Question

the lift sling shown in (figure 1) is used to hoist a container having a mass of 500 kg. the center of gravity of the container is located at g.
part a
determine the force in each of the cables ab and ac as a function of θ.
express your answer in kilonewtons in terms of θ using three significant figures.

part b
if the maximum tension allowed in each cable is 2.8 kn, determine the shortest lengths of cables ab and ac that can be used for the lift.
express your answer to three significant figures and include the appropriate units.

Explanation:

Part A

Step 1: Calculate the weight of the container

The mass of the container \( m = 500\space kg \), and the acceleration due to gravity \( g = 9.81\space m/s^2 \). The weight \( W = mg \).
\( W = 500\times9.81 = 4905\space N = 4.905\space kN \)

Step 2: Analyze the equilibrium in the vertical direction

Let the tension in each cable \( AB \) and \( AC \) be \( F_{AB} = F_{AC} = F \) (due to symmetry). The vertical component of each tension is \( F\sin\theta \). For equilibrium, the sum of vertical forces is zero: \( 2F\sin\theta = W \)

Step 3: Solve for \( F \)

From \( 2F\sin\theta = W \), we get \( F=\frac{W}{2\sin\theta} \). Substituting \( W = 4.905\space kN \), we have \( F=\frac{4.905}{2\sin\theta}=\frac{2.4525}{\sin\theta}\space kN \) (rounded to three significant figures, \( F = \frac{2.45}{\sin\theta}\space kN \) or \( F = 2.45\csc\theta\space kN \))

Step 1: Determine the angle when tension is maximum

We know that \( F = \frac{2.4525}{\sin\theta} \), and the maximum tension \( F_{max} = 2.8\space kN \). So, \( \sin\theta=\frac{2.4525}{F_{max}} \)

Step 2: Calculate \( \sin\theta \)

Substituting \( F_{max} = 2.8\space kN \), we get \( \sin\theta=\frac{2.4525}{2.8}\approx0.8759 \)

Step 3: Calculate \( \theta \)

\( \theta=\arcsin(0.8759)\approx61.1^\circ \)

Step 4: Calculate the length of the cable

From the diagram, the horizontal distance from \( A \) to the vertical line through \( B \) (or \( C \)) is \( 1.5\space m \). The length of the cable \( l \) can be found using the cosine of \( \theta \): \( \cos\theta=\frac{1.5}{l} \), so \( l = \frac{1.5}{\cos\theta} \)

Step 5: Substitute \( \theta \) to find \( l \)

\( \cos(61.1^\circ)\approx0.484 \), so \( l=\frac{1.5}{0.484}\approx3.10\space m \) (rounded to three significant figures)

Answer:

\( F_{AB} = F_{AC} = \frac{2.45}{\sin\theta}\space kN \) (or \( 2.45\csc\theta\space kN \))

Part B