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.

#### Solution:

Show me the final answers↓

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

The angles were found using the inverse of tan.

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.

(simplify)

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

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

(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.

(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}\,=\,270 N

#### Final Answers:

F_{AC}\,=\,270 N

How did you get the equation no.3

Equation 3 is just eq1 isolated for FAC. 🙂