Question

the crate has a weight of 540 lb. determine the force in each supporting cable. (figure 1) part a determine the force in cable ab. express your answer to three significant figures and include the appropriate units. part b determine the force in cable ac. express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Set up force - equilibrium equations

At point A, the sum of forces in the x - direction $\sum F_x = 0$ and the sum of forces in the y - direction $\sum F_y=0$. Let the force in cable AB be $F_{AB}$ and the force in cable AC be $F_{AC}$. The weight of the crate $W = 540$ lb acts in the negative y - direction.
In the x - direction: $F_{AB}\sin30^{\circ}-F_{AC}\frac{3}{5}=0$. In the y - direction: $F_{AB}\cos30^{\circ}+F_{AC}\frac{4}{5}-540 = 0$.

Step2: Solve the x - direction equation for $F_{AC}$

From $F_{AB}\sin30^{\circ}-F_{AC}\frac{3}{5}=0$, we can express $F_{AC}$ in terms of $F_{AB}$ as $F_{AC}=\frac{5}{6}F_{AB}$.

Step3: Substitute $F_{AC}$ into the y - direction equation

Substitute $F_{AC}=\frac{5}{6}F_{AB}$ into $F_{AB}\cos30^{\circ}+F_{AC}\frac{4}{5}-540 = 0$.
We have $F_{AB}\cos30^{\circ}+\frac{4}{5}\times\frac{5}{6}F_{AB}-540 = 0$.
$F_{AB}\cos30^{\circ}+\frac{2}{3}F_{AB}-540 = 0$.
Factor out $F_{AB}$: $F_{AB}(\cos30^{\circ}+\frac{2}{3}) = 540$.
Since $\cos30^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$, then $F_{AB}(\ 0.866+\frac{2}{3}) = 540$.
$0.866+\frac{2}{3}=\frac{2.598 + 2}{3}=\frac{4.598}{3}\approx1.533$.
$F_{AB}=\frac{540}{1.533}\approx352$ lb.

Step4: Solve for $F_{AC}$

Substitute $F_{AB}\approx352$ lb into $F_{AC}=\frac{5}{6}F_{AB}$.
$F_{AC}=\frac{5}{6}\times352\approx293$ lb.

Answer:

Part A: $F_{AB}=352$ lb
Part B: $F_{AC}=293$ lb