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 3.5 kN, determine the shortest lengths of cables AB and AC that can be used for the lift. The center of gravity of the container is located at G.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
Let us first draw the free body diagram showing cables AB, AC and force F.
Note that force F must support all of the weight as all the forces go through that point. Thus, we can write force F_1=500(9.81)=4905 N.
Now, let us assume that forces going \rightarrow^+ is positive and \uparrow+ is positive.
We will first write the equations of equilibrium for x-axis forces like so:
\sum \text{F}_\text{x}=0
F_{AC}\text{cos}\theta -F_{AB}\text{cos}\theta=0
(We can cancel out the \text{cos}\theta by dividing both sides of the equation by \text{cos}\theta) This gives us:
F_{AC}=F_{AB}=F
Now, we will write the equations of equilibrium for y-axis forces.
\sum \text{F}_\text{y}=0
4905-2F\text{sin}\theta=0
(Isolate for F)
F=\frac{2452.5}{\text{sin}\theta}
(remember, \frac{1}{\text{sin}\theta}=\text{csc}\theta)
F=2452.5\text{csc}\theta
Thus we can write:
F_{AC}=F_{AB}=F=2452.5\text{csc}\theta
The question states that the maximum tension allowed in a cable is 3.5 kN. Therefore, we can write the following:
F=2452.5\text{csc}\theta
3500=2452.5\text{csc}\theta
\text{csc}\theta=\frac{3500}{2452.5}
(remember, \frac{1}{\text{sin}\theta}=\text{csc}\theta)
\frac{1}{\text{sin}\theta}=\frac{3500}{2452.5}
(Inverse both sides of the equation)
\text{sin}\theta=\frac{2452.5}{3500}
\theta=\text{sin}^{-1}(\frac{2452.5}{3500})
\theta=44.48^0