (ie no acceleration), so we have:\n\nFcos30 = mgsin30\nF=mgtan30\n\nb) the reaction of the ramp on the block will be normal to the ramp and equal to the sum of the imposed forces resolved normal to the ramp, so:\n\nR = Fsin30 + mgcos30
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See the fig in the web site. Instead of lifting the load through a vertical height 'h' we are moving the load through a longer distace 'l' along the plane.\nIn any machine work will be the same.\nWork is the product of force and distance.\nIf the load is moved vertically we would have done a work Wxh.\nBut in inclined plane we move the effort P through a distance l . Work done is Pxl\n But P x l = W x h \nSince l is greater than h, P is less than w.\nTherefore \nAn inclined plane reduces the effort force by increasing the distance through which the force is applied\n\nIn your choise the second and fourth answer is  wrong\nas  no  (ideal) machine can increase or reduce our work. In an inclined plane effort distance (length of the plane) is more than the height. Therefore the first one is correct.
894578
Don't add the two. Only the vertical distance counts in determining work because only in lifting the block did you move it against the force of gravity.
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Here are the key things you need to know:\n\nForces acting on the small box:\n- gravity attracting its mass (pulling it downward)\n- large box (pushing it upward)\n- friction force between the two boxes (pushing it forward)\n\nForces acting on the large box:\n- gravity attracting its mass (pulling it downward)\n- the weight of the small box (= the force of gravity on the small box) (pushing it downward)\n- friction force between the two boxes (pushing it backward)\n- friction force between the large box and the underlying surface (pushing it backward)\n- the horizontal force (pushing it forward)\n\nFirst, treat the two boxes as a single mass, resting on the surface and being accelerated by the horizontal force. Calculate the force of friction between the large box and the surface (remembering that the weight of both boxes is acting to push it against the underlying surface). Calculate the force needed to both overcome this friction and to accelerate the mass of the two boxes at the specified rate (the second part of this requires the formula F = ma). That is the horizontal force.\n\nNow calculate the amount of force required to accelerate the small box at this rate. (Use F = ma again.) Divide this force by the weight of the small box to find the minimum coefficient of friction that will allow the large box to apply this force to the small box.\n\nIf you understand all this, you should be able to draw the two diagrams and answer the questions.\nGood luck!
667120
At the spring's original configuration, there is a force stretching it in the form of the mass.  The force and the amount of deflection are related through\n\nF = -kx\n\nSo the spring is stretched -F/k, or -mg/k, initially.\n\nWork to lift the weight is given by\n\nW = Fd\n\nWe have a force and a distance, so the work is then\n\nW = (mg)(-mg/k)\n= -(mg)^2/k
827110
In order to lift the water from ground level (or below) to a certain height it has to be supplied with some energy so that it moves against the gravitational pull of the earths  gravitational force.\n                                 \n                               So here a motor acts as a device which imparts energy to the fluid(water) so that it moves to a certain height.the two  most important pumps used to pump water are centrifugal pump and recipricating pump.in a reciprocating pump there will be  a piston and a cylinder where  in the piston moves inside the cylinder whereas in centrifugal pump there will be an impeller inside  a casing which rotates and inside it and imparts energy to the fluid so that it moves up.the impeller is rotated by connecting it to the motor.hence a pump is required to lift the water.\n\n                       But pump and motor are not required for drawing water because here,the water is moving in the direction of the earths gravitational   pull.here the weight of the fluid and the direction of gravitational pull are in the same direction.so the fluid will fall under its own weight.so a pump is not required.
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Need more information. In general, the answer is that the dynamic load depends on how quickly the payload decelerates. The dynamic load will be the mass of the payload times the peak deceleration rate. In a strength calculation, do not forget to include the static weight component of the load.\n\nThe deceleration depends on the elasticity of the cables supporting the load.\nIf the crane has a flexible supporting structure, then that could also affect the result. (You didn't say whether the "crane" is the traveling overhead hoist type usually seen in a factory - which typically has a stiff support - or the cantilevered truss type seen on construction sites - where the support might have a little flexibility.)
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Hydraulic machines work on Pascals principle.\nAccording to it the pressure exerted on a confined fluid is tranmitted equally and undiminshed in all directions.\nSo we can apply a small force on a large area and that pressure is transmitted throughout the fluid.\nThis pressure can be used to lift relatively large load(or force) by making it act on a small area of cross-section on the platform of the machine.\nPascals Principle is : P1=P2\nor\nF1/A1=F2/A2
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To give you a simple example...\n\nIf you want to move an object of 200kg it would require a force of 200kg.\n\nIf you use one pulley, it would require half of that force... 100kg\n\nIf you use a second pulley placed after the first one, you are reducing it by half again... and it requires a force of only 50kg\n\nUsing several pulleys is called "block and tackle".
597479
That the crane should lift, at the most, this weight.\n\nIf it lifts beyond its capacity, there could be a catastophic failure.
668137
According to Archimedes Principle, the upthrust force on a body totally or partially immersed in a fluid is equal to the weight of the fluid displaced by the body. \n\nImagine a ship that has a weight of 2,000,000 N.  A large crane is lowering it into the water.  When a part of the keel is submerged, it displaces, say 500,000 N of water.  There is still a net downwards force of 1,500,000 N.  If lowered further, more water is displaced, and therefore the upward force increases, while the net downward force decreases.  There comes a time when the ship, as it continues to be lowered, has displaced 2,000,000 N of water.  What then?  The upward force just balances the weight of the ship, and it sinks on further into the water.\n\nIncidentally, a needle does not always sink in water.  Smear it with oil and gently lower it in a bowl of water.  It floats!!!  Of course, this has nothing to do with Archimedes Principle but rather with the surface tension of the water molecules.
596883
Lift of WHAT?   Crane, Ballon, Brassier, Airplane wing....???\nGive us a CLUE
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.