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.

#### 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.

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.

(Simplify)

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

Substitute this value back into eq.2.

(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}\,=\,0.732W

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.

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

(Isolate for T_{BA})

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

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_{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:

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:

(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:

Super

Thanks!

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!!

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

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.