Determine the force in each cable and the force F needed to hold the 4-kg lamp in the position shown.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
We will analyze point B and C separately. First, we will draw a free body diagram for point B and write our equations of equilibrium.
Let us now write the equations for equilibrium for the x-axis forces and y-axis forces. Forces going right, \rightarrow^+ will be considered positive and forces going up,\uparrow+ will be considered positive.
T_{BC}\text{cos}\,(30^0)\,-\,T_{BA}\text{cos}\,(60^0)=0 —————-(eq.1)
\sum \text{F}_\text{y}=0
T_{BA}\text{sin}\,(60^0)\,-\,T_{BC}\text{sin}\,(30^0)\,-\,39.24=0 —————-(eq.2)
We will now solve for T_{BA} and T_{BC}.
Isolate for T_{BC} in eq.1.
(simplify)
T_{BC}=0.577T_{BA} —————-(eq.3)
Substitute this value into eq.2.
T_{BA}\text{sin}\,(60^0)\,-\,0.577T_{BA}\text{sin}\,(30^0)\,-\,39.24=0(solve for T_{BA})
T_{BA}=67.9 N
We can use this value to figure out T_{BC} by substituting it back into eq.3.
Now we will draw a free body diagram for point C.
As before, we will write our equations of equilibrium for x-axis forces and y-axis forces. Considerations for positive forces will be the same as before.
T_{CD}\text{cos}\,(30^0)\,-\,T_{BC}\text{cos}\,(30^0)=0 —————-(eq.4)
\sum \text{F}_\text{y}=0
T_{CD}\text{sin}\,(30^0)\,+\,T_{BC}\text{sin}\,(30^0)\,-\,F=0 —————-(eq.5)
Remember that we now know that T_{BC}=39.2 N.
Isolate for T_{CD} and substitute the value of T_{BC} in eq.4.
(Solve for T_{CD})
T_{CD}=39.2 N
Substitute the value of T_{CD} we found and the value of T_{BC} into eq.5.
(Solve for F)
F=39.2 N