Determine the maximum mass of the lamp that the cord system can support so that no single cord develops a tension exceeding 400 N.

#### Solution:

First, we will draw a free body diagram focusing on ring D.

Let us write our equations of equilibrium. We will start with the y-axis forces.

F_{DE}\text{sin}\,(30^0)\,-\,9.81m\,=\,0

(Isolate for F_{DE})

F_{DE}\,=\,\dfrac{9.81m}{\text{sin}\,(30^0)}(simplify)

F_{DE}\,=\,19.62m

Now, write an equation of equilibrium for the x-axis forces.

F_{DE}\text{sin}\,(30^0)\,-\,F_{DC}\,=\,0

(Substitute the value of F_{DE} we found)

19.62m\text{cos}\,(30^0)\,-\,F_{DC}\,=\,0(Isolate for F_{DC})

F_{DC}\,=\,16.99m

We can now switch our focus to ring C. Let us draw a free body diagram again.

Again, we will write our equations of equilibrium.

16.99m\,-\,F_{CA}\dfrac{3}{5}\,-\,F_{CB}\text{cos}\,(45^0)\,=\,0 (eq.1)

F_{CA}\dfrac{4}{5}\,-\,F_{CB}\text{sin}\,(45^0)\,=\,0 (eq.2)

Let us solve for F_{CA} and F_{CB}.

(Simplify)

F_{CA}\,=\,0.884F_{CB} (eq.3)

Substitute this value back into eq.1.

(Solve for F_{CB})

F_{CB}\,=\,13.7m

Substitute this value into eq.3 to figure out F_{CA}.

let us now look at the results we got:

F_{DC}\,=\,16.99m

F_{CB}\,=\,13.7m

F_{CA}\,=12.1m\,

From these forces, we can see thatÂ cord DE experiences the largest force.

Thus, knowing that the maximum tension a cord can handle is 400 N, we can write:

(solve for m)

m\,=\,20.38 kg

#### Final Answer: