There is one solution to this problem, and can be proven using compression measurements in the legs of any stool with an applied force down in the middle of the stool.\ni have received no correct answer yet(and I have posted question a few times). \n\nWRONG ANSWERS ALREADY RECEIVED:\n25 N in each leg\n0 N in each leg\n15 N in each leg\n\nClues:\nCan anything ever be measured exactly?\nHow many points make a plane?\nThe sum of all moments is zero (about any chosen point on stool - statics).\nAssume the stool's weight is negligible.\nAssume net force on stool is Zero (i.e. stool is not moving)\n\ncommon sense and a little math solves this one \n\n\n\nApr 24, 2006 at 7:47 pm\nMORE WRONG ANSWERS:\n100 N in each leg\n\nApr 24, 2006 at 7:48 pm\nMORE WRONG ANSWERS:\n25N + 1/4 the weight of stool\n\nApr 24, 2006 at 7:50 pm\nAssume point of pressure of applied force is an equal distance from each of legs (in center of top of stool)\n\nApr 24, 2006 at 7:51 pm\narbritage is on the right track but
207545
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.
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The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
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1) The angle of the rope needs to be considered to calculate horizontal and vertical components of the clerk's force. \nFh = 188*cos(25) = 170 N This is what pulls the box.\nFv = 188*sin(25) = 79.5 N This actually lifts the box slightly.\n\nThe symbol for coefficient of kinetic friction is µk. A lower number makes it easier to move the box - if you spilled some olive oil on the floor µk would perhaps go to 0.1. N is normal force. If this was on a slope, normal would be different from weight, but the floor here is level so N = W? \n\nActually no. Because the rope pulls up on the box slightly, the normal force pressing the box to the floor is less than the box's full weight. The box's weight = m*g = 36.9 kg*9.8 m/s^2 = 362 N. The vertical component of the tension in the rope relieves 79.5 N of that so N = 362 - 79.5 N = 282 N. \n\nSo Ff = .450*282 N = 127 N is the force that friction fights with. The horizontal portion of the clerk's pull is 170 N. So there is 43 N left over to accelerate the box. Newton's 2nd law says\nF = m*a\n43 N = 36.9 kg*a\nso a = 1.17 m/s^2\n\n2) Here the force is directed down at an angle so the vertical component of the force will add to the normal force used in calculating the friction.\nFh = 435 N*cos(35) = 356 N\nFv = 435 N*sin(35) = 250 N\n\nN = the weight of the box + Fv = 259 N + 250 N = 509 N\n\na) Ff = µk*N = .57* 509 N = 290 N This is not enough to hold the box stationary so 356 - 290 N = 66 N are available to accelerate the box. Newton's 2nd law again:\nBut first we need the box's mass. W = m*g\nm = W/g = 259/9.8 N = 26.4 kg. Now\nF = m*a\n66 N = 26.4 kg*a\na = 66/26.4 m/s^2 = 2.5 m/s^2\n\nThis is getting complicated. To move 9 m will take t seconds:\nx = (1/2)*a*t^2 Solve for t.\n\nb) Ff = µk*N = .75* 509 N = 381 N This is more than the horizontal portion of the clerk's push, so the box will resist. The answer is 0.
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Exclusively breastfed babies in the first few days of life should have one stool for every 24 hours that they are old (I have a diary for this is you are interested in keeping track.) In order for a stool to "count" it must be bigger than a fifty cent piece. \n\n The stool transitions from black meconium to green, tarry meconium to brown to yellow.  And if all is going well, then the baby should be passing yellow stool on day 4-5 of life.  The first milk that the breasts make, the colostrum, is a great laxative.  The more colostrum the baby gets,  the more poop they pass. That means that any breastfed baby who does not stool (in the first week of life) is not getting enough milk.
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Answering...\n\nExperiments have been done in places where air resistance is negligible...and it was found out that objects with different masses have the same acceleration as long as the test are done within a reasonable height...Therefore, the answer is C.\n\nThis can be shown in equation form, formulated by I. Newton:\n\nF = m x a ... where:\n\nF = force in newton\nm = mass in kg.\na = acceleration in m/sec^2\n\nSo:\n\na = F/m = 2 N/2 kg = 1 m/sec. squared
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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.
