Determine the magnitude and direction of the force P required to keep the concurrent force system in equilibrium.

#### Solution:

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We will first express each force in Cartesian vector form, including force P.

Using the diagram, we can write the forces in Cartesian vector form. We will start force force F_1

F_1\,=\,\left\{1.41i+1j-1k\right\} kN

To find force F_2, we will first find the position vector, then find the unit vector of that position vector, and finally multiply the magnitude of the force by the unit vector to write the force in Cartesian vector notation.

magnitude of r_{2}\,=\,\sqrt{(-1.5)^2+(3)^2+(3)^2}\,=\,4.5

u_2\,=\,\left(-\dfrac{1.5}{4.5}i+\dfrac{3}{4.5}j+\dfrac{3}{4.5}k\right)

F_2\,=\,0.75\left(-\dfrac{1.5}{4.5}i+\dfrac{3}{4.5}j+\dfrac{3}{4.5}k\right)

F_2\,=\,\left\{-0.25i+0.5j+0.5k\right\} kN

From the diagram, we see that force F_3 lies on the negative y-axis. Thus, we can write:

We will also write force P in Cartesian notation.

We can now write our equation of equilibrium. All forces added together must equal zero because the system is in equilibrium.

F_1+F_2+F_3+P\,=\,0

Since the system is in equilibrium, the x, y, z (i, j, k) components must also equal to zero when added together.

p_x\,=\,-1.16 kN

1+0.5-0.5+p_y\,=\,0

p_y\,=\,-1 kN

-1+0.5+p_z\,=\,0

p_z\,=\,0.5 kN

Now that we found each component of force P, we can write it as:

The magnitude of force P is:

The coordinate direction angles can be found by taking the cosine inverse of each component of force P divided by it’s magnitude.

\beta\,=\,\cos^{-1}\left(\dfrac{-1}{1.61}\right)\,=\,128^0

\gamma\,=\,\cos^{-1}\left(\dfrac{0.5}{1.61}\right)\,=\,72^0

#### Final Answers:

\alpha\,=\,136^0

\beta\,=\,128^0

\gamma\,=\,72^0