Relation between force, vectors and coordinates

Question

I have the following exercise in my math book:

A $1000\,\mathrm{kg}$ barrel hangs from a hoist. At the hoist $AC$ and $BC$ are rods and $DC$ is a steel cable. Determine the forces in the rods and the cable. The coordinates, expressed in meters are: $A(2,0,0), B(-2,0,0), C(0,2,3/2)$ and $D(0,-2,0)$.

I have an example so I do (kind of) know how to solve it, by transforming $\vec{CA}, \vec{CB}$ and $\vec{CD}$ into normed vectors (vectors with length 1) and decomposing F in the normed vectors. Which should give me a system of equations that I can solve.

However I don't understand the relation between a force and the other vectors ($\vec{CA}, \vec{CB}$ and $\vec{CD}$).

EDIT: Exercise Image

Answer 1

As I understand it, the hoist is located at $C$ and the barrel is connected to the rods at $A$ and $B$, and to the cable at $D$ (all at $z=0$).

There are forces along the rods and the cable. Let the forces acting on the barrel be as follows:

• in the $AC$ rod it's $\vec F_{CA} := \alpha (C-A) = \alpha(-2, 2, 3/2),$ where $\alpha$ is a scalar factor,
• in the $BC$ rod be $\vec F_{CB} := \beta (C-B) = \beta(2,2,3/2),$ where $\beta$ is a scalar factor,
• in the $DC$ cable be $\vec F_{CD} := \gamma (C-D) = \gamma(0, 4, 3/2),$ where $\gamma$ is a scalar factor,
• and of course gravity $\vec G := mg(0, 0, -1),$ where $m$ is the mass and $g$ the gravity acceleration.

Now the sum of these should vanish since the barrel is not accelerating. Thus, $$ \alpha(-2, 2, 3/2) + \beta(2,2,3/2) + \gamma(0, 4, 3/2) + mg(0, 0, -1) = 0. $$ From the above equation you should be able to determine $\alpha,\beta,\gamma$ and then $\vec F_{CA}, \vec F_{CB}, \vec F_{CD}.$