# Determine the maximum mass of the lamp

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

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

#### 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.

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

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

(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.

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

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

(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.

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

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

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

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

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

Isolate for $F_{CA}$ in eq.2.

$F_{CA}\,=\,\dfrac{F_{CB}\text{sin}\,(45^0)}{\dfrac{4}{5}}$

(Simplify)

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

Substitute this value back into eq.1.

$16.99m\,-\,0.884F_{CB}\dfrac{3}{5}\,-\,F_{CB}\text{cos}\,(45^0)\,=\,0$

(Solve for $F_{CB}$)

$F_{CB}\,=\,13.7m$

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

$F_{CA}\,=\,(0.884)(13.7m)$

$F_{CA}\,=12.1m\,$

let us now look at the results we got:

$F_{DE}\,=\,19.62m$

$F_{DC}\,=\,16.99m$

$F_{CB}\,=\,13.7m$

$F_{CA}\,=12.1m\,$

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:

$400\,=\,19.62m$

(solve for m)

$m\,=\,20.38$ kg