Recall last year we saw the amount of incremental work for N particles done for some virtual displacement (i.e. so small that the forces don't change) can be expressed:
d W = å N i (F i x d x + F i y d y + F i z d z )
We now consider the work done by a force as it moves a unit particle along a curved path. We consider this in the context of allowing the direction cosines (l, m, n) to define the orientation of some force vector F. Where:
l = cos a = x2 - x1 / ℓ m = cos b = y2 - y1 / ℓ And:
n = cos g = z2 - z1 / ℓ
With:
cos Θ = l1 l2+ m1 m2 + n1 n 2
If then Θ is the angle between the force vector and the path at some specified position of the particle, the component of force acting along the trajectory is F cos Θ . Then the total work done is:
W = òA B F cos Θ ds
Where ds is an element of the path. We term an integral of this form a line integral. More technically we may write:
òA B F cos Θ ds = òB A F cos Θ (- ds )= - òB A cos Θ ds
Then the work done in going from A to B is minus the work done in going from B to A. For convenience we may also write:
Fl1 = X, F m1 = Y and: F n1 = Z
For the force components parallel to the 3 coordinate axes. And then:
W = òA B (X dx + Y dy + Z dz)
Assume now there is a function V of the coordinates such that:
X = - ¶ V / ¶x Y = - ¶ V / ¶y Z = - ¶ V / ¶z
Then V is a potential function and we can also write:
dV = (¶ V / ¶x) dx + (¶ V / ¶y) dy + (¶ V / ¶z) dz
Or: (using previous coordinate equations):
dV = - (X dx + Y dy + Z dz)
And the work can be expressed more concisely:
W = òA B dV = - [V]A B = V A - V B
And: WAB = - WBA
Thus, the work done depends only on the limits and not the path followed. So the process is reversible and conservation follows naturally when forces are derived from a potential. A more detailed presentation is possible when we incorporate parametric equations (e.g. with x= x (t), y = y(t) etc. ) for the curve paths. In this case we consider the point of application of a force moving along a curve C, i.e.
F = i X (x,y,z) + j Y(x, y, z) + k Z (x,y,z)
Where X, Y, Z are as previously defined.
Then the work done by a force in the context:
W = òC F · dR
Where: R = ix + jy + kz
is the vector from the origin to the point (x,y,z) and:
dR = i dx + j dy + kdz
And if one calculates the dot product of the vectors F and dR then one can write:
W = òC X dx + Y dy + Z dz
Which may be compared with:
W = òA B (X dx + Y dy + Z dz)
Seen previously.
Example: Consider a force given by:
F = i(x 2 - y) + j (y 2 - z) + k (z 2 - x)
And let the point of application of the force move from the origin 0, to the point (1, 1, 1), along the straight line x= y = z and along the curve:
x = t, y = t 2 and z = t 3
Over: (0 < t < 1)
Then the integral for evaluation becomes:
W = òC (x 2 - y) dx + (y 2 - z) dy + (z 2 - x) dz
For the path over straight line: x = y = z:
W = ò0 1 3 (x 2 - x) dx = - 1/2
While along the parameterized curve:
W = ò0 1 2(t4 - t 3 ) t dt + 3 (t 6 - t ) t 2 dt = - 29/ 60
(Rem: dy = 2 t dt, dz = 3 t 2 dt)
So we see the work done is very nearly independent of the curve C connecting the points.
Problems:
1) Let G(x,y) = x 3 + 2xy + 2x and:
x = 2t, y = t 2 Over: (0 < t < 1)
Find the work done in going from A= 0 to B = 1 along the curve.
2) Two lines have direction cosines:
(1/2, Ö 6/ 4, - Ö 6/ 4 ) and (Ö 6/ 4, 1/4, 3/4)
Show they are orthogonal.
3) Given the force (vector function):
F = i (2xy) + j (x 2 + 3y 2 ln z) + k (y 3/z)
Find the potential function V from which F was derived.
4)Given a force: F = i (y + z) + j (z + x) + k (x + y)
Find the work done as the point of application moves from (0, 0, 0) to (1,1, 1):
(a) Along the straight line: x = y = z
(b) Along the curve: x = t, y = t 2 and z = t4
(c) Find the direction cosines of the line segment AB (e.g. from (0, 0, 0) to (1,1, 1) ) and the associated respective angles: a, b, g.
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