Determine the forces in cables AC and AB needed

Determine the forces in cables AC and AB needed to hold the 20-kg ball D in equilibrium.Take F = 300 N and d = 1 m.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

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Let us first draw a free body diagram around ring A.

The angles were found using the inverse of tan.

The orange angle was found by:

\text{tan}^{-1}\left(\dfrac{1}{2}\right)\,=\,26.56^0

The blue angle was found by: $90^0\,-\text{tan}^{-1}\left(\dfrac{1.5\,+\,1}{2}\right)\,=\,38.66^0$

(We subtracted the θ value from $90^0$ because we are looking for the angle between $F_{AB}$ and the positive y-axis, NOT the x-axis)

We can now write our equations of equilibrium starting with x-axis forces.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

300\,-\,F_{AC}\text{cos}\,(26.56^0)\,-\,F_{AB}\text{sin}\,(38.66^0)\,=\,0

(simplify)

$0.89F_{AC}\,+\,0.62F_{AB}\,=\,300$ (eq.1)

Now, we can write an equation of equilibrium for the y-axis forces.

+\uparrow \sum \text{F}_\text{y}\,=\,0

F_{AC}\text{sin}\,(26.56^0)\,+\,F_{AB}\text{cos}\,(38.66^0)\,-\,196.2\,=\,0

(simplify)

$0.45F_{AC}\,+\,0.78F_{AB}\,=\,196.2$ (eq.2)

We can now solve for $F_{AC}$ and $F_{AB}$ by first isolating for $F_{AC}$ in eq.1.

F_{AC}\,=\,337.07\,-\,0.7F_{AB}

(eq.3)

(Substitute this value into eq.2)

(0.45)(337.07\,-\,0.7F_{AB})\,+\,0.78F_{AB}\,=\,196.2

(solve for $F_{AB}$ )

$F_{AB}\,=\,96$ N

Substitute the value of $F_{AB}$ we just found into eq.3 to figure out $F_{AC}$ .

F_{AC}\,=\,337.07\,-\,(0.7)(96)

$F_{AC}\,=\,270$ N

F_{AB}\,=\,96

$F_{AC}\,=\,270$ N

Determine the forces in cables AC and AB needed 2