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
128262
Hey, you seem to know the answer. Why don't you keep it to yourself, or better still, give us YOUR answer: the RIGHT answer?
456083
The net force on the man is the sum of the force of gravity and the force on the ground. You've already calculated the net force, so simply subtract the man's weight from it. Remember that gravity is pulling him down, so you'll be subtracting a negative number (i.e. the final result should be greater, not less, than the net force).
536836
They specified a chain here for a reason.  Chains cannot sustain a compressive force.  The force on the table is produced only by the latest link to strike the table.  The part of the chain that has not yet hit the table is in freefall.\n\nSince they are asking for force over time, this really sounds like a momentum transfer problem.  Momentum has units like newton-seconds (force x time).  The force is then the momentum per second tansferred to the table.  Of course, the momentum is also mass x velocity.\n\nWhen the distance fallen is x, the link striking the table has fallen a distance of x.  It has a velocity based on the usual freefal relation:\n\n           v^2 = 2gx \n\nCall the total chain mass M, the length L and there are n links.  The mass and length of a link are M/n and L/n.\n\nThe momentum transferred by a link is P =  (M/n) * v \n\nThe time it takes from one link to hit until the next one is the link length divided by velocity:\n\n           t = (L/n)/v\n\nThe momentum is the product of force and time:\n\n P = F*t   so F= P/t\n\nSubstituting from above: \n\nF = ((M/n) * v) / (L/n)/v) = (M*v^2)/L\n\nSubstuting in the realtion between fall distance and velocity:\n\nF = 2Mgx/L\n\nAfter reading subsequent  answers, I notice that I completely forgot about the static potion of the problem, that is the dead weight of the chain on the table.  Since x length is on the table, that amounts to Mgx/L, which is exactly one half the static load.  Adding the two gives:\n\nF = 3Mgx/L\n\nYou can also work this via energy methods where you assume an elastic but stiff table.  The result is the same.
430532
I'm not going to give you the complete answer because you'll learn more by working through it yourself, but I'll give a few pointers.\n\nAny object travelling in a circle at a constant speed has a centripital acceleration that can be calculated from the velocity and the radius of the circle.\n\nNewton tells us that F[net] = ma, so the object must be experiencing a net centripetal force equal to its mass multiplied by its centripetal acceleration.\n\nWhere does this force come from?  In your problem, it sounds as though the only force that could be pulling the box toward the center of the circle (the centripetal force) is the friction between the box and the bed of the truck.  So, when you add up all the (horizontal) forces to get the net force, the olny one is going to be friction.  F[net] = F[friction] \n\nNow all you need to do set F[friction]  equal to the F[net] that you calculated above (remember that F[net] was equal to your centripetal force), and solve for the coefficient of friction (your textbook shold have the formula for static friction, which will include the coefficient of static friction).\n\nGood luck!
292500
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.
498767
at the moment there are the horizontal component of weight and the friction that opposes it are in equilibrium. we must apply a force that with the already present weight compontns will equal the kinetic friction. \nmeaning that \n\nWx + Fp = Friction.\n\n500cos20 + Fp = 0.6*500sin20 =>\n\nfp =367.24N
469263
Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
460956
The previous answerer has the right idea but the wrong details.\n\nThe four parts should weigh 1, 3, 9 and 27 pounds.  The logic is that the desired weight is written in "balanced ternary" notation, where the values of the digit positions are 1, 3, 9, 27, 81, . . . (that's "ternary"), but the allowed digit values instead of being 0, 1, 2 (that's regular ternary) are -1, 0, 1 (that's balanced ternary).  Then the digit values for any desired total weight written down in balanced ternary tell you whether the weight for that column is to go in one pan, go in the other pan, or stay on the floor.
523279
I expect that you have the data on stretch as a result of tension. \na) The object is just at rest. Tension, T = W = m*g.\n\nb) T = W + m*a\n\nc) Tmax = W + m*a
468571
Pins and needles are often mistaken as a circulatory problem but is generally not. When you hit your "funny bone" that is a similar feeling and that is a nerve you have hit. Sitting in one position like on a toilet bowl is a common time to get leg numbness. This generally occurs because of pressure on the sciatic nerve in the buttock. There is are less likely alternatives and some people do have conditions that predispose them to getting compression neuropathy. \n\nYour legs won't fall off from chronic compression neuropathy but you do run the risk of it become persistent, ie, one day the pins and needles may not go away. Get a cushion and don't cross your legs.
581836
A bowl should never move.  Your bowEl should though... ever other day at least... maybe you are anal retentive.
469860
a) As the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso w²= g/(L cosa)  \n\nso v = r sqrt(g/(Lcos a)) = 1.4 sin 30  sqrt( 9.8/1.4 cos30) = 1.99 m/s\n\nb) 4 = 1.4 sin a sqrt (9.8 /1.4 cos a)\n\nc) As mw²R = S sin a\n    and mg = S cos a\n\nw = sqrt (S²- m²g²) /mR
153038
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
63813
It's probably a wash. The weight of the body is distributed over twice the number of legs, so each leg has to do less work. More legs, though, mean that there's more energy required to move them all. The more legs thing could be cut down depending on the gait.
553884
For 4 a&b)\nThere would be three arrows.  The first would be a force in the upward direction (i.e. normal force) second would be force due to gravity(downward direction, with equal magnitude o that of the normal force).  The third arrow is an arrow that is in the same direction as the force applied to the large box.  Its magnitude is proportional to the normal force.  It s static friction that is applying that force.   \n\nAs far as part c goes It's been a while for me.  You've got the textbook, look up some formulas.  Besides, I think you'd need to include some masses in order for me to solve those.\n\nThe force diagram for the Large box is very similar to that of the small box, except that it has forces applied to it from the small box. (this is because the large box is supplying the normal force for the small box.)
643862
Did you take a stool softener or a laxative before? Are you on a special diet? Are you taking fat binders? I know models tend to have diets/take meds which other people don't, because they are under pressure to stay ultrathin and it is not always easy.\nDon't worry, it is something that can happen. The combination of the position (bending over) and laughing (generating pressure in your belly) can easily lead to what you described. Nothing is wrong with you. Hope you still kept the contract for that swimsuit store...
13334
Ok....the only components of the force that has any hope of moving the block (horizontally I'm assuming) are the....well....horzontal components of those two forces acting together.\n\nFriction is a force. It's given by...say...'f', where f = uN\n\nf also is mass times acceleration:  f = ma\n\nThe horizontal components of each force is found by resolving each one into 'x' and 'y' components. I'm going to let you figure out how you do this, but the results are:\n\nx1 = 6 N\nx2 = (4/37) N\n\nTheir sum is about 6.1 N\n\nSo you have:   uN = f\n\nu = coeff. of static friction (what you need to determine)\nn = "Normal" force = mg [Normally this is called "N", but I want to avoid confusion between that and the units for force...Newtons which I'm calling "N". Understand?\n\nm = object's mass = 5 kg\ng = gravitational const. = 9.8 m/s^2\n\n==> un = f\n==> un = 6.1 N\n==> u = (6.1 N)/n\n==> u = (6.1 N)/(mg)\n==> u = (6.1 N) = (5 kg*9.8 m/s^2)\n==> u = 0.125\n(done)\n\nHope that helps, but if not, lemme know.