If you apply a force of 100 N down into the middle of a circular stool with 4 legs,what is force in each leg?
473436
100 N divided by four is 25 newtons.\nHowever if one of the legs is shorter and there are only 3 legs, the stool will probably fall over.\nNothing can be ever measured accurately.. Check out Schroedingers Cat.
881560
If you lift a 100 pound weight with only 50 pounds of force, the weight will remain on the floor because of inertial (it is pulling down more by gravity than you are pulling up).  Although you may be greatly exerting yourself, you have added no energy to the weight.  If you now lift with an upward force of 101 pounds, you overcome the downward force of the weight and it rises.  Importantly you are now adding potential energy to the weight.  Lift the weight one foot and you have added 100 foot-pounds of potential energy by doing work (force times a distance) on the weight.  We know that a pendulum can convert potential energy (height) to kinetic energy (motion) continuously until stopped by friction (that converts the energy to heat and sound, etc.).  Therefore, a body will remain at rest until you add energy to it and it will then remain in motion until the energy is removed again.  Energy is added when a force acts on the weight for a distance and energy is removed when the weight acts as a force moving a resistance a distance.  This is simple in a frictionless (ideal) system but in fact friction often applies a force to the weight removing its energy in the form of heat.  Hope that helps a little.
891370
Get the unbalanced force from the first 2 #s in your ? :  2.8-2.6 = .2\n\nb)  F = ma---> m = F/a = 0.2/.11 = 1.818 kg
972973
well if it weighs 2 N and its being pulled down by 5 N. then theres a total downward force of 7 N, which means the normal force must also be 7N (upward). i think thats what you're asking..
875596
you count the legs and divide by 4
981506
a. The force of the gravitational attraction between two objects is given by the formula\n\nF= G*(M1*M2/R^2)\n\nwhere G is the gravitational constant, M1,M2 are the two masses and R is the distance between their center of masses.\n\nSince the distance between your center of mass and the Earth's center of mass is such a large distance, you can approximate this distance as being the mean, average, radius of the Earth.\n\nG can be easily found in any physics book and you are given everything else. Just plug everything in to the equation and that will give you the gravitational force between you and the Earth.\n\nb. How much you weigh is directly related to the force you just calculated. In fact, they are the same thing. Your weight is a measure of how hard the Earth is pulling you towards its center.\n\nMy old physics professor used to joke that if you want to loose weight just walk up some stairs. Since you can't loose any mass, if you are able to decrease the distance between you and the center of the Earth you will lessen the pull of gravity and thus loose weight.
757859
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
761976
1\n                                       1\n                                       v 83.1kg\n\n                                       T         Elevator floor\n                                       ^\n                                       1 Upward force, P=reading\n\nI hope the diagram doesn't get distorted when sent to you.\n\nSince the elevator is slowing down then the net force must be opposite to direction of the elevator, i.e. downward.  Therefore,\n\nF=83.1g-P\n\nBut F=ma\n\nSubstitute known values:\n\n83.1g-P=83.1*1.59\n814.4-P=132.1\nP=814.4-132.1\nP=682N upward. This is the reading on the scale or \n69.6kg
25616
the force needed is according to newton's second law F=m.a where m=900Kg \n& the acceleration a=v/t\nthe velocity should decrease from 10 to 0 then v=10m/s & t=5 seconds\na=v/t=2m/s^2\nF=900*2=1800newtons
621321
If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
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)
706663
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
725904
this is the amount of force.  Force equals mass times acceleration.  The mass is ur bodyweight in kilos and the acceleration is gravity 9.8m/s/s
757274
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
612424
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.
840333
You need to consider conservation of angular momentum\n\nI0*w0 = If*wf\n\nI0 is the initial moment of inertia, If is the final moment of inertia, w0 is the initial angular speed, and wf is the final angular speed.\n\nI = Iss + Io\n\nwhere Iss is the (constant) moment of inertia of the student and the stool, and Io is the moment of inertia of the objects. This general relation applies to both the initial and final moments of inertia. Io can be computed from\n\nIo = 2*m*r^2\n\nwhere m is the mass of the object and r is the distance from the axis of rotation. The factor of 2 accounts for the two different objects. The first equation then becomes\n\n(Iss + 2*m*r0^2)*w0 = (Iss + 2*m*rf^2)*wf\n\nwhere r0 is the initial radius (1.0 m) and rf is the final radius (0.40 m). Solving for the final angular speed you have\n\nwf = (Iss + 2*m*r0^2)*w0/(Iss + 2*m*rf^2)\n\nUsing this expression it is easy to compute the final angular speed.
1093836
Centrifugal force and gravity.