Question

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.

Answer 1

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