Need help in Physics concerning unit vectors...?

Question

The cable AB keeps the 8-kg collar A in place on the smooth bar CD. The y axis points upward. Determine the distance s from C to the collar A for which the tension in the cable is 150N.
location of:
B: (0m, 0.5m, 0.15m)
C: (0.4m, 0.3m, 0m)
D: (0.2m, 0m, 0.25m)

collar A's location is unknown. The required is the distance between C and collar A.
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http://img86.imageshack.us/img86/4710/392ij8.jpg

Answer 1

I finished the problem. The answer is s = 0.48517m. It was hard, although now that I've worked it, it doesn't seem as difficult anymore. Had to do some serious thinking though.

The collar A will tend to slide the bar CD. The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980]. The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].

We don't have to worry about the collar being supported by bar CD (the normal force). The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines. This force is 78.4 x .68376 = 53.60706N.

Let E be the point on the bar CD such that BE is perpendicular to CD. Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N. (I'm using numbers stored in the calculator and showing five decimal digits. All this can be rounded off at the end.)

Since the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB. We will get the distance BE.

Triangle BEA is in the plane defined by the points B, C, and D. Vector CB = [-0.4, 0.2, 0.15]. The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D. CB x CD = [0.095, 0.07, 0.16].

Now, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.

CD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].

Now we can get parametric equations of the lines CEAD and BE, and find the point of intersection E. Starting with point C and using vector CD, the first line is

x = 0.4 - 0.2t
y = 0.3 - 0.3t
z = 0 + 0.25t

Starting from point B and using the direction cosines from vector BE, the second line is

x = 0 + 0.75091u
y = 0.5 - 0.63914u
z = 0.15 - 0.16623u

Setting like terms equal to each other, we have three equations in two unknowns. The three equations are not linearly independent. Solving the x and y equations for t and u, we get

t = 0.29871 and u = 0.45313

(and these results are confirmed in the z equations). Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).

We know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.

Comparing the force and distance vectors in similar triangles BEA, we have

s/150 = 0.45313 / 140.09384

and the length s of the cable AB is 0.48517m. (Answer)

A difficult, but fun, problem.