A body initially is moving horizontally with the speed V along the inclined plane of angle A such as the friction coefficient is mu=tan(A). What is the final speed of the body (after a long time)?

Hmm, if you take the (as far as I know) standard definition of "co-efficient of friction" - meaning that friction opposes motion with a force of strength \mu*size of reaction between body and plane - and I've understood the question correctly so that the initial projection is along a contour of the plane, then the resulting vector ODE coming from Newton's second law is horribly non-linear (because it involves a unit vector in the direction of the velocity) and I can't solve it analytically.Have I misunderstood something?

Is it moving horizontally, or is it moving along an inclined plane?If the direction is a straight line, after a long time the speed will be either zero, infinity, or remain at the initial V.

QuoteOriginally posted by: joetHmm, if you take the (as far as I know) standard definition of "co-efficient of friction" - meaning that friction opposes motion with a force of strength \mu*size of reaction between body and plane - and I've understood the question correctly so that the initial projection is along a contour of the plane, then the resulting vector ODE coming from Newton's second law is horribly non-linear (because it involves a unit vector in the direction of the velocity) and I can't solve it analytically.Have I misunderstood something?joet, you understood everything right, initial velocity of the body is horizontal, so the path is not a straight line.

Soluation much easier to explain with picture (as most fresh physics questions are), but here goes. Change coordinates to parallel and perpendicular to the incline, call these x and y respectively. In the y direction you will have normal force pointing up and component of gravity pointing down which will offset each other: a_y(t) = g cos(A) - N/m = 0 N = mg cos(A)(m is the mass) and along the x direction you will initially have friction and gravity working against you: a_x(t) = -g(sin(A) + mu N/m) = -g(sin(A) + mu cos(A)) = -2g sin(A) in final step using mu = tan(A). Then, the velocity along the direction of incline will be v_x(t) = -2g sin(A) t + V/cos(A).so the object will proceed up the incline slowing down, will stop and turn around at time T = V /(2g sin(A) cos(A)). Once the object changes directions then friction will work in opposite direction, ie against gravity, and our new equation of motion is: a_x(t) = -g sin(A) + mu N/m = -g sin(A) + g tan(A) cos(A) = 0.At this point there is no net force on the object and its velocity in all directions is 0, so it will stop.

QuoteOriginally posted by: mabroeSoluation much easier to explain with picture (as most fresh physics questions are), but here goes. Change coordinates to parallel and perpendicular to the incline, call these x and y respectively. In the y direction you will have normal force pointing up and component of gravity pointing down which will offset each other: a_y(t) = g cos(A) - N/m = 0 N = mg cos(A)(m is the mass) and along the x direction you will initially have friction and gravity working against you: a_x(t) = -g(sin(A) + mu N/m) = -g(sin(A) + mu cos(A)) = -2g sin(A) in final step using mu = tan(A). Then, the velocity along the direction of incline will be v_x(t) = -2g sin(A) t + V/cos(A).so the object will proceed up the incline slowing down, will stop and turn around at time T = V /(2g sin(A) cos(A)). Once the object changes directions then friction will work in opposite direction, ie against gravity, and our new equation of motion is: a_x(t) = -g sin(A) + mu N/m = -g sin(A) + g tan(A) cos(A) = 0.At this point there is no net force on the object and its velocity in all directions is 0, so it will stop.mabroe, you've solved a different problem, which is very easy. Once again, initial velocity of the body is in the horizontal direction, not along the incline, so v0 would be in z direction in the system of coordinates you've chosen.

Quote Shyurik: A body initially is moving horizontally with the speed V along the inclined plane of angle A such as the friction coefficient is mu=tan(A). What is the final speed of the body (after a long time)?Your expression of the problem is poor. Do you mean: "A point mass is initially held on an inclined plane. It is then given an initial velocity that is purely horizontal, with component V parallel to a plane inclined at angle A. The coefficient of kinetic friction between the mass and the plane is mu=tan(A). What is the speed of the point mass as time t tends to infinity?"If the angle is positive (plane is inclined above the horizontal), then there's not enough information to answer the question: the point mass will eventually stop somewhere up the plane, and we have no way of telling if it will start moving again because we're not given the coefficient of static friction.If the angle is negative (plane is inclined below the horizontal), then you have to decide which solution region the angle A lies in: one that gives final velocity 0, constant, or infinity. You have to decide how long the point mass spends in the air initially, so that gravity acts to increase the parallel velocity component without friction dissipating any energy. Assuming the point mass doesn't bounce on hitting the inclined plane, this velocity becomes your new initial velocity. Now use conservation of energy, not Newton's 3rd Law. The rest is trivial.

