Calculate: a) the total force on the bottom of the bucket.\n           b) the weight of the full bucket.\n\ni hope you can help me with this.thx
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ANSWER A)\n\nUsing F=mg...and because 12 litres of water has a mass of 12kg.\n\nF = (12 * 9.81) = 117.72 N\n______________\n\nANSWER B)\n\nthe total weight of the full bucket is 127.72 N (you know why)
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This question practically answers itself. Divide the total weight (person & bike) by 2; divide the result of that by the pressure to get contact area in m^2.
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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.
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The force pulling the block down is m*g*sin(ø);  the force of friction is k*m*g*cos(ø),  The net force is the m*g*sin(ø) - k*m*g*cos(ø).  The resulting acceleration is this force divided by m, so\n\na = g*sin(ø) - k*g*cos(ø)\n\na = g[sin(ø) - k*cos(ø)]
512007
Torque.\n\nThe weight of each particle is mg. The torque caused by particle 1 is mg * 20, counterclockwise, and the torque from particle 2 is mg * 80, clockwise. Therefore the net torque is mg * 60, clockwise.\n\nSo the force on the left particle is (mg * 60) / 20 = 3 mg, and the force on the right particle is (mg * 60) / 80 = 3/4 mg. Both of these are clockwise, so the left particle goes up and the right particle goes down. To get acceleration, just cancel m.
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Homework?\nOK, only a hint. Consider a cylinder made of the circle and with a height dh. What is the volume of this cylinder? \n\nNow displace the cylinder along the path over dh, etc.
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a) F = ma\n    F = 0.0412 kg * 9.8m/s^2\n    F = 0.404 N\n\nb) F = 0.0370 kg * 9.8 m/s^2\n    F = 0.363 N\n\nc) The difference in "weight" is equivalent the weight of the displaced volume of water...  So...\n   0.404N - 0.363N = V * 1000 kg/m^3 * 9.8 m/s^2\n   0.041N = V * 9800N/m^3\n   V = 4.18 * 10^-6 m^3   or  4.18 cm^3\n\nd) Oh Juliet you are my density.... (bad joke!)\n     Density = m/V\n     D = 0.0412 kg / 4.18  * 10^-6 m^3\n     D = 9848 kg/m^3
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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
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Simply d1/ d2 = weight of metal / loss of weight in water = 10/ 2 = 5.\n\nd1 = 5x10^3 Kg/ m^3\n  \n\nDetails\n\nThe reduction in downward force is 2N.\n\nThis is due to the upward force given by equal volume of water. \n\nIf V is the volume and its mass is V.  10^3 ( density of water is 10^3 kg/m^3).\n\nThe weight of this volume of water is V.  10^3 g This is equal to 2N. \n\nThe weight of the metal is V. d . g ( d is the density of the metal) = 10N.\n\nTherefore,  V.  d . g/ V.  10^3 g ==10/ 2 = 5\n\nd = 5 x10^3 kg/m^3.
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Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
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You just need to convert the forces on each side.\n\nFor the 30kg mass  \nF = MA\nF = 30kg(G)     (G = Accelleration of gravity ~9.8 M/s2)\n\nFor the 20kg object (equation from the wiki link below)\nMG Sin(40) =  F  \n20(G) Sin(40) = F\n20(G) (.745)   =  F \n14.9G= F\n\nSo basically, you have 30kg(G) of force one side,  and 14.9kg(G) of force on the other side\nTotal net force = 15.1kg(G) force\nConverting for G  =  15.1(9.8)\nF = 147.98 KG/m/S2\n\n\nPart the question is a bit confusing because you'll have 2 accellerations.  One down (since the 30kg will go straight down), and one up at a 40 degrees.   I would read the problem as looking at things from the 20kg mass point of view and say that the accelleration is up at an angle of 40 degrees.\n\nAccelleration  is the net force against the total weight.\nF =  MA\n147.98 = 50kg(a)\n2.96 m/s2 = A\n\nTension = both forces combined\nT = f1 + f2\nT = 30kg(G) + 14.9G \nT = 44.9kg(G)
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Weigh 3 against 3.\n1. If they don't balance, weigh 1 against 1 from the heavy side. If they balance, it's the coin from that side that you didn't weigh.\n2. If they do balance, it's one of the other two - weigh them against each other.
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in reality, it is the surface tension (adherance of molecules to one another) and density of the fluid\nbecause the question does not state that the object is suspened in the fluid\n\notherwise the force acting to keep an object buoyant would need to equate and thus counteract the objects weight (which is the force of gravity pulling said object down)\n\nmass X volume = density which is a factor in determining a force acting upon the object; a force working to cause the object to disperse within the less dense fluid or conversely the fluid to compress and colapse the object\n\nSo the author of the question given is expecting the answer: weight
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(Since we can't draw here or attach a file, you will have to imagine.)\n\nThere are 3 forces acting on the brick:\na)  Weight of the brick which points vertically down;\nb)  A reaction force from the incline plane acting upwards perpendicular to the plane; and\nc)  Frictional force acting along the plane pointing up the slope.\n\n(a)  Weight of the brick = 10 * 9.81\n[9.81 m/s^2 is the acceleration of free fall.]\n\n(b)  Reaction force = 10 * 9.81 * cos60\n\n(c)  Frictional force = 10 * 9.81 * sin60
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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
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The net force is the difference between the two weights.\n\nHowever, there is not enough information in the question to answer without making some assumptions.\n\nYou are not given the mass of the man and/or the bricks. One can calculate the mass if given the force of gravity, however you aren't given that either.\n\nOne method of solving is:\n\nASSUMING that this system is on the surface of the Earth, you can find the two masses, using F=mg with g=9.8N/kg. Once you've got that you can find the acceleration using F=ma since you know the net force and the mass of the system.\n\nMake sure you indicate which direction the acceleration is in (in this case, which object is falling, and which is being raised.
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Imagine the stretching of the surface of the drum as the stretching of a spring. use hooke's law to do calculation. u will need to know about the strength of material and etc. very very very complicated