- #1

Northbysouth

- 249

- 2

## Homework Statement

The turnbuckle is tightened until the tension in the cable AB equals 2.9 kN. Determine the vector expression for the tension T as a force acting on member AD. Also find the magnitude of the projection of T along the line AC.

## Homework Equations

## The Attempt at a Solution

I was able to find the vector expression for T without any real trouble:

AB = <2.6, 1.7, -6.1>

|AB| = sqrt(2.6

^{2}+ 1.7

^{2}+ (-6.1)

^{2})

T*AB/|AB| = 0.42364<2.6, 1.7, -6.1>

T = 1.101i + 0.7201j - 2.584k

I know this part is correct but I'm having difficulty doing the projection.

I tried taking taking the dot product of the T vector with the unit vector AC:

AC = ,2.6, -2.8, -6.1>

|AC| = sqrt(2.6

^{2}+ (-2.8)

^{2}+ (-6.1)

^{2})

|AC| = 7.1979

AC/|AC| = 0.3612, -0.389, 0.8475

T°AC/|AC| = 1.101*0.3612 +(-0.389*0.7201)+(-0.8475*-2.584)

= 2.2746

But it says my answer is wrong and I'm not sure where I'm going wrong.

Advice would be greatly appreciated.