# If the tension developed in each of the four wires 5

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:

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

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

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

• • questionsolutions Post author

Thanks!

• Aidan

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!!

• Choi Joon

can you explain how do you solve substitution equation 3 into 2?

• questionsolutions Post author

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.