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.

Determine the maximum mass of the lamp

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

Solution:

Show me the final answer↓

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

Determine the maximum mass of the lamp

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.

Determine the maximum mass of the lamp

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 that 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

 

Final Answer:

Maximum mass of the lamp = 20.38 kg

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-29.

Leave a comment

Your email address will not be published.