Each of the four forces acting at e has a magnitude of 28 kn. Express each force as a cartesian vector and determine the resultant force.

Solution:

KN=10^3

we have to keep in mind nomancleature

f=28KN

a=4m, b=6m, c=12m

the position in a vector and then the forces

F_{EA}:

R_{EA} = (b,-a,-c)

F_{EA}=f/tfrac{r_{EA}

}{/left | R_{EA}

/right |}

F_{EA} = (12,-8,-24) KN

so, the cartesion for F_{EA} will be

F_{EA} = (12i-8j-24k)

F_{EB}:

R_{EB} = (-b, a,-c)

F_{EB}=f/tfrac{r_{EB}

}{/left | R_{EB}

/right |}

F_{EB} = (-12,8,-24) KN

so, the cartesion for F_{EB} will be

F_{EB} = (-12i+8j-24k)

F_{EC}:

R_{EC} = (-b,-a,-c)

F_{EC}=f/tfrac{r_{EC}

}{/left | R_{EC}

/right |}

F_{EC} = (12,8,-24) KN

so, the cartesion for F_{EC}

F_{EC} = (-12i+8j-24k)

F_{ED}:

R_{ED} = (-b,-a,-c)

F_{ED}=f/tfrac{r_{ED}

}{/left | R_{ED}

/right |}

F_{EC} = (-12,-8,-24) KN

so, the cartesion for F_{EC}

F_{EC} = (-12i-8j-24k)

Resultent force F_{r}=F_{EA}+F_{EB}+F_{EC}+F_{ED}

= (-96K) KN