the object has a initial velocity thats zero in x,y and some v(0) in z.the forces are acting like g in the x-y plane and a frictional force always opposite to the motion.Along the surface in X-Y plane--> force of gravity along the plane is gsin(a) = - (mu)gcos(a)....hence no unbalanced force in this direction.Perpendicular to the surface in X-Y plane --> gcos(a) but since its on an impenetrable solid wedge/incline...hence no unbalanced force in this direction toofrom the above discussion its clear that the object will move in the z-axis as if its moving on a plain floor....and since there is friction, the mass with have to come to a halt at some point of time....so the final velocity is zero......and its x-y co-ordinates ll be the same and only the z-co-ordinate would have changed.

oops Yes, then Legrangian is pretty ugly using corrdinate system x, y, z as defined below, but object would eventually come to rest in z direction.

QuoteOriginally posted by: INFIDELQuote Shyurik: A body initially is moving horizontally with the speed V along the inclined plane of angle A such as the friction coefficient is mu=tan(A). What is the final speed of the body (after a long time)?Your expression of the problem is poor. Do you mean: "A point mass is initially held on an inclined plane. It is then given an initial velocity that is purely horizontal, with component V parallel to a plane inclined at angle A. The coefficient of kinetic friction between the mass and the plane is mu=tan(A). What is the speed of the point mass as time t tends to infinity?"If the angle is positive (plane is inclined above the horizontal), then there's not enough information to answer the question: the point mass will eventually stop somewhere up the plane, and we have no way of telling if it will start moving again because we're not given the coefficient of static friction.If the angle is negative (plane is inclined below the horizontal), then you have to decide which solution region the angle A lies in: one that gives final velocity 0, constant, or infinity. You have to decide how long the point mass spends in the air initially, so that gravity acts to increase the parallel velocity component without friction dissipating any energy. Assuming the point mass doesn't bounce on hitting the inclined plane, this velocity becomes your new initial velocity. Now use conservation of energy, not Newton's 3rd Law. The rest is trivial.Sorry, I think I didn't make the description clear enough - a point mass initially at rest on an infinite inclined plane of angle A with the coefficient of kinetic friction mu = tan(A) is given a horizontal (in z direction in mabroe's system of coordinates) velocity V parallel to the plane (it never leaves the plane). What's the velocity of the point mass in the limit t -> infinity?

QuoteOriginally posted by: mabroeoops Yes, then Legrangian is pretty ugly using corrdinate system x, y, z as defined below, but object would eventually come to rest in z direction.Yes, so the question is - what's the final velocity in x direction?

QuoteOriginally posted by: ShyurikQuoteOriginally posted by: mabroeoops Yes, then Legrangian is pretty ugly using corrdinate system x, y, z as defined below, but object would eventually come to rest in z direction.Yes, so the question is - what's the final velocity in x direction?is my answer right?

Quoteis my answer right?Think your answer is right in that the object will eventually come to rest in all directions, but the object will move along the x-direction because initally the net force along the x-axis is gravity only. The direction of fricitional force is opposite direction the instantanous directional vector, so initially friction will point negative z-direction (assuming we push object along positive z-direction). As object traverses down the infinite-incline, assuming gravity stays constant g of course , at some point the object will move completely along x-dir, at which point it will stop because forces cancel out. Of course this is Shyurik's question so he is final authority and i could be blowing smoke .....To find exact trajectory wrote out legrangian and got some nasty equations. any hint as to better generalize coordinate system, more elegant solution?

QuoteOriginally posted by: mabroeQuoteis my answer right?Think your answer is right in that the object will eventually come to rest in all directions, but the object will move along the x-direction because initally the net force along the x-axis is gravity only. The direction of fricitional force is opposite direction the instantanous directional vector, so initially friction will point negative z-direction (assuming we push object along positive z-direction). As object traverses down the infinite-incline, assuming gravity stays constant g of course , at some point the object will move completely along x-dir, at which point it will stop because forces cancel out. Of course this is Shyurik's question so he is final authority and i could be blowing smoke .....To find exact trajectory wrote out legrangian and got some nasty equations. any hint as to better generalize coordinate system, more elegant solution?lemma's answer is not correct. The final velocity is not zero, the final acceleration is zero though. Hint: Choose the instantaneous system of coordinates with one axis (q) along the instantaneous direction of the body's velocity and another along the incline (x in mabroe's convention). What's the relation between dVq and dVz?

Sorry, I meant between dVq and dVx