How do I use Bernoulli's equation to find the upward force on an airplane's wing?
821783
Bernoulli's equation can be used to estimate the pressure difference on the upper surface versus the lower surface, using the velocities you are given. Then recall that pressure is force per unit area, so take the pressure difference times the total area of the wings (remember you have two wings), and you have force!
899525
Using the example....from the problem you can tell that the tail wind helps the plane move faster.  Let's say the speed of the plane in still air is x, so the speed of the plane with the tail wind is equal to x+25.  The distance the plane travels is equal to (x+25)5, which is the speed of the plane times the number of hours the plane flew.  Now for the return flight, we know the wind is now working against the plane, so it reduces the plane's speed.  Therefore the speed on the return flight is x-25.  The distance traveled can be calculated by (x-25)6.  Since the flight to the destination and the return flight must be the equal, we can set the  distances equal to each other.\n\n(x+25)5 = (x-25)6\n5x + 125 = 6x - 150\n\nFrom there you can solve the problem.
593602
The answer is to draw an outline of the box and sketch all forces acting upon it.\n\nThen, resolve those forces into their horizontal and vertical components.\n\nThe box is not required to accelerate in order for constant-velocity motion to occur.  So, it is simpler than F = ma.  Just use Sum (F horizontal) = 0, and sum (F vertical) = 0.\n\nSolution:\n\nSum of horizontal forces:\n\nF - 0.3 * N * cos(12) * N * sin (12) = 0\n\nSum of Vertical forces:\n\n- mass * gravity - 0.3 * N * sin(12) + N * cos(12) = 0\n\nBe careful with pounds as a unit of mass.  Better to convert to slugs by dividing by 32.17 ft/s^2 (gravity).\n\nFirst solve for N (the normal force perpendicular to the plane acting upward on the box) using the vertical force eqn:\n\n-(35/32.17) * (32.17) - 0.3 * N * sin(12) + N*cos(12) = 0\n\nThen plug N into the Horizontal Force sum equation to solve for F.\n\nWatch your units!
123215
The well-known Bernoulli equation is obtained from conservation of momentum and energy applied to an inviscied flow. \nIts simplest form is:\n\np1/ρ + (u1)^2/2= p2/ρ + (u2)^2/2\n\nwhere,\n p1 : initial pressure\n p2 : final pressure\n u1 : initial velocity\n u2 : final velocity.\n\nAnd it implies that total energy of the flow through an stream line is constant and is exchanged between kinemaical and statical pressure.
848443
when u will draw ur inclined plane and mass on it there r different forces acting in it \nfirst force will be directly perpendicular to the ground level i.e. Fg=mg\nsecond force acting is mgcos(theta) which is perpendicular to hypotenuse of the inclined and makes up Fn=mgcos(theta)\nthen another component parallel to the hyp. of the inclined \nwhich equals mgsin(theta) which goes down the ramp\nand  then u can\nequal the force going down with the frictional force\nmgsin(theta)-(mew)mgcos(theta)=ma\n(as we know frictional force is (mew)Fn=(mew)mgcos(theta) )\nsolve by putting values\nmew=0.146    ans
481398
sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
830483
1) Billy Boy's answer is exactly not the right answer. It is a common misconception that lift is generated due to the rounded "airfoil" shape, forcing fluid to travel faster on top, and thus generating a pressure gradient through incompressible Bernoulli's equation. That is INCORRECT. A flat board can generate lift if you tilt it and swing it through the air.\n\n2) I don't think you really want a mathematical explanation. Because the mathematical explanation involves complex partial differential equations, or, if you prefer, integral equations. These are known as the Navier-Stokes equations, and they govern 3D fluid flow. A fully mathematical answer would be: the wings generate lift because the solution to the Navier-Stokes equations around the wings shows an upward momentum deficit in the flow, thus generating lift on the wings.\n\n3) But really, it's quite simple. When the plane is flying, the wing is pushing air down. Not straight down, but slightly down, and by Newton's 3rd law, the wing is pushed up. (Who's doing the work? The engines are pushing the plane forward, which forces the wings to push air down.)\n\nIn fact, if you actually follow the link Billy Boy gave, NASA would tell you the same - in different and probably more confusing terms.
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.)
