- #1
Ak94
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Not exactly homework, but this is a problem that a guy posed me.. and we seem to be getting different answers. I'm not entirely sure where either of us is going wrong.
Homework Statement
A uniform rod of mass M and length L lies on the x-axis, with its ends at x = 0 and x = L. A varying force of magnitude F is pulling the rod in the right direction, such that the rod is moving with constant velocity. The floor beneath the rod is rough only in the region from x = 0 to x = L, with the coefficient of friction between the rod and floor being 'kx', where k is a constant and x is the x-coordinate of the point. Find the magnitude of the work done by the frictional force in moving the entire rod out of the rough region.
The attempt at a solution
Here's what I came up with:
I take a length 'dx' of the rod at a distance 'x' from the origin. Its mass is (M/L)dx, force on it [(M/L)dx * gkx], and work done by friction on it [(M/L)dx * gkx * (L-x)]. I integrate x from 0 to L to get [itex] \frac{kMgL^{2}}{6} [/itex].
A person who gave me the question has this solution:
At an instant, let the left end of the rod have an x-coordinate "y". Consider a part of the rod with coordinate "x" of length dx. Force on it:[itex] \int \frac{Mgkx(dx)}{L} [/itex] from "y" to "L". Then dW = (That integral) * dy, and he then integrates from 0 to L to get [itex] \frac{kMgL^{2}}{3} [/itex].
I think his solution is wrong because he's assumed a constant force to be displacing the entire rod by "dy" to get "dW", which won't make sense because different forces are acting on each part of the rod. I'm not fully confident about my own solution either, so it'd be helpful if someone could clarify.
Homework Statement
A uniform rod of mass M and length L lies on the x-axis, with its ends at x = 0 and x = L. A varying force of magnitude F is pulling the rod in the right direction, such that the rod is moving with constant velocity. The floor beneath the rod is rough only in the region from x = 0 to x = L, with the coefficient of friction between the rod and floor being 'kx', where k is a constant and x is the x-coordinate of the point. Find the magnitude of the work done by the frictional force in moving the entire rod out of the rough region.
The attempt at a solution
Here's what I came up with:
I take a length 'dx' of the rod at a distance 'x' from the origin. Its mass is (M/L)dx, force on it [(M/L)dx * gkx], and work done by friction on it [(M/L)dx * gkx * (L-x)]. I integrate x from 0 to L to get [itex] \frac{kMgL^{2}}{6} [/itex].
A person who gave me the question has this solution:
At an instant, let the left end of the rod have an x-coordinate "y". Consider a part of the rod with coordinate "x" of length dx. Force on it:[itex] \int \frac{Mgkx(dx)}{L} [/itex] from "y" to "L". Then dW = (That integral) * dy, and he then integrates from 0 to L to get [itex] \frac{kMgL^{2}}{3} [/itex].
I think his solution is wrong because he's assumed a constant force to be displacing the entire rod by "dy" to get "dW", which won't make sense because different forces are acting on each part of the rod. I'm not fully confident about my own solution either, so it'd be helpful if someone could clarify.