balanced force example?
968407
Everywhere you look you can see this.  An object on a table, a curtain hanging from a window sill, a sealed and shaken soda bottle, etc.
487370
* spring balance\n    * pivot balance\n    * load cell
826484
Bernoulli Effects\n\nSee how to demonstrate it here --> http://www.physics.umn.edu/outreach/pforce/Bernoulli.html
4764
You can't balance it because you don't know what it yields.\n\n\nP.S. This needs to be in the chemistry category. You really must not know much about chemistry and physics.
879813
you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
641452
opposite force being applied
783617
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".
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
832550
Bernoulli's principle can be tried.\nLaws of forces can be demonstrated.\nThermodynamics-transfer of heat.....etc
459581
Your question is unclear about what set of assumptions you want to use. You seem to imply a perfectly constructed stool, with a perfectly centered force, using incompressible material. You want the stool to have no mass, and you imply that you want to use statics to solve this problem. Strictly applying statics, this system is indeterminate, NOT the ridiculous answer of 0 [N] in two legs and W/2 [N] in the others. There are only 3 equations of equilibrium here, 1 for vertical forces and 2 for moments. Since you have four legs, this system cannot be solved by statics alone. Treating the stool as a deformable structure, you can use geometric compatibility to get a fourth equation. I challenge you to find a competently built stool that has 0 [N] in any leg (a physical stool built of real materials), or an incompetently built stool with 0 [N] in any two legs (unstable system that is obviously not static). The answer indicated as "correct" requires an incorrect application of statics. A large part of studying physics is learning when to use a particular tool, and the physical limits of that tool.\n\nedit: \nI am perfectly aware of what a zero force member is, and there are no zero force members here. You can only find zero force by statics if the structure is determinate, which is not true in this case.\n\nI completely agree with the PowerPoint presentation (definitely not a proof BTW). "But if one leg does not touch the floor, the force in it is zero. If one leg is zero, then the opposite leg is also zero by moment equilibrium. The two remaining legs carry all of the load = 90 pounds/leg. All four legs must be designed to carry the 90 pounds (since any two legs could be loaded)." only applies if the premise is satisfied, namely "if one leg does not touch the floor". I assure you, the normal idea of a four-legged stool requires all four legs to touch. Notice they finish off the calculation using a 45 lb/leg design load, not 90 lb/leg.\n\nI did not call you ridiculous; I rarely resort to name-calling. Please use the built-in spell checker.
432088
*copied from previous post\n\nYour question is unclear about what set of assumptions you want to use. You seem to imply a perfectly constructed stool, with a perfectly centered force, using incompressible material. You want the stool to have no mass, and you imply that you want to use statics to solve this problem. Strictly applying statics, this system is indeterminate, NOT the ridiculous answer of 0 [N] in two legs and W/2 [N] in the others. There are only 3 equations of equilibrium here, 1 for vertical forces and 2 for moments. Since you have four legs, this system cannot be solved by statics alone. Treating the stool as a deformable structure, you can use geometric compatibility to get a fourth equation. I challenge you to find a competently built stool that has 0 [N] in any leg (a physical stool built of real materials), or an incompetently built stool with 0 [N] in any two legs (unstable system that is obviously not static). The answer indicated as "correct" requires an incorrect application of statics. A large part of studying physics is learning when to use a particular tool, and the physical limits of that tool.
975783
The force balance says that\n\nSFx = 0 = -2*g*sin(q) + m*v^2/r\nSFy = 0 = 2*m*g*cos(q) - m*g\n\nwhere m is the mass, v is the speed, g is the gravitational acceleration, q is the angle that the ropes makes with vertical, and r = L*sin(q) is the radius of the orbit. SFx and SFy represent the sum of the forces in the x- and y-directions, respectively.\n\nSFy indicates that\n\n2*m*g*cos(q) = m*g\n\nor\n\ncos(q) = 0.5\n\nHaving the angle, you can solve SFx for the velocity\n\n2*m*g*sin(q) = m*v^2/r\n\nor\n\nv^2 = 2*g*L*sin(q)^2\n\nGiven the velocity, the angular velocity is\n\nw = v/r.
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.
784877
How are they applied? Tangentially? In opposite direction? Or same?\nLet us assume oppo direc and on the border.\nThen net force=35N.\nTorque = r x F = I w\n=35x0.341= ''0.5'' m R^2\nR radius m mass\n ''Check if it is 0.5 only for a solid disk or cyliner about its axis.\n\nHope it helps
878201
Generally, the force is applied at the point of "input force" - that point is determined by what class of lever you're using.\n\nFor example, if you're using scissors (First class lever), the fulcrum is at the pivot between the two handles and blades, the input force is at the handle, and the output force is on the blade.\n\nIf you're using, say, a wheelbarrow (2nd class lever), the fulcrum is at the wheel, the input is on the handle (end), and the output force is in the middle - in the bucket.\n\nFor a baseball bat (third class levers), the input force is in the middle (your hands), fulcrum at one end (you), and the output force is at the other end (the end of the bat).
791491
The maximum possible net force is 125N, if the two forces are pushing in the exact same direction.\n\nForce is a vector quantity, which means that any measurement of force includes both a magnitude (size, the number) and a direction (the way that the force is pushing).  The simplest way to indicate direction is with a positive or negative sign, which only gives us two options: left/right, forward/backward, up/down, etc.\n\nIn your question, no direction is specified for either of the forces, but we can figure out the minimum and maximum possible net force rather easily.  The minimum net force would result when the two forces are pushing directly against one another:  100 -25 = 75N (toward the direction that the 100N was pushing).  Similarly, the maximum net force would result when the two forces are pushing together in the same direction:  100 +25 = 125N.\n\nIn general, then, the maximum possible net force (in cases where you are not given directions with the forces) can be found by adding the magnitudes of the forces together, as if they were all pushing in the same direction.