1119653
Ok, lets see here...\n\n1st step is to break velocity into horizontal and vertical components (think right triangle here):\n\nvertical velocity = 25.5 * sin(32.1) = 13.55 m/s\nhorizontal velocity = 25.5 * cos(32.1) = 21.60 m/s\n\n\nPart 1:\nFigure out how many seconds before the rock's vertical velocity gets to zero (peak of arc):\ntime up = 13.55 / 9.8 = 1.383 sec\n\nFind the average vertical velocity during that time with linear acceleration from 13.55 m/s to zero: \nVavg up = 13.55 / 2 = 6.775 m/s\n\nFind the distance based on average speed and time:\ndistance up = 6.775 * 1.383 = 9.368 meters above the roof\n\n\nPart 2:\nSo we know that the rock is at 22.368 meters above the ground with zero vertical velocity.  Now, figure out what it's vertical velocity will be when it hits the ground.  \n\nVavg = average vertical velocity on the way down\nVgnd = vertical velocity when it hits\nt = time from peak to hitting ground\n\nWe know:\nVavg = Vgnd / 2\nVgnd = 9.8 * t\nt    = 22.368 / Vavg    = 44.736 / Vgnd\n\nSo...\nVgnd^2 = 9.8 * 44.736\nVgnd = 20.938 m/s\n\nNow convert to a total velocity using the original horizontal component which isn't changed (right triangle again):\n\nVtotal = sqrt(20.938^2 + 21.60^2) = 30.08 m/s\nAngle = inverse tan(20.938/21.60) = 44.11 deg (below the horizontal)\n\n\nPart 3:\nThis is easy.  Horizontal velocity is a constant for the flight (21.60 m/s).  Just multiply by flight time to get distance:\nFrom part2:  time down = 44.736 / Vgnd = 44.736 / 20.938 = 2.137 sec\nflight time = time up + time down = 1.383 + 2.137 = 3.52 sec\nd = 21.60 * 3.52 = 76.02 meters
1038155
Bernoulli principle if you planning create a sailboat.
152180
Lift force depends on wing area, angle of attack, air density, airspeed, and several other parameters.  Google "aerodynamic lift" and see.  Happy landing.
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.
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)
510692
Try: allaboutphysics.com I think that might help.
681873
They perform that feat by two methods. The classical method that was been taught since the mid 1700's is called the Bernoulli effect discovered by Daniel Bernoulli, a Swiss mathematician. This effect is that a fluid flowing with a greater velocity has a lower pressure than a fluid flowing with a lesser velocity. This pricipal has been put into effect with the construction of the aircraft wing. The upper surface of the wing is curved and has a longer surface over which air must flow than the lower surface. Therefore the air must flow faster over the upper surface than the lower surface. This lowers the air pressure over the upper surface of the wing and causes a lift by the pressure of the air on the lower surface.\n\nThis however is not the only force acting to keep the aircraft afloat though it has been taught as the force for many years. \n\nThe second force that is used, is cause by the angle of attact of the wing on the air. As the aircraft is in level flight the wing is tilted slightly upward. This causes the air to push against the bottom of the wing at a slight angle. \n\nTake your left hand and hold it at a slight angle to the horizontal in front of your eyes with the fingers pointing toward the right. Then with the fingers of the other hand pointing opposite to the fingers of the hand held at an angle, push horizontally against the palm of the hand.\n\nWhen you do this you will notice that the left hand tends to rise as you push against it. That is the second force that helps keeps the plane afloat.
772033
Draw a picture of it with arrows representing the vertical and horizontal velocities as 2 legs of a right triangle.  This triangle will show you that the angle with the vertical is arctan 3.5/5.5 = theta\n\nThen the flight path V will be Vv/cos(theta) or Vh/sin(theta)\n\nSince you're ruling out pythagoras, you really need to find the angle first, then the speed.......
682473
Cars are aerodynamic to reduce wind resistance but this does not create the Bernoulli effect necessarily.\n\nIn fact cars, especially race cars, are designed like upside down aircraft.  Instead of lift, they use the Bernoulli principle to create a down force.\n\nIf cars were round on the bottom and flat on the top then the Bernoulli principle would kick in and create lift.  Probably not enough lift to make the car fly but probably enough lift to make it uncontrollable on the ground.