The weight needed to balance a lever varies inversely with the distance from the fulcru,. A 50-IB weight is placed 4ft. from a fulcrum. What weight should you put 5 ft. from the fulcrum to balance the lever?
122456
It varies inversely with the distance from the fulcrum meaning if the distance is higher the weight would be lower.  So to balance the 50lb weight placed 4ft on one side would need 40lb weight placed at 5ft on other side to balance it.
172341
force = constant * distance\nF=kx\nforce = mass* acceleration\nF=ma\n\nx=F/k\nx=m*a/k = 4.0*9.8/800= about .05m, or 5cm
649367
I don't know the name since I saw it in a western only now I can't remember which one it was. Anyways, what you do is get a bar about a foot in length and tie a piece of rope in the middle of it long enough that it almost hits the floor when held at mid waist. At the end of the rope put a weight or something that weighs about 3-4lbs on it. Then by rolling your wrists and turning the bar, roll the rope on the bar and then let it down again by doing the same motion only backward.
449637
the correct answer is not listed ( 32 would be correct if box 2 did not move ) \n\nsince the friction of the surface is not given the correct answer is unknown - it would be the amount of force required to overcome the friction OF THE SECOND BOX ONLY ( you said BETWEEN the boxes and when in motion force is proportional to resistance )\n\nor you did not phrase the ? properly ( likely )
351161
Take the value of the height* and multiply the height to itself.\n \n(height times height) = height squared\n\n*Make sure that the heigh has been converted  all to inches,\n OR all to feet, first.\n\nREMINDER: 1 foot = 12 inches\n\nex.  5 ft. 2 in.  = ? inches  \n\n5 ft. x (12 in. / 1 ft.) + 2 in.  \n60 in. + 2 in. \n62 in.\n\n5 ft. 2 in.  =  62 inches\n\nAND,\n\n(inches times inches) = inches squared \n\n(62 in. X  62 in.) = 3,844 inches squared\n\nTherefore,  BMI=weight divided by ( inches times inches)\n\n\nGood Luck!
461081
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
134206
the rule is 100lbs at 5' then add 5 lbs for every inch after...so ideally (though most of us arent ideal) you should weigh 155.
51921
Torque is force*distance.\n\nFind the weight of the person (M*g) = 89kg *9.81m/s^2 \nM*g= 873.09 Newtons\n\nthere is a 1:3 lever arm there--> 4cm/12cm\nThe muscle is having to exert a force of 3*873.09N or 2619N\n\nThe torque around the pivot is 2619N @ 4cm (or .04m)\n2619N * .04m = 104.77N*m\n\nYour question says you want to "find the torque" yet you ask for the "forces pushing up".  Make sure you know which one you need.
462167
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.
486169
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.)
386032
I agree with the first answerer...where is the figure? But let's see if we can solve it just the same. We have to guess a little regarding the location of the supports.\n\nGiven: length of board=5.0m\n           mass of board=54kg\n    Hence,  weight of board =54g N.\n    Place this at the middle of the board because it is a uniform board as given in the problem.\n\nFirst support is 3.4m from the end, and the second is 4.6m form the end. \n\nDiver stands at end of the board. He weighs 65g N.\n\nFind: F1 at first support and F2 at second support.\n\nTake torque about F1:\n\n54g*(2.5-0.4)+65g*4.6=\nF2*(1.6-.4)\n\n54g*2.1+65g*4.6=1.2F2      Call this equation 1.\n113.4g+299g=1.2F2\nF2=(1111.3+2930.2)/1.2\n    =3368N upward.\n\nTake torque about F2:\n\n54g*(3.4-2.5)+65g*3.4=\nF1*(1.6-.4)\n\n54g*0.9+65g*3.4=1.2F1       Call this equation 2.\n\n48.6g+221g=1.2F1\nF1=2642/1.2\n    =2201.7N downward\n\nCheck your answer:\n\nF1+F2=65*9.8+54*9.7\n-2201.7+3368=637+529.2\n\n3368=637+529.2+2201.7\n       =3367.9\n0.1 difference is acceptable. so our solution is correct. (Assuming my assumptions as to locations of supports are same as in your figure.)
357718
centre of gravity of both pans along with arms of the balance lies just below the fulcrum,when it is tilted ,the centre of gravity tries to balance the weight hence stops with a little tilt.
476991
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.
483631
It is often a problem in gyms with machines using pulleys.  If the pulley is a single one, then whatever weight is on the stack is what you're lifting, but many machines have some sort of mechanical advantage, so there might be 100kg on the stack, but you aren't actually exerting 100kg force (kgf).  On Earth a kilogram mass exerts a kilogram force downward, so we can ignore the "force" term.\n\nI keep a little retractable tape in my gym bag so every time I come across a new machine, I do the following (it's easier if you have someone to help you):\n\nMeasure the distance the weights actually rise during the full range of the exercise, and multiply that by the total weight.  That gives you total work in kg-metres.  So you might have 50kg on stack, and your shoulder press raises it 0.45m. So that's 22.5kg-m of work.\n\nNow measure the distance your hands move through the exercise (usually significantly further than than the weights). This might be 0.65m.  Divide your work (22.5kg-m) by this distance (0.65m) to get the force at your hands, in this case 34.6kg. So if you did a free weight shoulder press, around 35kg would give you the same effort as 50kg on the machine.\n\nA special case is the 45 degree sliding leg press machine.  Your legs may push the slide about 0.5m, but the weights have only risen vertically by 0.5 x sine 45 deg, or 71% of the slide distance i..e. 0.35m.  So pushing the 200kg on the rack is only 141 kg force at your feet, and your total work is 71kg-m.  That doesn't count the platform weight of course.\n\nI always record work as well as weight, as a heavy weight is not necessarily heavy work.  As an extreme example, anyone can lift a wheel of a 1-ton truck off the ground with one hand - using a jack!
539692
5N * 3 = 15N\n\nYour arm is a lever in weight training. The input force is at your shoulder, the output force is on your hand, and the fulcrum is your elbow.
68855
If it is a fixed pulley, you need the same force as the weight of the load which is 9.8*0.2 Newtons (9.8 is the acc. due to gravity and 0.2kg is the mass of the load) or simply 200gf (where gf is gram force). Now you can tell one half of this number.\n\nAlternately, if you have a movable pulley, the force required is half the weight of the load, i.e., 100gf.  You can tell one half of this number too.\n\nIf you need to find the difference between fixed and movable pulleys, look up a physics text (or science text), I am not totally familiar with the schooling system here.   Otherwise, Google should show you pictures of fixed and movable pulleys.
444393
This one isn't too bad.  The difference between the bags on the scales is 25 - 4 = 21.  The difference between the bags to be added is 9 - 2 = 7.  \n\n21 / 7 = 3, so, by adding three sets of the 9 lb/2 lb bags, the scales will balance.\n\nTo check this answer: 25 + 3 * 2 = 25 + 6 = 31 and 4 + 3 * 9 = 4 + 27 = 31 --- same result