Determine the force in each cable and the force F


Determine the force in each cable and the force F needed to hold the 4-kg lamp in the position shown.

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.

Determine the force in each cable and the force F needed to hold the 4-kg lamp in the position shown.

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.

\sum \text{F}_\text{x}=0

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.

T_{BC}=\frac{T_{BA}\text{cos}\,(60^0)}{\text{cos}\,(30^0)}

(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.

T_{BC}=(0.577)(67.9)\,=\,39.2 N

 

Now we will draw a free body diagram for point C.

Determine the force in each cable and the force F needed to hold the 4-kg lamp in the position shown.

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.

\sum \text{F}_\text{x}=0

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.

 

T_{CD}=\frac{39.2\text{cos}\,(30^0)}{\text{cos}\,(30^0)}

(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.

39.2\text{sin}\,(30^0)\,+\,39.2\text{sin}\,(30^0)\,-\,F=0

(Solve for F)

F=39.2 N

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-12.

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