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find the force of her weight. \nthat should be the answer
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Hope this helps...\nBreak this question down into four independent events.  1) Bob and Joe falling, 2) Keg with nails falling, 3) Joe falling, 4) Keg falling.\n\nIf you look at one specific event, then you can assume that both sides of the pulley start at rest (since either it starts that way, or some poor guy smacks the ground or the pulley).  Therefore: look for Fnet on the total masses on the rope.  Take for instance, the first episode, where Bob and Joe (Total mass 180kg) is against the keg and nails (120kg).  Since F=ma, then the individual forces on Bob and Joe = 180(10) and the keg and nails 120(10).  So, Fnet would be 180(10)-120(10) for this specific case.  Then, you would look at the total Force, the total mass, and the acceleration of the entire system.  Once you have the acceleration, then you can easily figure out time, since you have a 16m pulley height (you have to assume Bob and Joe are starting at pulley height, otherwise the problem is unsolvable).  Just use the -1/2at^2+Vot+do=D equation, and since there is no Vo, this should be easy.\n\nDo this for each subsequent episode with the varying masses, and that should give you your total time.  Oh, and the rope has to be massless and the pulley frictionless, too.\n\nAnd unless I'm totally missing something, just plug and chug.  Good luck!
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Well we need two equatons here:\nF = ma : this will give us the mass\nF = uN : this is just needed\n\nSet these two equal to each other.\nu = .30\nN = 80N\nm = whats needed\na = acceleration due to gravity on earth 9.8m/s^2\n\nThus setting those two equations equal to each other results as follows:\nForce = Force\nuN = ma\n(.3)*(80N) = m * 9.8m/s^2\nm = (.3*80)/9.8\nm = 2.44kg\n\nI have no clue where that above guy created his equation from, but it's sooo wrong.  I mean who weighs items in Newtons.  Newtons is a measure of Force, not weight!
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WORK IS EQUAL TO ZERO (READ ALL FOR EXPLAINATION)\nmass of body=30kg\nits weight=30kg X 9.8m/s^2\nweight= 294N\n\nBUT HERE WE DON'T HAVE TO DEAL WITH ITS WEIGHT.\n\nAs the body is moving at constant speed, it means that it's not accelerating and thus no force is being applied to it.\n\nOr in other way\n\nthe frictional force which is 20N is canceling the force with which the boy is moving the body. Thus the boy is also exerting a 20N force (20N - x = 0, of course x=20N)\n\nTHE WORK IS ZERO BECAUSE THE FORCE IS ZERO TOO.\n\nsince the frictional force is cancelling the force of the boy.\n\n20N(frictional force)-20N(force applied)=0\nResultant force = 0\n\nFIND WORK NOW\n\nwork =  net force X distance\nwork =  0  X  5m\nwork =  0 joule\nNo work is done\n\nLISTEN\nThe person below me by the name of sblogs is incorrect as I think. Let me prove this by giving some EXAMPLES\nPUSH THE WALL BUT THE WALL WILL NOT MOVE FROM ITS PLACE THOUGH YOU ARE WASTING ENERGY BUT THE WORK DONE IS ZERO.\nBECAUSE THE WALL DOESN'T MOVE THROUGH SOME DISTANCE.\nWORK= FORCE X DISTANCE\nWORK= (FORCE APPLIED ON THE WALL) X (ZERO METERS)\nWORK= F X 0\nWORK= 0 JOULE
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I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
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haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
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I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.
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Well, whilst she is sliding, ignoring air resistance, the only force acting on her is the force of friction. That's the most important part of this problem. As such, you know that Fnet = Ffriction. Ffriction = Fnormal (Uk), so you can solve for Ffriction. That's Fnet, and Fnet = ma. You know Fnet and her weight, so divide her weight by 9.8 m/s to find her mass. Solve that equation for a, and then plug it into one of the handy-dandy kinematic equations. :)
250419
I assume this is an analysis of Archimedes' famous quote "Give me a place to stand, and I will move the Earth."  Unfortunately, while this is mathematically sound, it is practically impossible!\n\nLet me explain.  Say that the pole has a length L.  We'll place the Earth at one end, a person at the other end, and a fulcrum at a distance D from the Earth.  If the Earth exerts a force F_E and the person exerts a force F_0, then we have the equation\n\nF_E * D = F_0 * (L-D)\n\nThis was proved by Archimedes, of course.  We know that an object of mass M exerts a gravitational force of F = M g, where g = 9.81 m/s^2.  The Earth has a mass M_E = 5.9742 x 10^24 kg, and an average person has a mass M_0 = 70 kg.  This gives the the equation\n\n5.9742 x 10^24 * D = 70 * (L-D)\n\nso that L/D = 8.53457 x 10^22.  The length of the pole L depends on the location D where the fulcrum is placed.\n\nUnfortunately, this ratio is impractically large: If we place the fulcrum at a distance of D = 2 micrometers (which is slightly smaller than a strand of spider web), then the pole must be a length of L = 16.3 light-years.  This is farther in distance than the fartest known star!
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dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
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If as described the entire meter stick is in balance, then the center of its mass/gravity would be where the support is, at the 22.4cm mark\n\naha ... here we go then,\n\n142.3 * (22.4-9)  = 134 * (x - 22.4)\n\nsolve this for x to get the answer\n( think of the entire stick as a ball placed at position x)