if the force is always parallel to the slope,what is the coefficient of static friction between the slope and thebox?
206136
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
348531
A body in motion stays at that same velocity unless acted on by an outside force (Newton's laws of motion). So the sum of the forces acting on the book is zero since the velocity is constant. Since you are applying a one newton force in the horizontal direction, and friction is the only other force acting on it in the horizontal direction, the force due to friction must be one newton in a direction opposite to the force you are applying.
505057
They all involve friction, but I think you mean which has the most force due to friction, right? :-&#92;  If so, I believe the answer is that they are all the same.\n\nIn simple situations, the frictional force created by sliding two surfaces against eachother is given by:\n\nf = u * N\n\nWhere u (mu) is the frictional coefficient (an experimentally-determined constant) and N is the normal force, which is mg (mass times accelleration of gravity) if the surface isn't tilted.\n\nSo, assuming the frictional coefficient is the same for all sides of the book, there is no reason why changing the sides would make a difference.
375864
slope should be zero as you say that the mirror image was right in front of it ( parallel in other words )\n\nintercept should be +n as that should be the distance between the centre point and the images.
9908
look up stuff like static friction, and your coefficient of friction...
399907
You don't need the coefficients of friction for this problem.  All you need to do is recognize that there are 2 forces acting on the block parallel to the incline:\n\n1) Parallel component of gravity: Fg(par) = m·g·sin(Θ) = 93.19 N\n2) Frictional force (pointing UP the incline)\n\nSince the block is at rest, these two forces must be equal and opposite:\n\nF(fric) = 93.19 N (up the incline)
515446
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
537696
if acceleration is on the x axis, the slope represents 1/mass\n\nas\n\nslope = x/y = acc/force\nand\nforce=mass x acc\nso\nacc/force=1/mass\n\nif force is on the x axis, the slope represents mass\n\nas\n\nslope = x/y = force/acc\nand\nforce=mass x acc\nso\nforce/acc=mass
441951
Since the first incline is frictionless, there is no energy loss in coming down. Hence KE at the foot of the incline = PE at the start. This gives the velocity at the foot of the first incline as\n1/2 m*v^2 = m*g*h.\nThe frictional force on the horizontal surface = Uk*m*g\nThe work done in overcoming the fricional force =Uk*m*g*L\nThus the available energy at the foot of the second incline = m*g*h - Uk*m*g*L = m*g*(h - Uk*L)\nIf the height to which the block rises is h1,\nm*g*h1 = m*g*(h - Uk*L). Hence \nh1 = h - Uk*L
493517
The frictional force acting against the motion is F\n\n= (mu)x mg x cos a. \n\n= 0.2 x 22 x 9.8 x cos 25 = 39 N. \n\nThe net force acting along the inclined plane\n\n= (mg sin a +39) Since both the forces are acting opposite to the motion.\n\n= 22 x 9.8 x sin 25 + 39\n\n= 130. 12 N. \n\nIf x is the distance moved along the plane, the work done by this net force is Fx.\n\nThis work done is used to remove the kinetic energy of the body = 0.5 m v v.\n\nFx = 0.5 m v v \n\nx = 0.5 x 22 x 3.9 x 3.9 / 130.12 = 1.28m.
339467
Consider a force diagram at the point that the 499 N load is applied.  You have 499 N in the negative "y" direction, and two equal Tension forces (call them T), each at an angle to the horizontal.  Thus, the tension forces have each an "x" and "y" component.  One tension force goes to the tree, the other to the car.  You've got the angle from the horizontal from trig.\n\nNow do a force balance on those three forces, in the "y" direction. There is no acceleration, so sum of forces = ma = 0\n\nT*sin(4.76) + T*sin(4.76) - 499 = 0\n\nOnly one unknown here, solving.......\n\nT = 3006.67 N\n\nThat's a lot of force!\n\nCheers,
198584
If the line is parallel then it must have an equal slope. Therefore the parallel line would also have a slope of 0.
427434
The forces on the person are\n\nSFx = 0 = N - Fc\nSFy = 0 = Fs - m*g\n\nwhere SFx and SFy are the sums of the forces in the x- and y-directions, respectively, N is the normal (to the wall) force, Fc = m*v^2/r is the centripetal force, Fs = u*N is the friction force, m is the mass, and g is the gravitational acceleration.\n\nThese equations say that\n\nN = Fc = m*v^2/r = m*w^2*r\nu*N = u*m*w^2*r = m*g\n\nwhere w = v/r is the angular speed. Solving for the coefficient of static friction gives\n\nu = g/(r*w^2)
48071
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
379160
For both A the crate is moving at constant velocity, so there is no acceleration.  That means that all forces are balanced on the crate.\n\nWhat are the forces?\n\nGraivity directly downward\n1400 N\nForce F, 300N pushing downward at an angle of 20 degrees\nand friction = kN\nwhere k is the coefficient of kinetic friction\nand N is the normal force\n\nN=1400+sin(20)*300\nso friction is\nk*1502.6 N\n\nThe friction must exactly balance the horizontal force since we have constant velocity\nCos(20)*300=k*1502.6\nk=cos(20)*300/1502.6\n=.188\n\nb) Now that we know k, reevaluate the forces on the crate under the conditions stated:\nsince f=m*a\na=\n(cos(20)*F-k*(mg-sin(20)*F))/m\na=\n9.8*(cos(20)*300-.188*(1400-sin(20)*300))/1400\na=.266 m/s^2\n\nj
56740
I think "µs" and "Ms" are both the coefficient of static friction; they're just in different notations. "~" means "is similar to". The first equation is just saying that the coefficient of static friction and coefficient of kinetic friction are similar quantities: static is for non-moving contact surfaces, and kinetic is for moving contact surfaces.\n\n"Mk = f (v,m)" just means that the coefficient of kinetic friction is a function of both mass and velocity. This means that the coefficient is dependent of both mass and velocity (if either of them change, the coefficient will change also). It's important to note that mass and velocity are not dependent on each other.
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 )