# Homework Help: Vectors with appl of dot product

1. Oct 10, 2006

### teknodude

I'm trying to solve this problem with different approaches. Link is here

http://img138.imageshack.us/img138/2268/problem1eb1.th.png [Broken]
the answer is 0i + 0.0927j + 0.0232 k lb

* Means DOT

Method one: Find the force in scalar components. Then Take the cross product of two position vectors to get a normal vector (V) to the surface. THen take the unit vector of V. Then F(normal) = (V*F)V
THis method worked F(normal)= 0i + 0.0927j + 0.0232 k lb

Method 2: Find the force in scalar components. Draw a perpendicular line to the surface and by similar triangles the normal vector V is 8j + 2k. Take the units vector of V. Then F(normal) = (V*F)V
THis method worked F(normal)= 0i + 0.0927j + 0.0232 k lb

Method 3: This one i tried taking a position vector along the surface like from point A to point J. Position vector AJ = 0i-1j+4k. THen getting of units vector, unit vector eAJ = 0i - 0.2425j + 0.9701k.

THen F(parallel)= (eAJ*F)eAJ

F(normal) = F- F(parallel)
F(normal) = -0.1231 i + 0.0927j + 0.0232 k lb

THe only difference is the x-component, which differs a little. Is this answer still right? I didn;t round any numbers when calculating.

Last edited by a moderator: May 2, 2017
2. Oct 13, 2006

### Staff: Mentor

The normal to that surface would by definition have no x-component, so the Method 3 answer is wrong. I'm not understanding what you are trying to do in that method, so I'm not able to figure out what is wrong.

3. Nov 5, 2006

### teknodude

Here's the work that I did for method 3.

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4. Nov 7, 2006

### doodle

The manner in which you obtain the F (parallel) is incorrect. The direction of F (parallel) is not simply AJ just because AJ lies on the surface of the slope.

To visualize F (parallel), try imagining the shadow of the string on the sloped surface if a light source is at some point above (though not exactly) the string.