Force exerted by a table on a falling chain?
961750
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.
799062
All you have to do is write an equation for the displacement of the springs as a function of x.  When the the particle is pulled to any x, the springs form 2 right triangles with the x and y axes:\n\n&#92;\n &#92;\n  @\n /\n/\n\nso lets call the length of the stretched springs h and define the displacement d = h-L:\nh^2 = x^2 + y^2\ny = L\nh^2 = x^2 + L^2\nh = sqrt(x^2 + L^2)\nd+L = sqrt(x^2 + L^2)\nd = sqrt(x^2 + L^2) - L\n\npotential in a spring is equal to 1/2 times the constant times the displacement squared.  So for one spring:\n\nPE = 0.5*k*d^2\nand for both springs, double it:\nPE = k*d^2\nPE = k*(sqrt(x^2 + L^2) - L)^2\n\nTo find the forces you have to do some more trig.  The total force of one spring equals k times the displacement.  In the y direction, you know that the net forces will always be 0 because the two springs cancel out.  In the x direction, the x components will be added.  x components are found be multiplying the magnitude times cos(theta) where theta is the angle formed. So:\n\nF_y = 0\nF_x = k*d*cos(theta) + k*d*cos(theta)\nF_x = 2*k*d*cos(theta)\nF_x = 2*k*(sqrt(x^2 + L^2) - L)*cos(theta)\n\ncos(theta) = adjacent/hypotenuse\ncos(theta) = x/h\ncos(theta) = x/(d+L)\ncos(theta) = x/(sqrt(x^2 + L^2) - L +L))\ncos(theta) = x/sqrt(x^2 + L^2)\n\nF_x = 2*k*(sqrt(x^2 + L^2) - L)*x/sqrt(x^2 + L^2)
770662
Let m be the mass of the block and M be the mass of the ball.\n\nThe force pulling the block downward mg sin 35 = 60g sin 35 = 34.41 g\n\nThe frictional force opposing this force is 0.22 x 60 g cos 35 = 10.81 g\n\nThe weight of the ball is also opposing this force = Mg\n\nSince the block is moving with constant speed the net force is zero,\n\n34.41 g - 10.81 g - Mg = 0\n \n M = 23.6 kg.
612059
Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.
909402
Well, it's Wednesday morning.  I hope this answer still has some value to you.\n\nI started by summing the torques at the hinge.  There are three:\nT(light)=5*9.8*7.5*cos(37)\nT(cable)=-tension*6.31*sin(37)\nT(pole)=8*97.5/2)*cos(37)\n\nThese all sum to zero.\n\nI computed tension=140.5\n\nThe sum of the horizontal forces must be zero.\nThese are the tension in the cable and the horizontal reaction force at the hing.  So the horizontal component is 140.5\nThe vertical forces are the weights of the pole and the light offset by the vertical reaction force at the hinge\nthe vertical component is (8+5)*9.8\n=127.4\n\nI checked the answer by computing the magnitude of the reaction force two ways.  Using Pythagorean theorem \nsqrt(140.5^2+127.4^2)\n=189\n\nI then summed the forces parallel to the pole:\n5*9.8*sin(37)\n8*9.8*sin(37)\n140.5*cos(37)\n=189\n\nj
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.
892916
I think the key here is that the objects will be rolling down the plane; thus the gravitational potential energy they have is not only converted into translational kinetic energy, but also into rotational kinetic energy.\n\nAnyway, I would approach this from energy considerations.  Start by calculating the amount of potential energy the objects will lose by falling a distance of 3 m.  (For both, this will simply be PE = m g h).  Equate this to the total amount of kinetic energy the objects have at the bottom of the ramp (translational plus rotational, or 0.5 m v^2 + 0.5 I w^2 where I is the moment of inertia, and w is the angular velocity.)\n\nSo, you need to know the moment of inertia for a disk and a hoop.  The hoop I know off the top of my head, I(hoop) = m R^2.  The disk, I looked up to find I(disk)=0.25 m R^2.\n\nFinally, recognize that as they do not slip, you can relate the angular velocity w, to the translational velocity v:  v=2 Pi R w.\nOnce you do this, you find that the kinetic energy depends only on the velocity v, so we have one equation with one unknown.  Just solve for v with the two different values of I, and I think you ought to find that the disk will be moving faster than the hoop.\n\nSorry for being long winded.... but I hope it helps.
