If the tension developed in each of the four wires

If the tension developed in each of the four wires is not allowed to exceed 600 N , determine the maximum mass of the chandelier that can be supported.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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Let us first draw a free body diagram and assume that the weight of the chandelier is W.

Now, we will write our equations of equilibrium.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

$T_{DC}\text{cos}\,(30^0)\,-\,T_{DB}\text{cos}\,(45^0)\,=\,0$ (eq.1)

$+\uparrow \sum \text{F}_\text{y}\,=\,0$

$T_{DC}\text{sin}\,(30^0)\,+\,T_{DB}\text{sin}\,(45^0)\,-\,W\,=\,0$ (eq.2)

We will now write $T_{DC}$ and $T_{DB}$ in terms of W. To do so, we will isolate for $T_{DC}$ in eq.1.

T_{DC}\,=\,\dfrac{T_{DB}\text{cos}\,(45^0)}{\text{cos}\,(30^0)}

(Simplify)

$T_{DC}\,=\,0.816T_{DB}$ (eq.3)

Substitute this value back into eq.2.

0.816T_{DB}\text{sin}\,(30^0)\,+\,T_{DB}\text{sin}\,(45^0)\,-\,W\,=\,0

(isolate for

T_{DB}

)

T_{DB}\,=\,0.897W

We can now substitute this value back into eq.3 to write $T_{DC}$ in terms of W.

T_{DC}\,=\,(0.816)(0.897W)

T_{DC}\,=\,0.732W

Now, we will focus on ring B and draw a free body diagram.

(Remember, we found $T_{DB}\,=\,0.897W$ )

Again, we will write our equations of equilibrium, however, we will first write our equation of equilibrium for the y-axis forces.

+\uparrow \sum \text{F}_\text{y}\,=\,0

$T_{BA}\text{sin}\,(30^0)\,-\,0.897W\text{sin}\,(45^0)\,=\,0$

(Isolate for $T_{BA}$ )

T_{BA}\,=\,1.27W

Now, write an equation of equilibrium for x-axis forces. Note that we just found $T_{BA}\,=\,1.27W$ .

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

0.897W\text{cos}\,(45^0)\,+\,T_{BC}\,-\,1.27W\text{cos}\,(30^0)\,=\,0

(Isolate for

T_{BC}

)

T_{BC}\,=\,0.465W

We now have all of the tensions in the ropes in terms of W. They are the following:

T_{DB}\,=\,0.897W

T_{DC}\,=\,0.732W

T_{BA}\,=\,1.27W

T_{BC}\,=\,0.465W

From these values, we can see that rope BA, or $T_{BA}$ experiences the most tension. The question also states that the maximum tension in each rope is 600 N, thus we can write the following:

T_{BA}\,=\,1.27W

600\,=\,1.27W

(solve for W)

$W\,=\,472.4$ N

We just found the maximum weight of the chandelier that can be hung, but the question asks us for the mass. We can use this following formula to find the mass:

W=mg

(Where $W$ is weight, $m$ is mass and $g$ is the force of gravity)

$m\,=\,\dfrac{W}{g}$

$m\,=\,\dfrac{472.4}{9.81}$

$m\,=\,48.1$ kg

Final answer:

Maximum mass of the chandelier that can be hung = 48.1 kg

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-32.

5 thoughts on “If the tension developed in each of the four wires”

Reply ↓

Zaher May 5, 2021 at 12:14 pm

Super
- Reply ↓
  
  questionsolutions Post author May 27, 2021 at 4:54 am
  
  Thanks!
Reply ↓

Aidan September 26, 2021 at 9:32 am

Thank you so much for this very detailed explanation for a long question. I appreciate your effort very much. It has helped me a lot!!
Reply ↓

Choi Joon February 5, 2022 at 12:00 am

can you explain how do you solve substitution equation 3 into 2?
- Reply ↓
  
  questionsolutions Post author April 5, 2022 at 1:08 pm
  
  You are taking the value you isolated for, and using that value in eq 2. So you are literally plugging in the value you isolate for.