Can someone answer a physics problem concerning static friction?
818860
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!
746965
You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
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
885634
using conservation of energy the kinetic energy at the bottom is \n1/*m*v^2\nThe frictional force times the distance is the work done by friction\nand the displacement in the vertical direction is the potential energy that gets converted to kinetic and friction\n\n1/2*m*v^2+f*L=m*g*h\nwhere h /L= sin(15.9)\n\nPlug in the numbers and solve using algebra\n(.5*6500*3.8*3.8)+4300*L=(6500*9.8*sin(15.9))*L\n46930+4300L=17451*L\n46930/(17451-4300)=L\n=3.6m\n\nj
390283
coefficient of friction and specific gravity
924331
a) 1. friction coefficient\n2. whether the bead is tilted as a whole or some portion of it. if a portion then you need to know mass distribution of patient and where the bent in the bead is.\n\nb) assuming the bead is tilted as whole then F_friction=mu*mg*cos(32deg) ,ie a projection of force due to gravity onto the normal axis.\nthe friction force is balanced by the projection of force of gravity to surface: mg*sin(32deg),\nequate the two, then you get mu=tan(32deg)=0.625
882649
First solve for the force of friction between the block and the table using the formula,\n\nf=uR, where f is the force of friction, u the coefficient of static friction, and R is the force normal to the table. R therefore is the weight of the block equal to 31g.\n\nf=0.425*31*9.8\n =129.1N\n\nThe weight of the sand added to the bucket plus the weight of the bucket itself should be just slightly equal to 129.1N so that the system will begin to move. At that point, however, the block will begin to accelerate because the force of friction acting on it will no longer be static friction but kinetic friction, which is lower in magnitude. Thus there will be a net force equal to 129.1 minus the force of kinetic friction on the block. Thus,\n\nNet force, F=129.1-0.320*31*9.8\n                 =129.1-97.2\n                 =31.9N\n\nF=ma, where F is our net force of 31.9N, m the total mass of block plus mass of sand and bucket, and a is the acceleration of the system.\n\nMass of sand and bucket=129.1/9.8\n                                    =13.2kg\nSubstitute known values:\n\n31.9=(31+13.2)a\na=31.9/44.2\n  =0.72m/s^2
981852
You can solve this problem by resolving the forces along the ramp for part a) and normal to the ramp for part b).\n\na) The net force along the ramp must be zero for the crate to move at constant velocity (ie no acceleration), so we have:\n\nFcos30 = mgsin30\nF=mgtan30\n\nb) the reaction of the ramp on the block will be normal to the ramp and equal to the sum of the imposed forces resolved normal to the ramp, so:\n\nR = Fsin30 + mgcos30
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
845875
Ah, time to dig out the physics from last year.  If I remember correctly:\n\nStatic friction is when the force of friction is greater than the force applied; thus basically what amount of force it takes to move the object from rest.\n\nKinetic friction is when you apply a force for an object to keep sliding.  In this, it "breaks free" of your push and continues sliding.\n\nAnd the force needed to initially break the static friction (to get it moving) is much more than what is needed to keep it going.\n\nHope this helped a bit.  Just remember that static friction is associated with objects at rest and kinetic is objects in motion, and don't get the two mixed up. :*P\n\nHave fun with physics!
897942
You should only ask one question at a time.\n\nI will answer the first:\n\nConsider the forces on the box:\n\nGravity vertically downward\nfriction pulling backwards\nThe force of gravity parallel to the ramp \nminus the force of friction\n=m*a=m*1.22\nso\nm*sin(34)*g-f=m*1.22\nfriction can be expressed as\nf=m*u*cos(34)*g\nnote that mass divides out\nsin(34)*g-u*cos(34)*g=1.22\n\nu=.52\n\nj
67664
friction is F=uN where u is the coefficient and N is the normal force.  In this case 260N is the normal force. So:\nu*260N=150N* cos 40\nYou must subtract off (150 * sin 40), also because you are counteracting part of the normal force.\nSo the final eqn should be\nu = (150N* cos 40) / (260N - (150* sin 40))\nLet me know how it goes\nemail: ke7alw@yahoo.com
35148
Frictional force, F = u*m*g\nHere, F = 300 N, u = ?, m = 80 kg, g = 10 m/s/s (assume)\nTherefore, 300 = u*80*10\n=> u = 3/8
824753
For someone walking, it is STATIC friction that is acting, not KINETIC friction since the person's feet are not in motion relative to the ground while they are in contact with the ground (unless the person is slipping). So, in terms of the static friction, as a person walks, they are bushing backwards against the ground and the static friction is what is pushing against them, pushing them forward, so again it IS in the opposite direction to the force being applied.
703117
here u go!\n a):\nuse the principal of F=ma to find the frictional force.  \nIn order to do this, you need to determine acceleration, so use the kinematic formula: v^2=vo^2+2a(x-xo).  \nwith this equation, you should find a to be equal to -1.5m/s^2.  \nPlug this answer into the F=ma equation, with m equal to .090kg.  Since Friction is the only force in the horizontal direction, F is equal to the frictional force.\nSolving for F=(.090)(-1.5)\nF=-.135N, so its magnitude is .135N\n\nb.):\nOne physics principal states that a frictional force is equal to mew (the coefficient of kinetic friction) times the normal force.\nSince you now know the frictional froce, just solve that equation.\nThe normal force in this problem is equal to the opposite of the gravitational force, which is (9.8)(.090)=.882\nSo, when you have the equation \n(.882)(coefficient of friction)=.135\njust divide .135 by .882 and you will get the coffecient of friction to be .153!!!!!!! yay!!!
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
135785
Ive done that before, it happens from the friction, and the static electricity!!