Chapter 3.3, Problem 12P

The lift sling is used to hoist a container having a mass of 500 kg. Determine the force in each of the cables AB and AC as a function of θ. If the maximum tension allowed in each cable is 5 kN, determine the shortest length of cobles AB end AC that can be used for the lift. The center of gravity of the container is located at G.

Prob. 3–12

To determine

The force in each of the cables AB and AC as a function of $\theta$.

The shortest length of cables AB and AC, which is used for the lift.

Answer to Problem 12P

The force in each of the cables AB and AC as a function of $\theta$ (F) is $\underset{\_}{2.45\cos \theta \text{kN}}$.

The shortest length of cables AB and AC which is used for the lift (l) is $\underset{\_}{1.72\text{m}}$.

Explanation of Solution

Given information:

• The mass of the lift sling is 500 kg.
• The maximum tension allowed in each cable is 5 kN.

Show the free body diagram of the lift sling with the container as in Figure 1.

Observation:

The force F has to support the entire weight of the container.

Determine the weight of the lift sling.

$F=mg$ (I)

Here, m is the mass of the lift sling and g is the acceleration due to gravity.

Determine the force in each of the cables AB and AC by applying the equation of equilibrium.

Along the horizontal direction:

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{AC}\cos \theta -F_{AB}\cos \theta =0\\ F_{AC}\cos \theta =F_{AB}\cos \theta \\ F_{AC}=F_{AB}=F\end{array}$ (II)

Here, the force acting on cable AC is $F_{AC}$ and the force acting on cable AB is $F_{AB}$.

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F-F_{AC}\sin \theta -F_{AB}\sin \theta =0\end{array}$ (III)

Determine the shortest lengths of cables AB and AC using geometry.

$l_{AB}=l_{AC}=\frac{l_{BG}}{\cos \theta }$ (IV)

Here, the length between the point B and G is $l_{BG}$.

Formula used:

$\cos \theta =\frac{1}{\sin \theta }$

Conclusion:

Substitute 500 kg for m and $9.81\text{m}/{\text{s}}^{\text{2}}$ for g in Equation (I).

$\begin{array}{c}F=500\times 9.81\\ =4,905\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\times \frac{1\text{N}}{1\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}}\\ =4,905\text{N}\end{array}$

Substitute 4,905 N for F and F for $F_{AB}$ and $F_{AC}$ in Equation (III).

$\begin{array}{l}4,905-F\sin \theta -F\sin \theta =0\\ \sin \theta \left(2F\right)=4,905\\ F=\frac{4,905}{2\sin \theta }\times \frac{\cos \theta }{\frac{1}{\sin \theta }}\\ F=2,452.5\cos \theta \text{N}\times \frac{1\text{kN}}{1,000\text{N}}\end{array}$

$F=2.45\cos \theta \text{kN}$

Determine the tension force angle using the formula.

$F=2.45\cos \theta \text{kN}$

Here, the maximum allowable tension force in the cable is F.

Substitute 5 kN for F.

$\begin{array}{l}5=2.45\cos \theta \times \frac{\frac{1}{\sin \theta }}{\cos \theta }\\ 5\sin \theta =2.45\\ \theta ={\sin }^{-1}\left(\frac{2.45}{5}\right)\\ \theta =29.37°\end{array}$

Substitute 1.5 m for $l_{BG}$ and $29.37°$ for $\theta$ in Equation (IV).

$\begin{array}{c}{l}_{AB}=l_{AC}=\frac{1.5}{\cos 29.37°}\\ =1.72\text{m}\end{array}$

Thus, the force in each of the cables AB and AC as a function of $\theta$ is $\underset{\_}{2.45\cos \theta \text{kN}}$.

Thus, the shortest length of cables AB and AC which is used for the lift is $\underset{\_}{1.72\text{m}}$.