Homework help on "Forces in Fluids"?
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#1 is 400N :)\n#3 is 350N
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You need to sum the components. Let Fr, Fl, Fu, and Fd be to right, left, upward, and downward directed forces, respectively. Then\n\nFx = Fr - Fl\n\nand\n\nFy = Fu - Fd.\n\nThe right and upward directions are taken as the positive direction so that the minus sign indicates that the forces (Fl and Fd) are directed to the left and downward.\n\nThe magnitude of the resulting force is then computed as\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the direction can be found from\n\ncos(q) = Fx/F\n\nUse some reasoning to be sure that the direction you compute is consistent with the orientation of the force components.
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1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
30290
draw the free body diagram. ok, there's a force of 21N going down and an unknown force going up. you know that the net force is 12N. so that means how much of the 21N force is being cancelled out by the buoyant force to produce the new weight? so it's just 21N-12N which is 9N
561732
water displacement and boyles law of bouyancy
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The upthrust force equals the weight of the water displaced.\n\nIf d = density of the water,\nV = volume displaced,\ng = acceleration due to gravity,\n\nThe force of upthrust is,\nF = d*V*g
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1. The bouyant force of the water on the object is the difference in the reading, i.e. BF=(.0554-.0370)g\n                                   =0.0184gN\n\nBased on Archimedes' Principle, BF =Volume*Density of water displaced.\n \n0.018g=Volume*1000g\nVolume=.018/1000\n            =1.8*10^-5m3\n\n2. Density of Object=Mass/Volume\n                                =.0554/1.8*10^-5\n                                =3078kg/m3
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The following link is a pdf slide show that gives examples of buoyancy calculations and theory of floating objects.\n\nhttp://www.colorado.edu/physics/phys1010/phys1010_fa05/lectures/class12-09-29.pdf
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http://en.wikipedia.org/wiki/Pressure_gradient_force\nhttp://en.wikipedia.org/wiki/Temperature_gradient\nHope this helps, good luck!
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You must already understand the displacement description so here's an attempt to physically understand the process.\n\nGravity pulls everything with mass  towards the center of the Earth.  This includes water and even air.  This is what creates air pressure and water pressure within a column of the fluid.  The force towards the center of the Earth is called the Weight.  Weight is simply the mass of an object multiplied by the acceleration of gravity (32.2 feet per second).\n\nMolecules in fluids can move around very easily.  They are constantly being pulled downward by gravity.  Generally, they are blocked by other fluid molecules already below them but they pile up and apply a pressure to the poor suckers below them.  The molecules at the bottom of the stack are under tremendous pressure in very deep portions of the ocean because they have to support the weight of the molecules above them.  (This also creates a sea level air pressure of 14.7 pounds per square inch)\n\nShips weigh less than the water they displace.  Gravity is pulling that ship down, trying to get it to the center of the Earth.  Once the weight of the water equals the weight of the ship, the water pushes back with enough force to keep it afloat.  Water molecules can flow easily.  When they begin to weigh more than the effective weight of the ship, they will flow beneath it and push the ship upward.
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READ THIS CAREFULLY\nAs the block just slides on the table the net force just equals the limiting friction acting on the body\n\nhence,\nF=coeff. of friction * Nor,mal rxn\n\nnormal rxn= weight of body on a plane\n\nhence F= 20*9.8*0.4\n              =78.4N\n\nACC \nit is just begg to slide and the given data is just unsufficient 2 find the acc. of the box
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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.)
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If i understand your scenario, the force is pushing downward on the block.  The horizontal component of the force is 15 N times cosine of 40 degrees.  The vertical component of the force is 15 N times sine of 40 degrees.  However, there is also the force of gravity, which is (3.5 kg * 9.8 m/s2)  (units are N, like the other forces).  So if you add the force of gravity to the vertical component, you have the total downward force.  Multiply this by the coeficient of friction, and you have the force of friction.  \n\nThe force of friction is directed against the direction of motion.  Subtract the frictional force from the horizontal component, and you have the total horizontal force.  Divide the horizontal force by the mass and you should get the acceleration.\n\nfriction force = [(15*sine 40)+(3.5*9.8)]*.25\nacceleration = [(15*cosine 40)-friction force]/3.5
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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
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p=F/A \n\nSo, you have the pressure of the atmosphere pluss the weight of the water above you.  Which is usually figured as a cylinder.  Hope that helps.
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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!!!!
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The forces on the person are\n\nSFx = 0 = N - Fc\nSFy = 0 = Fs - m*g\n\nwhere SFx and SFy are the sums of the forces in the x- and y-directions, respectively, N is the normal (to the wall) force, Fc = m*v^2/r is the centripetal force, Fs = u*N is the friction force, m is the mass, and g is the gravitational acceleration.\n\nThese equations say that\n\nN = Fc = m*v^2/r\nu*N = u*m*v^2/r = m*g\n\nSolving for the coefficient of static friction\n\nu = g*r/v^2\n\nRemember to convert the frequency (rev/s) into speed (v)\n\nv = w*r = 2*pi*f*r\n\nwhere w is the angular speed and f is the frequency.