794040
This is really a calc question.\n\nAn item will fall at 32ft per sec per sec from a rest.\nh - the free fall = t\n\n2. As it is stated the axle is frictionless, therefore the cyinder and axle will move and there is no force between the two. However there is a force between the rope and or bucket and the cylinder.\n\ndouble check the 32ft/s in your book, there may be more to the equation.\n\nCarl
652894
the amount of pressure exerted on an object
937470
Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
747946
It's a bit hard to tell without the drawing, but I think I can see how this is set up.  Start by calculating the applied force, based on what you know about the spring.  F = -kx, so the applied force is 0.053 m, the compression of the spring, multiplied by 470 N/m, the spring constant; the negative sign can be ignored in this case, because we're dealing with magnitudes, not vectors.  Now, if the block doesn't slip, the upwards friction force must equal the downwards gravitational force.  Call the spring force F_spring.  The friction force, then, would be F_friction = u*F_spring, where u is the unknown coefficient of static friction.  The force of gravity, the weight, would be W = mg, where m is the 1.9 kg mass of the block and g is gravity, 9.8 m/s^2.  So F_friction = W ==> u*F_spring = mg ==> ukx = mg, and u is your only unknown.
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
920693
For equilibrium of the board-and-boulder system, two conditions must be satisfied:\n\n!) Sum of vertical forces must be zero and \n\n2) The turning effects of the forces of Will and Cliff (about the center-of-mas of the system) must total up to zero.\n\nNow considering the vertical forces: \nWeight of the boulder is 71 x 9.8 N = Fwill + Fcliff\n\nFor the balancing of turning effect condition:\n(By the principle of moments) The distances Dcliff and Dwill from the CM of the two forces should be in inverse proportion  of the two forces .: \n\nDcliff = 3.15m and  \nDwill = (3.50-3.15)\nleading to: \nFcliff  = (71 x 9.8) 3.15/3.50 =. 626.22 N and\nFwill  =  (71 x 9.8) (3.50 - 3.15 ) /3.50 =.69.58 N\n\nAnswer:\nCliif applies 626.22N and Will applies 69.58 N..Got it?
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..
111044
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!?
965191
This isn't very hard - you just need to do each of the little pieces. I'll give some hints.\n1) When the block is at the extreme from the rest position all the system energy is stored in the spring. The stored energy is spring force (k*x) times spring deflection distance (x). So it's k*x^2 where x is 8.5. \n\nThe block has maximum speed when the block is in the center of the oscillation range (the spring is fully relaxed at that point), so all the energy is in kinetic energy. Energy is conserved, so\nMax spring energy = Max KE\nk*x^2 = (1/2)*m*v^2\n\n2) Best hint here: think, no math req'd.\n\n3) Spring force F = k*x. Acceleration = F/m\n\n4) Acceleration is max (see #3) at x = 8.5 and zero at x = 0. In the formula\nx = (1/2)*a*t^2,\nyou can use the average of the accelerations those 2 points for a.
731589
Well To solve this problem, you need to break up forces in x and y components. Once you find both x and y components, you can find magnitude of force and direction ken must pull. \n\nTo keep the bear from moving, some of the forces (in both x and y direction) equal to zero. \n\nSo, sum of the forces in X-direction = 350sin(60) - 400sin(45) - F(in x direction) = 0\nSum of the forces in y-direction = 400cos(45) + 350cos(60) - F(in y direction) = 0\n\nSolve for F(in x direction) and F(in y direction), you will get these: F(in x direction) = 20.266 lb east and F(in y-direction) = 457.84 lb south\nTo find magnitude = ||F||= sqrt((F(x))^2+(F(y))^2) = 458.29 lb\n\nTo find angle= tan(theta) =(F(y))/(F(x)) where theta is the angle\ntheta = arctan (F(y)/F(x))= 87.5 degree east south\n\nSo Ken must pull the rop with force 458.29 lb in the direction East 87.5 degree South to keep the bear from moving. \nI think this is right. Good Luck.!!!